The second algebra-based physics course (thermodynamics, electric fields and circuits, magnetism and induction, optics, waves, and modern physics) taught unit by unit to the current College Board framework. Like Physics 1, the exam grades the reasoning as much as the arithmetic. Exam format from 2027: this is a hybrid digital exam: multiple choice in Bluebook, free response handwritten in a paper booklet. Work the practice problems on paper and check your steps against the worked answer, the way you will sit the free-response section in May.
3H EXAM40 MCQ4 FRQ34 LESSONS70 PRACTICE PROBLEMSPREREQ: PHYSICS 1
Course overview
What this course covers, and how the exam weights it.
AP Physics 2 follows the seven units of the College Board course framework. The College Board numbers them 9 through 15, continuing from the eight units of Physics 1, and this course keeps that numbering so it matches the official materials. Thermodynamics, electric fields, and circuits are the three largest units; together they're roughly half the exam.
U9Thermodynamics15–18%
U10Electric Force, Field, and Potential15–18%
U11Electric Circuits15–18%
U12Magnetism and Electromagnetism12–15%
U13Geometric Optics12–15%
U14Waves, Sound, and Physical Optics12–15%
U15Modern Physics12–15%
All seven units are open, 34 lessons in all. Every
lesson pairs a short explanation with worked examples and a problem
to try yourself, the same problem types that show up on the exam.
Each unit closes with a short video walk-through and a ten-problem
practice set with hidden answers.
Free preview: open any 5 lessons, or watch one unit video, without
an account. The counter on the left keeps track.
Lesson 9.1 · Unit 9 · CED topics 9.1–9.2
Thermodynamic systems, pressure, and thermal equilibrium
Thermodynamics is Newton's laws applied to more particles than you could
ever track, so we stop tracking them and describe the crowd instead.
Three crowd-level variables carry the whole unit: pressure, volume, and
temperature.
Definition
A system is whatever you choose to analyze (usually the
gas in a container); everything else is the surroundings.
A system is closed if only energy crosses its boundary and
isolated if nothing does.
Pressure is force per unit area, \(P = \dfrac{F}{A}\),
measured in pascals (1 Pa = 1 N/m²); it comes from molecules colliding
with the walls. Two objects are in thermal equilibrium
when no net energy flows between them: they have the same temperature.
If A is in equilibrium with C and B is too, then A and B are in
equilibrium with each other (the zeroth law); that's why a thermometer works.
Worked example · Force from a pressure difference
A vertical cylinder is sealed by a frictionless piston of area
0.0050 m². The gas below the piston is at 1.50 × 10⁵ Pa; the air above
it is at 1.01 × 10⁵ Pa. Find the net force the two gases exert on the piston.
Each gas pushes on its side of the piston with \(F = PA\), in opposite directions, so
\[F_{\text{net}} = (P_{\text{in}} - P_{\text{out}})A
= (1.50\times10^{5} - 1.01\times10^{5}\ \text{Pa})(0.0050\ \text{m}^2)
= 245\ \text{N},\]
directed upward. If the piston is at rest, its weight plus any load on
it must total 245 N.
Worked example · Explain in words
A sealed, rigid can of gas is warmed. Explain at the particle level why
the pressure rises.
Warming the gas raises the average kinetic energy of its molecules, so
they move faster. Faster molecules hit the walls harder (each collision
transfers more momentum) and more often (they cross the can more times
per second). Force on the wall is the momentum delivered per second, so
both effects raise the force on the same area, and pressure is force
per area. That two-part sentence, harder and more often, is
what graders look for in a "particle-level explanation."
Exam tip: pressure is a scalar. A gas pushes outward on every wall equally.
Try it
A 2.0 kg textbook, 0.20 m by 0.25 m, lies flat on a table. What pressure
does it exert on the table? Use g = 9.8 m/s².
One equation links the three crowd variables, and one more links
temperature to what individual molecules are doing. Together they let
you go back and forth between the macroscopic gas and the microscopic
particles: the central skill of this unit.
Formula
Ideal gas law: \(PV = nRT = Nk_BT\), with \(n\) in moles,
\(N\) the number of molecules, \(R = 8.31\ \text{J/(mol·K)}\), and
\(k_B = 1.38\times10^{-23}\ \text{J/K}\). \(T\) must be in kelvin.
Kinetic theory: the average translational kinetic
energy of one molecule is \(K_{\text{avg}} = \tfrac{3}{2}k_BT\), so the
root-mean-square speed is
\[v_{\text{rms}} = \sqrt{\frac{3k_BT}{m}} = \sqrt{\frac{3RT}{M}},\]
where \(m\) is the mass of one molecule and \(M\) the molar mass in kg/mol.
Temperature fixes the kinetic energy; heavier molecules therefore move
slower at the same \(T\).
Worked example · A gas problem
0.250 mol of helium is sealed in a rigid 2.00 L container at 300 K.
Find the pressure, and then the pressure after heating to 450 K.
\(V = 2.00\ \text{L} = 2.00\times10^{-3}\ \text{m}^3\).
\[P = \frac{nRT}{V} = \frac{(0.250\ \text{mol})(8.31\ \text{J/(mol·K)})(300\ \text{K})}{2.00\times10^{-3}\ \text{m}^3}
= 3.12\times10^{5}\ \text{Pa}.\]
With \(n\) and \(V\) fixed, \(P \propto T\):
\(P_2 = P_1\dfrac{T_2}{T_1} = (3.12\times10^{5})\dfrac{450}{300} = 4.67\times10^{5}\ \text{Pa}\).
Setting up the ratio is faster than recomputing and avoids a second
rounding.
Worked example · rms speed and mass
Air at 300 K: find \(K_{\text{avg}}\) and \(v_{\text{rms}}\) for N2
(28 u; 1 u = 1.66 × 10⁻²⁷ kg), then compare helium (4 u) at the same temperature.
\(K_{\text{avg}} = \tfrac{3}{2}(1.38\times10^{-23}\ \text{J/K})(300\ \text{K}) = 6.21\times10^{-21}\ \text{J}\): the same for every gas at 300 K. For N2,
\(m = 28(1.66\times10^{-27}) = 4.65\times10^{-26}\ \text{kg}\):
\[v_{\text{rms}} = \sqrt{\frac{3(1.38\times10^{-23})(300)}{4.65\times10^{-26}}} \approx 517\ \text{m/s}.\]
Helium is 7 times lighter, so its speed is \(\sqrt{7} \approx 2.65\)
times greater: about 1370 m/s. Same energy, less mass, more speed.
Try it
(a) A gas is heated from 200 K to 800 K. By what factor does \(v_{\text{rms}}\) change?
(b) At what temperature does O2 (32 u) have the same \(v_{\text{rms}}\) as N2 at 300 K?
Show answer
(a) \(v_{\text{rms}} \propto \sqrt{T}\), and \(T\) quadruples, so speed doubles (factor 2).
(b) Equal speeds need equal \(T/m\): \(T = 300\ \text{K}\times\dfrac{32}{28} \approx 343\ \text{K}\).
Lesson 9.3 · Unit 9 · CED topic 9.5
Heat and energy transfer
Heat is not a substance an object contains; it's energy in transit
because of a temperature difference. A hot object doesn't "have heat":
it has internal energy, and it transfers some as heat when it touches
something cooler. Getting that vocabulary right earns points on its own.
Definition
Heat \(Q\) is energy transferred between objects at
different temperatures. It moves three ways: conduction
(molecule-to-molecule collisions through a material),
convection (bulk motion of a fluid carrying energy),
and radiation (electromagnetic waves, needing no medium).
For conduction through a slab of thermal conductivity \(k\), area \(A\),
and thickness \(L\):
\[\frac{Q}{\Delta t} = \frac{kA\,\Delta T}{L}.\]
When heat enters an object without a phase change, its temperature
changes by \(Q = mc\,\Delta T\), where \(c\) is the specific heat.
Worked example · Conduction through a wall
An insulated wall has area 10 m², thickness 0.10 m, and
\(k = 0.040\ \text{W/(m·K)}\). Inside is 20 °C, outside −5 °C. Find the
rate of heat loss, and state what doubling the thickness would do.
\(\Delta T = 25\ \text{K}\) (a temperature difference is the same in °C and K).
\[\frac{Q}{\Delta t} = \frac{(0.040\ \text{W/(m·K)})(10\ \text{m}^2)(25\ \text{K})}{0.10\ \text{m}} = 100\ \text{W}.\]
The rate is inversely proportional to \(L\), so doubling the thickness
halves the loss to 50 W. Doubling the area or the
temperature difference would double it.
Worked example · Reaching thermal equilibrium
A 0.500 kg aluminum block (\(c = 900\ \text{J/(kg·K)}\)) at 80.0 °C is
dropped into 1.20 kg of water (\(c = 4186\ \text{J/(kg·K)}\)) at 20.0 °C in an
insulated cup. Find the final temperature.
Heat lost by the block equals heat gained by the water:
\(m_{\text{Al}}c_{\text{Al}}(80.0 - T_f) = m_w c_w (T_f - 20.0)\).
\[(0.500)(900)(80.0 - T_f) = (1.20)(4186)(T_f - 20.0)\]
\[450(80.0 - T_f) = 5023(T_f - 20.0)
\;\Rightarrow\; 36{,}000 + 100{,}464 = 5473\,T_f
\;\Rightarrow\; T_f \approx 24.9\ ^\circ\text{C}.\]
The water barely warms because its heat capacity \(mc\) is eleven times
the block's. Temperature measures average energy per molecule; heat is
the total transferred: a bathtub of lukewarm water holds far more
thermal energy than a cup of boiling water.
Try it
How much energy must a kettle deliver to bring 0.250 kg of water from 20 °C to 100 °C?
A gas can trade energy with its surroundings two ways: as heat, and as
work when its volume changes against a pressure. The PV diagram makes
the work visible, it's an area, and, for an ideal gas, it also tells
you the internal energy at every point.
Formula
When a gas changes volume at constant pressure, the work done
on the gas is \(W = -P\,\Delta V\). In general the
magnitude of the work is the area under the PV curve.
Sign: expansion (\(\Delta V \gt 0\)) means the gas does positive work
on its surroundings and \(W_{\text{on gas}} \lt 0\); compression means
\(W_{\text{on gas}} \gt 0\). The internal energy of an ideal monatomic gas is
\[U = \tfrac{3}{2}nRT = \tfrac{3}{2}PV,\]
so \(U\) depends only on the state (the point on the diagram), not on
the path taken to reach it. Work and heat depend on the path.
Worked example · Constant-pressure expansion
A gas expands at a constant 2.0 × 10⁵ Pa from 1.0 L to 3.0 L. How much
work is done, and by whom?
\(\Delta V = 2.0\ \text{L} = 2.0\times10^{-3}\ \text{m}^3\).
\[W_{\text{by gas}} = P\,\Delta V = (2.0\times10^{5}\ \text{Pa})(2.0\times10^{-3}\ \text{m}^3) = 400\ \text{J},\]
so \(W_{\text{on gas}} = -400\ \text{J}\). On the diagram this is a
rectangle 2.0 × 10⁵ Pa tall and 2.0 × 10⁻³ m³ wide. Always convert
liters to cubic meters before multiplying: 1 Pa·m³ = 1 J.
Worked example · Reading a PV diagram
A monatomic ideal gas moves along a straight line from state A
(V = 1.0 L, P = 3.0 × 10⁵ Pa) to state B (V = 4.0 L, P = 1.0 × 10⁵ Pa).
Find the work done on the gas and the change in internal energy.
The area under the slanted segment is a trapezoid:
\[|W| = \tfrac{1}{2}(P_A + P_B)\,\Delta V = \tfrac{1}{2}(3.0\times10^{5} + 1.0\times10^{5})(3.0\times10^{-3}) = 600\ \text{J}.\]
The gas expands, so \(W_{\text{on gas}} = -600\ \text{J}\). For the
internal energy use the endpoints only:
\(U_A = \tfrac{3}{2}(3.0\times10^{5})(1.0\times10^{-3}) = 450\ \text{J}\) and
\(U_B = \tfrac{3}{2}(1.0\times10^{5})(4.0\times10^{-3}) = 600\ \text{J}\), so
\(\Delta U = +150\ \text{J}\). The gas got hotter even while doing work:
it must have absorbed heat (next lesson).
Exam tip: on a PV diagram, a higher isotherm \(PV = \text{const}\) means
a higher temperature. A point with larger \(PV\) is a hotter state.
Try it
A gas is compressed at a constant 1.2 × 10⁵ Pa from 5.0 L to 2.0 L. Find the work done on the gas.
Show answer
\(\Delta V = -3.0\times10^{-3}\ \text{m}^3\), so
\(W_{\text{on}} = -P\,\Delta V = -(1.2\times10^{5})(-3.0\times10^{-3}) = +360\ \text{J}\).
Positive: the surroundings did 360 J of work on the gas.
Lesson 9.5 · Unit 9 · CED topic 9.8
The first law of thermodynamics and special processes
The first law is conservation of energy for a gas: its internal energy
changes by exactly the heat that flows in plus the work done on it.
Everything else in this lesson is that one sentence applied to four
special paths on the PV diagram.
Rule
\[\Delta U = Q + W\]
with the AP sign convention: \(Q \gt 0\) when heat flows
into the gas, and \(W\) is the work done on the gas
(\(W \gt 0\) for compression, \(W \lt 0\) for expansion). Textbooks that
write \(\Delta U = Q - W\) define \(W\) as work by the gas; the
physics is identical, so always state which convention you're using.
Isothermal (constant \(T\)): \(\Delta U = 0\), so \(Q = -W\). A hyperbola on the diagram.
Isobaric (constant \(P\)): \(W = -P\,\Delta V\). A horizontal line.
Isochoric (constant \(V\)): \(W = 0\), so \(\Delta U = Q\). A vertical line.
Adiabatic (\(Q = 0\)): \(\Delta U = W\). Steeper than an isotherm; compression heats the gas.
Worked example · Isochoric heating
2.0 mol of a monatomic ideal gas in a rigid tank is heated from 300 K to 400 K. How much heat was added?
Rigid means \(\Delta V = 0\), so \(W = 0\) and \(Q = \Delta U\):
\[Q = \tfrac{3}{2}nR\,\Delta T = \tfrac{3}{2}(2.0\ \text{mol})(8.31\ \text{J/(mol·K)})(100\ \text{K}) \approx 2.49\times10^{3}\ \text{J}.\]
Worked example · A complete cycle
A monatomic ideal gas runs clockwise around a rectangle: A (2.0 L, 1.0 × 10⁵ Pa)
→ B (2.0 L, 3.0 × 10⁵ Pa) → C (6.0 L, 3.0 × 10⁵ Pa) → D (6.0 L, 1.0 × 10⁵ Pa) → A.
Find the net work and net heat for one cycle.
Work happens only on the horizontal legs. B→C is an expansion at 3.0 × 10⁵ Pa:
\(W_{BC} = -P\,\Delta V = -(3.0\times10^{5})(4.0\times10^{-3}) = -1200\ \text{J}\).
D→A is a compression at 1.0 × 10⁵ Pa: \(W_{DA} = +(1.0\times10^{5})(4.0\times10^{-3}) = +400\ \text{J}\).
Net work on the gas: \(W = -800\ \text{J}\), the gas does 800 J
of net work, equal to the enclosed area \((2.0\times10^{5})(4.0\times10^{-3})\).
Around any closed cycle the gas returns to its starting state, so
\(\Delta U = 0\) and \(Q_{\text{net}} = -W = +800\ \text{J}\): the gas
absorbed 800 J more heat than it expelled, and turned it into work.
Clockwise cycles are engines; counterclockwise cycles are refrigerators.
Check one leg: \(U_B = \tfrac{3}{2}PV = 900\ \text{J}\), \(U_C = 2700\ \text{J}\), so
\(Q_{BC} = \Delta U - W = 1800 - (-1200) = 3000\ \text{J}\) flows in during the expansion.
Try it
0.10 mol of a monatomic ideal gas is compressed adiabatically, with 250 J
of work done on it. Find \(\Delta U\) and the temperature change.
Show answer
Adiabatic: \(Q = 0\), so \(\Delta U = W = +250\ \text{J}\). Then
\(\Delta T = \dfrac{\Delta U}{\tfrac{3}{2}nR} = \dfrac{250}{1.5(0.10)(8.31)} \approx 201\ \text{K}\).
The gas heats up with no heat added: the work did it.
Lesson 9.6 · Unit 9 · CED topic 9.9
The second law of thermodynamics and entropy
The first law allows a cold cup of coffee to spontaneously heat up by
chilling the room: energy would be conserved. It never happens. The
second law explains why, and it puts a hard ceiling on how much of any
heat flow can be turned into useful work.
Definition
Entropy \(S\) measures how dispersed a system's energy
is: how many microscopic arrangements are consistent with what you
see. When heat \(Q\) enters an object held at temperature \(T\), its
entropy rises by \(\Delta S = Q/T\). Second law: the
total entropy of an isolated system never decreases; spontaneous
processes move toward more dispersed, more probable states, and are
therefore irreversible. A heat engine
takes in \(Q_H\), expels \(Q_C\), and does work \(W = Q_H - Q_C\); its
efficiency is
\[e = \frac{W}{Q_H} = 1 - \frac{Q_C}{Q_H},\]
which is always less than 1 because some heat must be expelled.
Worked example · Why heat flows hot to cold
500 J of heat flows from a block at 400 K to a block at 300 K, both large enough that their temperatures barely change. Show that total entropy increases.
Hot block: \(\Delta S_H = \dfrac{-500\ \text{J}}{400\ \text{K}} = -1.25\ \text{J/K}\).
Cold block: \(\Delta S_C = \dfrac{+500\ \text{J}}{300\ \text{K}} = +1.67\ \text{J/K}\).
Total: \(+0.42\ \text{J/K} \gt 0\). The same joules count for more
entropy at low temperature, so the cold block gains more than the hot
block loses. The reverse flow would give \(-0.42\ \text{J/K}\): forbidden.
Worked example · Engine efficiency
Each cycle, an engine absorbs 1200 J from a hot reservoir and exhausts 900 J to the cold one. Find the work per cycle and the efficiency.
\(W = Q_H - Q_C = 1200 - 900 = 300\ \text{J}\), so
\(e = \dfrac{W}{Q_H} = \dfrac{300}{1200} = 0.25\), or 25%.
An engine claiming to convert all 1200 J to work (\(Q_C = 0\)) would
lower the hot reservoir's entropy with nothing to compensate: it
violates the second law, not the first.
In words: a shattered glass doesn't reassemble, gas doesn't rush back
into one corner of a room, and hot never spontaneously gets hotter,
because each of those would move from a very probable arrangement to a
spectacularly improbable one. Energy is conserved either way; entropy
picks the direction.
Try it
An engine takes in 4000 J per cycle and exhausts 2400 J. Find its efficiency and its work output. Could an engine with these reservoirs be 100% efficient?
Show answer
\(W = 4000 - 2400 = 1600\ \text{J}\); \(e = 1 - \dfrac{2400}{4000} = 0.40\), or 40%.
No: 100% efficiency would mean \(Q_C = 0\), decreasing total entropy, which the second law forbids.
Unit 9 practice · 10 problems
Unit 9 practice: Thermodynamics
Ten problems covering the whole unit, in roughly exam order. Work each one on paper
before revealing the answer: the reveal shows the key steps, not just the number.
Constants: \(R = 8.31\ \text{J/(mol·K)}\), \(k_B = 1.38\times10^{-23}\ \text{J/K}\),
g = 9.8 m/s². Treat every gas as monatomic and ideal. Calculator allowed.
A vertical cylinder is sealed by a frictionless piston of area 0.012 m². The gas inside is at 2.5 × 10⁵ Pa; the atmosphere above the piston is at 1.0 × 10⁵ Pa. Find the net force the two gases exert on the piston, and the largest mass the piston could support at rest (ignore the piston's own weight).
A sealed sample of gas is heated so that its absolute temperature doubles. The rms speed of its molecules
Temperature is a measure of the average kinetic energy of the molecules, so it cannot rise without the molecules moving faster. In a sealed container the added energy goes entirely into molecular motion.
\(\tfrac{3}{2}k_BT = \tfrac{1}{2}mv_{\text{rms}}^2\) gives \(v_{\text{rms}} = \sqrt{3k_BT/m} \propto \sqrt{T}\), so doubling \(T\) doubles the average kinetic energy but multiplies the speed by only \(\sqrt{2}\).
Doubling the speed would quadruple the kinetic energy (since \(K \propto v^2\)) and therefore quadruple the temperature. The speed scales as the square root of \(T\), not linearly with it.
Quadrupling comes from applying the \(v^2\) relationship in the wrong direction. \(T\) is proportional to \(v^2\), so \(v\) is proportional to \(\sqrt{T}\), and the speed changes by less than the temperature does, not more.
A single-pane window has area 1.5 m², thickness 4.0 mm, and thermal conductivity 0.80 W/(m·K). The inside surface is 15 K warmer than the outside surface. Find the rate of heat conduction through the glass and the energy lost in one hour. What would doubling the glass thickness do to the rate?
Show answer
\(\dfrac{Q}{\Delta t} = \dfrac{kA\,\Delta T}{L} = \dfrac{(0.80)(1.5)(15)}{4.0\times10^{-3}} = 4.5\times10^{3}\ \text{W}\). In one hour, \(Q = (4500\ \text{W})(3600\ \text{s}) = 1.6\times10^{7}\ \text{J}\). The rate is inversely proportional to \(L\), so doubling the thickness halves it to \(2.3\times10^{3}\ \text{W}\).
A 0.200 kg copper block (\(c = 385\ \text{J/(kg·K)}\)) at 150 °C is dropped into 0.500 kg of water (\(c = 4186\ \text{J/(kg·K)}\)) at 20.0 °C in an insulated cup. Find the final temperature, and explain in one sentence why it is so close to the water's starting temperature.
Show answer
Heat lost by copper = heat gained by water: \(m_c c_c(150 - T_f) = m_w c_w(T_f - 20.0)\), so \(77(150 - T_f) = 2093(T_f - 20.0)\). Then \(11{,}550 + 41{,}860 = 2170\,T_f\), giving \(T_f \approx 24.6\ ^\circ\text{C}\). The water's heat capacity \(mc = 2093\ \text{J/K}\) is 27 times the copper's 77 J/K, so the same heat changes its temperature 27 times less.
On a PV diagram a gas moves along a straight line from state A (V = 2.0 L, P = 4.0 × 10⁵ Pa) to state B (V = 5.0 L, P = 1.0 × 10⁵ Pa). Find the work done on the gas, the change in internal energy, and the heat transferred.
Show answer
The area under the slanted segment is a trapezoid: \(|W| = \tfrac{1}{2}(P_A + P_B)\Delta V = \tfrac{1}{2}(5.0\times10^{5})(3.0\times10^{-3}) = 750\ \text{J}\). The gas expands, so \(W_{\text{on}} = -750\ \text{J}\). Internal energy from the endpoints: \(U_A = \tfrac{3}{2}P_AV_A = \tfrac{3}{2}(4.0\times10^{5})(2.0\times10^{-3}) = 1200\ \text{J}\), \(U_B = \tfrac{3}{2}(1.0\times10^{5})(5.0\times10^{-3}) = 750\ \text{J}\), so \(\Delta U = -450\ \text{J}\). First law: \(Q = \Delta U - W = -450 - (-750) = +300\ \text{J}\) flows into the gas.
1.5 mol of gas expands at a constant pressure of 2.0 × 10⁵ Pa from 3.0 L to 5.0 L. Find the work done on the gas, the temperature change, the change in internal energy, and the heat added.
Show answer
\(W = -P\,\Delta V = -(2.0\times10^{5})(2.0\times10^{-3}) = -400\ \text{J}\). From \(P\,\Delta V = nR\,\Delta T\): \(\Delta T = \dfrac{400}{(1.5)(8.31)} \approx 32\ \text{K}\). \(\Delta U = \tfrac{3}{2}nR\,\Delta T = \tfrac{3}{2}(400) = 600\ \text{J}\). \(Q = \Delta U - W = 600 - (-400) = 1000\ \text{J}\). For an isobaric expansion, only 3/5 of the heat raises the temperature; the rest leaves as work.
A gas runs clockwise around a rectangle on a PV diagram: A (1.0 L, 1.0 × 10⁵ Pa) → B (1.0 L, 4.0 × 10⁵ Pa) → C (3.0 L, 4.0 × 10⁵ Pa) → D (3.0 L, 1.0 × 10⁵ Pa) → A. Find the net work done by the gas per cycle, then rank the four legs by the heat added to the gas, greatest to least.
Show answer
Net work by the gas is the enclosed area: \((3.0\times10^{5}\ \text{Pa})(2.0\times10^{-3}\ \text{m}^3) = 600\ \text{J}\). Per leg, using \(U = \tfrac{3}{2}PV\) and \(Q = \Delta U - W\): AB (isochoric, \(W = 0\)): \(Q = \Delta U = +450\ \text{J}\). BC (isobaric expansion, \(W = -800\ \text{J}\)): \(\Delta U = +1200\ \text{J}\), \(Q = +2000\ \text{J}\). CD (isochoric): \(Q = \Delta U = -1350\ \text{J}\). DA (isobaric compression, \(W = +200\ \text{J}\)): \(\Delta U = -300\ \text{J}\), \(Q = -500\ \text{J}\). Ranking: BC > AB > DA > CD. Check: \(450 + 2000 - 1350 - 500 = +600\ \text{J} = -W_{\text{net, on gas}}\). ✓
A gas is compressed rapidly enough that no heat is exchanged with the surroundings. Explain in words, using the first law and a particle-level picture, why its temperature rises.
Show answer
Adiabatic means \(Q = 0\), so \(\Delta U = W\); compression is positive work on the gas, so \(U\), and therefore \(T\), must rise. At the particle level, molecules that bounce off the inward-moving piston rebound faster than they arrived, like a ball hit by a moving bat. Each collision adds kinetic energy, and with no heat leaving, the average kinetic energy, which is temperature, goes up. This is why a bicycle pump gets warm.
Each cycle, a heat engine absorbs 3000 J from a reservoir at 600 K and exhausts 2100 J to a reservoir at 300 K. (a) Find the work per cycle and the efficiency. (b) Find the entropy change of each reservoir and show that the total entropy increases. (c) Explain why an engine with \(Q_C = 0\) is forbidden.
Show answer
(a) \(W = Q_H - Q_C = 900\ \text{J}\); \(e = \dfrac{W}{Q_H} = \dfrac{900}{3000} = 0.30\), or 30%. (b) \(\Delta S_H = \dfrac{-3000}{600} = -5.0\ \text{J/K}\); \(\Delta S_C = \dfrac{+2100}{300} = +7.0\ \text{J/K}\); total \(+2.0\ \text{J/K} \gt 0\). (c) With \(Q_C = 0\) the hot reservoir's entropy drops by 5.0 J/K and nothing compensates, so total entropy would decrease: the second law forbids it even though energy would be conserved.
Lesson 10.1 · Unit 10 · CED topics 10.1–10.2
Electric charge, conservation, and charging
Everything in Units 10 and 11 starts with one fact: matter carries
charge, and charge is never created or destroyed; only moved around.
Every "the balloon sticks to the wall" question is a bookkeeping
question about where electrons went.
Definition
Charge is quantized: every object's net charge is an
integer multiple of the elementary charge, \(q = ne\) with
\(e = 1.60\times10^{-19}\ \text{C}\). Electrons carry \(-e\), protons \(+e\).
Charge is conserved: the total charge of an isolated
system never changes. In a conductor some electrons
move freely; in an insulator they stay put, though the
atoms can polarize (stretch so one side is slightly
positive). Like charges repel; unlike attract.
Worked example · Counting electrons
A balloon rubbed on hair acquires a charge of −2.4 × 10⁻⁹ C. How many electrons did it gain, and what happened to the hair?
\[n = \frac{|q|}{e} = \frac{2.4\times10^{-9}\ \text{C}}{1.60\times10^{-19}\ \text{C}} = 1.5\times10^{10}\ \text{electrons}.\]
Only electrons move during rubbing. By conservation, the hair lost
exactly those electrons and is left with +2.4 × 10⁻⁹ C.
Charging by friction never creates charge; it separates it.
Worked example · Explain in words: charging by induction
A neutral metal sphere sits on an insulating stand. Describe how to give it a net positive charge using only a negatively charged rod, without touching it with the rod.
Bring the rod near one side. Free electrons in the sphere are repelled
to the far side, leaving the near side positive: the sphere is still
neutral overall, just polarized. Now touch the far side with a finger
(a ground): the crowded electrons escape into your body.
Remove the finger first, then the rod. The sphere is left
short of electrons, so it's positive. Order matters: lift the rod
first and the electrons rush back before you can trap the deficit.
Worked example · Charging by conduction
Identical conducting spheres carrying +6.0 μC and −2.0 μC touch and are separated. Find each final charge.
Total charge is conserved: \(+6.0 + (-2.0) = +4.0\ \mu\text{C}\). Identical
conductors share it equally, so each ends with +2.0 μC.
Equal sharing works only for identical spheres: say so on the exam.
Try it
A sphere holding +8.0 nC touches an identical neutral sphere. Find each sphere's final charge and the number of electrons that moved between them.
Show answer
Each gets +4.0 nC. The neutral sphere became positive by losing
electrons to the other one: \(n = \dfrac{4.0\times10^{-9}}{1.60\times10^{-19}} = 2.5\times10^{10}\) electrons.
Lesson 10.2 · Unit 10 · CED topic 10.3
Coulomb's law
Coulomb's law looks just like Newton's gravitation with charge in place
of mass: an inverse-square law along the line joining two objects. The
one new wrinkle is that it can push as well as pull, so you must reason
about direction separately from magnitude.
Formula
The magnitude of the force between point charges \(q_1\) and \(q_2\) a distance \(r\) apart is
\[F = \frac{k\,|q_1 q_2|}{r^2}, \qquad k = 8.99\times10^{9}\ \text{N·m}^2/\text{C}^2.\]
The force acts along the line joining the charges: repulsive for like
signs, attractive for unlike. The two charges feel equal and opposite
forces (Newton's third law). With several charges, the net force on
one is the vector sum of the forces from each of the others
(superposition): find each magnitude, assign each a direction, then add.
Worked example · Three charges in a line
On the x-axis: \(q_1 = +2.0\ \mu\text{C}\) at \(x = 0\), \(q_2 = -3.0\ \mu\text{C}\)
at \(x = 0.30\ \text{m}\), and \(q_3 = +4.0\ \mu\text{C}\) at \(x = 0.50\ \text{m}\). Find the net force on \(q_2\).
Directions first. \(q_1\) and \(q_2\) have opposite signs,
so \(q_2\) is attracted toward \(q_1\): that force points in the −x
direction. \(q_3\) and \(q_2\) also have opposite signs, so \(q_2\) is
attracted toward \(q_3\): +x direction. The two forces oppose each other.
Magnitudes. From \(q_1\) (0.30 m away):
\[F_{12} = \frac{(8.99\times10^{9})(2.0\times10^{-6})(3.0\times10^{-6})}{(0.30)^2} = 0.599\ \text{N}.\]
From \(q_3\) (0.20 m away):
\[F_{32} = \frac{(8.99\times10^{9})(3.0\times10^{-6})(4.0\times10^{-6})}{(0.20)^2} = 2.70\ \text{N}.\]
Net: \(F = 2.70 - 0.60 = 2.10\ \text{N}\) in the +x direction
(toward \(q_3\)). Notice that \(q_3\) wins easily: it's both larger
and closer, and the distance enters squared.
Worked example · Proportional reasoning
If one charge is doubled and the separation tripled, how does the force change?
\(F \propto \dfrac{q_1 q_2}{r^2}\), so the new force is \(\dfrac{2}{3^2} = \dfrac{2}{9}\) of the original. Ratio questions like this are the most common Coulomb MCQ, no constant needed.
Try it
\(+4.0\ \mu\text{C}\) sits at \(x = 0\) and \(+1.0\ \mu\text{C}\) at \(x = 0.60\ \text{m}\). Where between them would a third charge feel zero net force?
Show answer
Set the magnitudes equal: \(\dfrac{k(4.0)q}{x^2} = \dfrac{k(1.0)q}{(0.60 - x)^2}\), so
\(\dfrac{2}{x} = \dfrac{1}{0.60 - x}\), giving \(x = 0.40\ \text{m}\): closer to the smaller charge, as it must be. The result doesn't depend on the third charge's sign or size.
Lesson 10.3 · Unit 10 · CED topics 10.4–10.5
Electric fields
Coulomb's law needs two charges. The field idea splits the job: a charge
sets up a condition in the space around it, and any other charge that
shows up feels a force from that condition. Fields let you describe the
space itself, before you know what will be placed in it.
Definition
The electric field at a point is the force per unit charge on a small positive test charge placed there:
\(\vec{E} = \dfrac{\vec{F}}{q}\), in N/C. So a charge \(q\) in a field feels
\(\vec{F} = q\vec{E}\): along \(\vec{E}\) if \(q\) is positive, opposite if negative.
A point charge \(Q\) produces \(E = \dfrac{k|Q|}{r^2}\), pointing
away from a positive charge and toward a negative one.
Fields from several charges add as vectors.
Field-line rules: lines start on positive charges and end on
negative ones (or at infinity); the number of lines is proportional to the
charge; closer lines mean a stronger field; lines never cross; at a
conductor's surface they meet it perpendicularly; between two large,
closely spaced parallel plates the lines are evenly spaced and straight:
a uniform field.
Worked example · Field of a point charge
Find the field 0.20 m from a −5.0 nC charge, and the force on a proton placed there.
\[E = \frac{k|Q|}{r^2} = \frac{(8.99\times10^{9})(5.0\times10^{-9})}{(0.20)^2} = 1.12\times10^{3}\ \text{N/C},\]
directed toward the negative charge. The proton is positive, so its
force is along \(\vec{E}\): \(F = eE = (1.60\times10^{-19})(1.12\times10^{3}) = 1.80\times10^{-16}\ \text{N}\),
toward the charge. An electron at the same spot would feel the same
magnitude of force pointing away.
Worked example · Superposition of fields
Two +3.0 nC charges sit at \(x = 0\) and \(x = 0.40\ \text{m}\). Find the field at \(x = 0.10\ \text{m}\) and at the midpoint.
At \(x = 0.10\): the charge at the origin is 0.10 m away and pushes the
field away from itself, in +x:
\(E_1 = \dfrac{(8.99\times10^{9})(3.0\times10^{-9})}{(0.10)^2} = 2697\ \text{N/C}\).
The other charge is 0.30 m away and its field points away from it, in −x:
\(E_2 = \dfrac{(8.99\times10^{9})(3.0\times10^{-9})}{(0.30)^2} = 300\ \text{N/C}\).
Net: \(2697 - 300 \approx 2.40\times10^{3}\ \text{N/C}\) in the +x direction.
At the midpoint the two contributions are equal and opposite, so
\(E = 0\), even though the potential there (next lesson) is not zero.
Try it
A proton (\(m = 1.67\times10^{-27}\ \text{kg}\)) is released in a uniform field of 2.0 × 10⁴ N/C between two plates. Find its acceleration.
Show answer
\(a = \dfrac{F}{m} = \dfrac{eE}{m} = \dfrac{(1.60\times10^{-19})(2.0\times10^{4})}{1.67\times10^{-27}} \approx 1.9\times10^{12}\ \text{m/s}^2\), in the direction of \(\vec{E}\). Gravity (9.8 m/s²) is utterly negligible here.
Lesson 10.4 · Unit 10 · CED topics 10.6–10.7
Electric potential energy and electric potential
Forces and fields are vectors, and adding vectors is work. Energy is a
scalar. Whenever a problem asks how fast a charge is moving after it has
been pushed around by other charges, potential energy is the shortcut,
no directions to track.
Formula
Potential energy of two point charges (zero at infinite separation):
\(U = \dfrac{k q_1 q_2}{r}\); keep the signs; like charges store positive
energy, unlike charges negative. Potential is potential
energy per unit charge: for a point charge, \(V = \dfrac{kq}{r}\), in volts
(J/C). Potentials from several charges add as ordinary numbers.
Moving a charge through a potential difference changes its energy by
\[\Delta U = q\,\Delta V,\]
and energy conservation gives \(\Delta K = -q\,\Delta V\).
Equipotential lines are always perpendicular to \(\vec{E}\), and
\(\vec{E}\) points from high potential to low. In a uniform field,
\(|E| = \dfrac{|\Delta V|}{d}\), so N/C and V/m are the same unit.
Worked example · Potential is a scalar
A +2.0 nC charge and a −2.0 nC charge are 0.40 m apart. Find \(V\) at a point 0.10 m from the positive charge and 0.30 m from the negative one (on the line between them).
\[V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2} = (8.99\times10^{9})\left[\frac{2.0\times10^{-9}}{0.10} - \frac{2.0\times10^{-9}}{0.30}\right] \approx 120\ \text{V}.\]
No directions, just signed numbers. At the midpoint the two terms cancel and \(V = 0\), though \(\vec{E}\) there is strong and points toward the negative charge.
Worked example · An accelerated charge
A proton (\(m = 1.67\times10^{-27}\ \text{kg}\)) starts from rest and moves through a potential drop of 500 V. Find its final speed.
A positive charge speeds up moving to lower potential, so \(\Delta V = -500\ \text{V}\) and
\[\Delta K = -q\,\Delta V = -(1.60\times10^{-19}\ \text{C})(-500\ \text{V}) = 8.0\times10^{-17}\ \text{J}.\]
\[v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2(8.0\times10^{-17})}{1.67\times10^{-27}}} \approx 3.1\times10^{5}\ \text{m/s}.\]
An electron through the same 500 V would gain the same energy but,
being 1836 times lighter, would move about 43 times faster.
Worked example · Parallel plates
Two plates 3.0 mm apart are connected to a 12 V battery. Find the field between them.
\(E = \dfrac{\Delta V}{d} = \dfrac{12\ \text{V}}{3.0\times10^{-3}\ \text{m}} = 4.0\times10^{3}\ \text{V/m}\), from the positive plate toward the negative plate. The equipotentials are planes parallel to the plates, spaced evenly.
Try it
Two +3.0 μC charges are held 0.050 m apart. Find their potential energy, and the work needed to push them to 0.025 m apart.
Show answer
\(U_1 = \dfrac{(8.99\times10^{9})(3.0\times10^{-6})^2}{0.050} = 1.62\ \text{J}\).
Halving \(r\) doubles \(U\): \(U_2 = 3.24\ \text{J}\). Work by you \(= \Delta U = +1.62\ \text{J}\).
Lesson 10.5 · Unit 10 · CED topic 10.8
Capacitors
Two conductors with opposite charges store energy in the field between
them. That's a capacitor, and the parallel-plate version is the one the
exam loves, because every quantity about it can be predicted from
geometry, and because "what changes if…" questions test whether you
know which variable is being held fixed.
Formula
Capacitance is charge stored per volt: \(C = \dfrac{Q}{\Delta V}\), in farads (C/V).
For parallel plates of area \(A\), separation \(d\), filled with a
dielectric of constant \(\kappa\) (\(\kappa = 1\) for air):
\[C = \frac{\kappa\,\varepsilon_0 A}{d}, \qquad \varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/(\text{N·m}^2).\]
Energy stored: \(U = \tfrac{1}{2}Q\,\Delta V = \tfrac{1}{2}C(\Delta V)^2 = \dfrac{Q^2}{2C}\).
The field between the plates is uniform, \(E = \Delta V/d\). A dielectric
polarizes and weakens the field inside, raising \(C\) by the factor \(\kappa\).
Worked example · A parallel-plate capacitor
Plates of area 0.020 m² are separated by 0.50 mm of air and connected to 12 V. Find \(C\), \(Q\), and the stored energy.
The plates of a charged capacitor are pulled to twice their separation. What happens to \(C\), \(Q\), \(\Delta V\), \(E\), and \(U\) if the battery has been disconnected? If it stays connected?
Either way, \(C = \varepsilon_0 A/d\) halves. Then pick the formula whose other variable is fixed:
Doubling d
C
Q
ΔV
E = ΔV/d
U
Isolated (Q fixed)
× ½
same
× 2 (V = Q/C)
same
× 2 (U = Q²/2C)
Battery connected (ΔV fixed)
× ½
× ½ (Q = CV)
same
× ½
× ½ (U = ½CV²)
Isolated, the energy doubles because you did work pulling the
attracting plates apart. Connected, the energy halves because half the
charge flowed back into the battery. Inserting a dielectric runs the
same logic in reverse: \(C\) rises by \(\kappa\); with the battery
connected \(Q\) and \(U\) rise by \(\kappa\), while isolated \(\Delta V\) and \(U\) fall by \(\kappa\).
Try it
A 2.0 μF capacitor is charged to 9.0 V and disconnected. Find \(Q\) and \(U\). The plates are then pulled to three times their separation; find the new \(\Delta V\) and \(U\).
Show answer
\(Q = C\,\Delta V = 18\ \mu\text{C}\); \(U = \tfrac{1}{2}(2.0\times10^{-6})(9.0)^2 = 81\ \mu\text{J}\).
Isolated, \(Q\) stays 18 μC while \(C\) drops to 0.667 μF, so \(\Delta V = Q/C = 27\ \text{V}\) and
\(U = \tfrac{1}{2}Q\,\Delta V = 243\ \mu\text{J}\): triple, matching the work done pulling the plates apart.
Unit 10 practice · 10 problems
Unit 10 practice: Electric Force, Field, and Potential
Ten problems covering the whole unit, in roughly exam order. Work each one on paper
before revealing the answer. Constants: \(k = 8.99\times10^{9}\ \text{N·m}^2/\text{C}^2\),
\(e = 1.60\times10^{-19}\ \text{C}\), \(\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/(\text{N·m}^2)\),
\(m_e = 9.11\times10^{-31}\ \text{kg}\). Calculator allowed.
Identical conducting sphere A carries +6.0 nC and sphere B carries −4.0 nC. They touch and are separated. Find each final charge, and the number of electrons that moved from one sphere to the other. Which direction did they move?
Show answer
Charge is conserved: \(+6.0 + (-4.0) = +2.0\ \text{nC}\), shared equally by identical spheres, so each ends with +1.0 nC. Sphere B went from −4.0 nC to +1.0 nC, so it lost 5.0 nC of electrons to A: \(n = \dfrac{5.0\times10^{-9}}{1.60\times10^{-19}} \approx 3.1\times10^{10}\) electrons moved from B to A.
A +3.0 μC charge and a −5.0 μC charge are 0.20 m apart. Find the magnitude of the force on each, and state whether it is attractive or repulsive.
Show answer
\(F = \dfrac{k|q_1q_2|}{r^2} = \dfrac{(8.99\times10^{9})(3.0\times10^{-6})(5.0\times10^{-6})}{(0.20)^2} = 3.4\ \text{N}\). Opposite signs: attractive. Each charge feels 3.4 N toward the other (Newton's third law), regardless of which is larger.
Both charges in a pair are doubled and their separation is halved. The electric force between them becomes
A factor of 2 would come from doubling only one charge and leaving the distance alone. Here both charges double (a factor of 4 in the numerator) and the distance halves (another factor of 4 from the inverse square).
4 accounts for the two doubled charges (\(2 \times 2\)) but ignores the halved separation, or accounts for the separation but forgets the charges. Each change contributes its own factor of 4.
8 is what you get by treating the distance as inverse-linear, as if halving \(r\) merely doubled \(F\). Coulomb's law has \(r^2\) in the denominator, so halving \(r\) multiplies \(F\) by 4, not 2.
\(F \propto \dfrac{q_1q_2}{r^2}\): the numerator grows by \(2\times2 = 4\) and the denominator shrinks by \((\tfrac{1}{2})^2 = \tfrac{1}{4}\), so \(F\) grows by \(4 \times 4 = 16\).
On the x-axis: \(q_1 = +2.0\ \mu\text{C}\) at \(x = 0\), \(q_3 = -1.0\ \mu\text{C}\) at \(x = 0.10\ \text{m}\), and \(q_2 = +2.0\ \mu\text{C}\) at \(x = 0.40\ \text{m}\). Find the net force on \(q_3\).
Show answer
Directions first: \(q_3\) is attracted toward \(q_1\) (−x) and toward \(q_2\) (+x). Magnitudes: from \(q_1\), 0.10 m away, \(F_1 = \dfrac{(8.99\times10^{9})(2.0\times10^{-6})(1.0\times10^{-6})}{(0.10)^2} = 1.80\ \text{N}\); from \(q_2\), 0.30 m away, \(F_2 = \dfrac{(8.99\times10^{9})(2.0\times10^{-6})(1.0\times10^{-6})}{(0.30)^2} = 0.20\ \text{N}\). Net: \(1.80 - 0.20 = 1.6\ \text{N}\) in the −x direction (toward \(q_1\), the closer charge).
Find the electric field 0.30 m from a +6.0 nC point charge. An electron is released there; find the force on it and its initial acceleration, with directions.
Show answer
\(E = \dfrac{k|Q|}{r^2} = \dfrac{(8.99\times10^{9})(6.0\times10^{-9})}{(0.30)^2} = 6.0\times10^{2}\ \text{N/C}\), pointing away from the positive charge. The electron is negative, so \(\vec{F}\) is opposite \(\vec{E}\): \(F = eE = (1.60\times10^{-19})(600) = 9.6\times10^{-17}\ \text{N}\) toward the charge. \(a = \dfrac{F}{m_e} = \dfrac{9.6\times10^{-17}}{9.11\times10^{-31}} \approx 1.1\times10^{14}\ \text{m/s}^2\), toward the charge.
A +4.0 nC charge is at \(x = 0\) and a +9.0 nC charge at \(x = 0.50\ \text{m}\). Find the point between them where the net electric field is zero, and the electric potential at that point.
Show answer
Set the magnitudes equal: \(\dfrac{k(4.0)}{x^2} = \dfrac{k(9.0)}{(0.50 - x)^2}\), so \(\dfrac{2}{x} = \dfrac{3}{0.50 - x}\), giving \(1.0 - 2x = 3x\) and \(x = 0.20\ \text{m}\) (closer to the smaller charge). Potential is a scalar sum: \(V = k\left(\dfrac{4.0\times10^{-9}}{0.20} + \dfrac{9.0\times10^{-9}}{0.30}\right) = (8.99\times10^{9})(5.0\times10^{-8}) \approx 4.5\times10^{2}\ \text{V}\). Zero field, but not zero potential.
Rank the electric potential energy of these pairs of point charges, greatest to least: (i) \(+2q\) and \(+2q\) separated by \(r\); (ii) \(+q\) and \(-4q\) separated by \(r\); (iii) \(+q\) and \(+q\) separated by \(r/2\); (iv) \(-2q\) and \(-2q\) separated by \(2r\).
Show answer
\(U = \dfrac{kq_1q_2}{r}\), signs included. In units of \(kq^2/r\): (i) \(\dfrac{4}{1} = 4\); (ii) \(-4\); (iii) \(\dfrac{1}{1/2} = 2\); (iv) \(\dfrac{4}{2} = 2\). Ranking: (i) > (iii) = (iv) > (ii). Like charges store positive energy; the only negative entry is the unlike pair, which sits at the bottom no matter its size.
Two parallel plates 2.0 cm apart are held at a potential difference of 400 V. (a) Find the field between them. (b) An electron released from rest at the negative plate reaches the positive plate; find its speed there. (c) How much work is needed to push a +3.0 μC charge from the negative plate to the positive plate?
Show answer
(a) \(E = \dfrac{\Delta V}{d} = \dfrac{400}{0.020} = 2.0\times10^{4}\ \text{V/m}\), from the positive plate toward the negative one. (b) The electron moves from low to high potential, so it speeds up: \(\Delta K = -q\,\Delta V = -(-e)(+400\ \text{V}) = e\,\Delta V = (1.60\times10^{-19})(400) = 6.4\times10^{-17}\ \text{J}\), and \(v = \sqrt{\dfrac{2K}{m_e}} = \sqrt{\dfrac{2(6.4\times10^{-17})}{9.11\times10^{-31}}} \approx 1.2\times10^{7}\ \text{m/s}\). (c) Pushing a positive charge toward the positive plate raises its energy: \(W = q\,\Delta V = (3.0\times10^{-6})(400) = 1.2\times10^{-3}\ \text{J}\).
A field-line sketch shows two charges, A and B. Twelve lines leave A; six of them end on B and six run off the page. Explain in words (a) the sign of each charge, (b) the ratio \(|q_A|/|q_B|\), (c) why field lines never cross, and (d) how you would compare the field strength at two points using only the sketch.
Show answer
(a) Lines leave positive charges and end on negative ones, so A is positive and B is negative. (b) Line count is proportional to charge: 12 lines on A versus 6 ending on B, so \(|q_A| = 2|q_B|\). (c) The field at any point has one direction; a crossing would give two. (d) The field is stronger where the lines are closer together, so compare the spacing of the lines around each point.
A parallel-plate capacitor has plate area 0.040 m² and separation 0.20 mm (air). It is charged to 50 V and then disconnected from the battery. Find \(C\), \(Q\), and the stored energy. A dielectric with \(\kappa = 3.0\) is then slid between the plates; find the new voltage and energy, and explain where the energy went.
Show answer
\(C = \dfrac{\varepsilon_0 A}{d} = \dfrac{(8.85\times10^{-12})(0.040)}{2.0\times10^{-4}} = 1.8\times10^{-9}\ \text{F}\); \(Q = C\,\Delta V = 8.9\times10^{-8}\ \text{C}\); \(U = \tfrac{1}{2}C(\Delta V)^2 = 2.2\times10^{-6}\ \text{J}\). Disconnected, \(Q\) is fixed while \(C\) triples to \(5.3\times10^{-9}\ \text{F}\), so \(\Delta V = Q/C = 50/3 \approx 17\ \text{V}\) and \(U = \dfrac{Q^2}{2C}\) falls to one third, \(7.4\times10^{-7}\ \text{J}\). The field pulls the dielectric in; the lost energy became work done on the slab.
Lesson 11.1 · Unit 11 · CED topics 11.1–11.2
Current, resistance, and resistivity
A circuit is a closed path along which charge flows. Two questions
organize the whole unit: how much charge passes per second (current),
and how hard the path makes that flow (resistance). Both have clean
microscopic pictures the exam wants you to be able to describe.
Definition
Current is the rate at which charge passes a point:
\(I = \dfrac{\Delta Q}{\Delta t}\), in amperes (1 A = 1 C/s).
Conventional current points in the direction positive
charge would move, from the + terminal through the circuit to the −
terminal, even though in a metal it's electrons drifting the other
way. The resistance of a uniform wire is
\[R = \frac{\rho L}{A},\]
where \(\rho\) is the resistivity (a property of the
material, in Ω·m), \(L\) the length, and \(A\) the cross-sectional area.
Resistivity of metals rises with temperature.
Worked example · Charge and current
A phone charger delivers 2.0 A for 30 s. How much charge passes, and how many electrons is that?
\(\Delta Q = I\,\Delta t = (2.0\ \text{A})(30\ \text{s}) = 60\ \text{C}\).
Number of electrons: \(\dfrac{60\ \text{C}}{1.60\times10^{-19}\ \text{C}} = 3.8\times10^{20}\).
Sixty coulombs of charge moved, but the wire stays neutral throughout: charge flows through it, it doesn't pile up.
Worked example · Resistance of a wire
Find the resistance of 25 m of copper wire (\(\rho = 1.72\times10^{-8}\ \Omega\text{·m}\)) with diameter 1.6 mm.
Radius 0.80 mm, so \(A = \pi r^2 = \pi(0.80\times10^{-3}\ \text{m})^2 = 2.01\times10^{-6}\ \text{m}^2\).
\[R = \frac{\rho L}{A} = \frac{(1.72\times10^{-8}\ \Omega\text{·m})(25\ \text{m})}{2.01\times10^{-6}\ \text{m}^2} = 0.21\ \Omega.\]
Small, which is why we usually treat connecting wires as resistance-free.
Worked example · Explain in words
Why does a longer, thinner wire have more resistance?
Electrons drifting through a metal keep colliding with the lattice of
ions. A longer wire means more lattice to get through: more collisions
along the way, so more opposition per volt applied; that's the
\(L\) upstairs. A thicker wire offers more side-by-side channels for
charge to move through at once, like adding lanes to a highway, so the
same voltage moves more charge per second; that's the \(A\) downstairs.
Stretching a wire to double its length at fixed volume halves its area,
so \(R\) goes up by a factor of four.
Try it
Nichrome has \(\rho = 1.10\times10^{-6}\ \Omega\text{·m}\). What length of nichrome wire with cross-section 0.50 mm² makes a 5.0 Ω heating element?
Push harder and more flows: for many materials the current through a
component is proportional to the voltage across it. Combine that with
the definition of power and you can predict which bulb glows brighter:
a question the exam asks in some form almost every year.
Formula
Ohm's law: \(\Delta V = IR\). A component is ohmic
if \(R\) is constant, so its \(I\)–\(V\) graph is a straight line through
the origin with slope \(1/R\). A light-bulb filament is non-ohmic: it
heats up as current grows, its resistance rises, and the graph bends
toward the \(V\) axis. Power delivered to a component is
\[P = I\,\Delta V = I^2 R = \frac{(\Delta V)^2}{R}.\]
Choose the form whose two quantities you actually know. Energy over time
\(\Delta t\) is \(E = P\,\Delta t\).
Worked example · Bulb brightness
A 60 W bulb and a 100 W bulb are both rated for 120 V. Which is brighter when they're connected (a) in parallel across 120 V, and (b) in series across 120 V?
First find each bulb's resistance from its rating, using \(R = (\Delta V)^2/P\):
\(R_{60} = \dfrac{120^2}{60} = 240\ \Omega\) and \(R_{100} = \dfrac{120^2}{100} = 144\ \Omega\).
The brighter bulb is the one dissipating more power.
(a) In parallel each bulb gets the full 120 V, so each runs at its
rating: the 100 W bulb is brighter (lower \(R\), more current).
(b) In series the same current flows through both:
\(I = \dfrac{120\ \text{V}}{240 + 144\ \Omega} = 0.3125\ \text{A}\). Then
\(P_{60} = I^2R_{60} = (0.3125)^2(240) = 23\ \text{W}\) and
\(P_{100} = (0.3125)^2(144) = 14\ \text{W}\). The "60 W" bulb is now
brighter, and both are dim. Same current, so
\(P = I^2R\) favors the larger resistance. Rule of thumb: parallel
favors small \(R\); series favors large \(R\).
Worked example · Energy and current draw
A 1500 W space heater runs on 120 V for 3.0 h. Find the current it draws and the energy used.
\(I = \dfrac{P}{\Delta V} = \dfrac{1500\ \text{W}}{120\ \text{V}} = 12.5\ \text{A}\): close to a 15 A household breaker.
\(E = P\,\Delta t = (1500\ \text{W})(3.0\ \text{h}) = 4500\ \text{Wh} = 4.5\ \text{kWh}\), or \(1.6\times10^{7}\ \text{J}\).
Try it
A 12 V battery drives 0.50 A through a resistor. Find the resistance, the power dissipated, and the energy converted to heat in 10 minutes.
Series and parallel resistors and Kirchhoff's rules
Most circuits can be collapsed step by step into a single equivalent
resistor. When they can't (two batteries in different loops, say) Kirchhoff's two rules take over. Both rules are conservation laws you
already believe: charge doesn't pile up, and energy is conserved
around a loop.
Rule
Series (same current through each): \(R_{\text{eq}} = R_1 + R_2 + \cdots\);
voltage divides in proportion to \(R\).
Parallel (same voltage across each): \(\dfrac{1}{R_{\text{eq}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \cdots\);
current divides inversely to \(R\), and \(R_{\text{eq}}\) is smaller than the smallest branch.
Junction rule: current into a junction equals current out (charge conservation).
Loop rule: the sum of potential changes around any closed loop is zero
(energy conservation). Walking through a battery from − to + is \(+\varepsilon\);
walking through a resistor in the direction of its current is \(-IR\).
Worked example · A mixed network
A 12 V battery feeds \(R_1 = 4.0\ \Omega\), which then connects to \(R_2 = 6.0\ \Omega\) and \(R_3 = 3.0\ \Omega\) wired in parallel, returning to the battery. Find every current.
Collapse the parallel pair: \(\dfrac{1}{R_p} = \dfrac{1}{6.0} + \dfrac{1}{3.0} = \dfrac{1}{2.0}\), so \(R_p = 2.0\ \Omega\).
In series with \(R_1\): \(R_{\text{eq}} = 6.0\ \Omega\) and \(I = \dfrac{12}{6.0} = 2.0\ \text{A}\).
Now expand back out. Across \(R_1\): \(V_1 = (2.0)(4.0) = 8.0\ \text{V}\), leaving
\(12 - 8.0 = 4.0\ \text{V}\) across the parallel pair. So
\(I_2 = \dfrac{4.0}{6.0} = 0.67\ \text{A}\) and \(I_3 = \dfrac{4.0}{3.0} = 1.33\ \text{A}\).
Check the junction: \(0.67 + 1.33 = 2.0\ \text{A}\). ✓
Worked example · Two loops with two batteries
Three branches join a top node to a bottom node. Left branch: a 12 V
battery (+ terminal up) in series with \(R_1 = 2.0\ \Omega\). Middle branch:
\(R_2 = 3.0\ \Omega\). Right branch: a 9.0 V battery (+ terminal up) in series
with \(R_3 = 6.0\ \Omega\). Find the three currents.
Guess directions: \(I_1\) up through the left branch, \(I_3\) up through
the right, \(I_2\) down through \(R_2\). Junction (top node):
\(I_1 + I_3 = I_2\).
Left loop, clockwise from the bottom (up through the battery, down through \(R_2\)):
\(+12 - 2.0I_1 - 3.0I_2 = 0\).
Right loop, counterclockwise (up through the 9 V battery, down through \(R_2\)):
\(+9.0 - 6.0I_3 - 3.0I_2 = 0\).
From the loop equations, \(I_1 = \dfrac{12 - 3I_2}{2}\) and \(I_3 = \dfrac{9 - 3I_2}{6}\).
Substitute into the junction rule and multiply by 6:
\[3(12 - 3I_2) + (9 - 3I_2) = 6I_2 \;\Rightarrow\; 45 = 18I_2 \;\Rightarrow\; I_2 = 2.5\ \text{A}.\]
Then \(I_1 = 2.25\ \text{A}\) and \(I_3 = 0.25\ \text{A}\), both positive, so
the guessed directions were right. (A negative answer would simply mean
that current runs the other way: don't restart.)
Check: \(2.25 + 0.25 = 2.5\). ✓
Try it
Resistors of 2.0 Ω, 3.0 Ω, and 6.0 Ω in parallel are connected in series with a 5.0 Ω resistor across an 18 V battery. Find the current through the 6.0 Ω resistor.
Show answer
\(\dfrac{1}{R_p} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = 1\), so \(R_p = 1.0\ \Omega\) and \(R_{\text{eq}} = 6.0\ \Omega\).
Total current \(I = 18/6.0 = 3.0\ \text{A}\); voltage across the parallel group \(= (3.0)(1.0) = 3.0\ \text{V}\);
\(I_6 = 3.0/6.0 = 0.50\ \text{A}\).
Lesson 11.4 · Unit 11 · CED topic 11.6
Capacitors in circuits
Put a capacitor in a circuit and its behavior changes over time: current
flows while the capacitor charges, then stops. The exam rarely asks for
the exponential curve itself. It asks about two snapshots, the instant
the switch closes and a long time later, and a capacitor is simple in each.
Rule
Parallel capacitors share a voltage and add: \(C_{\text{eq}} = C_1 + C_2 + \cdots\).
Series capacitors carry the same charge and combine as
\(\dfrac{1}{C_{\text{eq}}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \cdots\): the
opposite of resistors. In an RC circuit with an initially uncharged capacitor:
At t = 0, the capacitor has no charge and no voltage, so it behaves like a bare wire.
At steady state, no more charge flows onto it, so it behaves like an open switch; its voltage is whatever the rest of the circuit puts across that gap.
In between, charge and capacitor voltage rise toward their final values
while current decays toward zero, with time constant \(\tau = RC\)
(63% of the way after one \(\tau\)). Discharging reverses the curves.
Worked example · Combining capacitors
A 6.0 μF and a 3.0 μF capacitor are in series across 12 V. Find the charge on each and the voltage across each. Then find the total charge if they're instead placed in parallel across 12 V.
Series: \(\dfrac{1}{C_{\text{eq}}} = \dfrac{1}{6.0} + \dfrac{1}{3.0} = \dfrac{1}{2.0}\), so \(C_{\text{eq}} = 2.0\ \mu\text{F}\) and
\(Q = C_{\text{eq}}\,\Delta V = 24\ \mu\text{C}\) on each capacitor. Voltages:
\(V_6 = \dfrac{24\ \mu\text{C}}{6.0\ \mu\text{F}} = 4.0\ \text{V}\), \(V_3 = 8.0\ \text{V}\): they sum to 12 V, and the smaller capacitor takes more of it.
Parallel: \(C_{\text{eq}} = 9.0\ \mu\text{F}\), so \(Q_{\text{total}} = (9.0\ \mu\text{F})(12\ \text{V}) = 108\ \mu\text{C}\).
Worked example · An RC circuit at t = 0 and steady state
A 12 V battery is in series with \(R_1 = 4.0\ \Omega\), which leads to \(R_2 = 8.0\ \Omega\) and an uncharged 5.0 μF capacitor connected in parallel with each other. The switch closes at t = 0. Find the battery current at t = 0 and at steady state, and the final charge on the capacitor.
t = 0: the capacitor acts as a wire, shorting out \(R_2\), so
only \(R_1\) limits the current: \(I_0 = \dfrac{12}{4.0} = 3.0\ \text{A}\).
Steady state: the capacitor acts as an open switch, so the
current runs through \(R_1\) and \(R_2\) in series:
\(I_\infty = \dfrac{12}{4.0 + 8.0} = 1.0\ \text{A}\). The capacitor sits across
\(R_2\), so \(V_C = I_\infty R_2 = 8.0\ \text{V}\) and
\(Q = C\,V_C = (5.0\times10^{-6})(8.0) = 4.0\times10^{-5}\ \text{C} = 40\ \mu\text{C}\),
storing \(U = \tfrac{1}{2}CV_C^2 = 1.6\times10^{-4}\ \text{J}\).
Try it
A 12 V battery, a 100 Ω resistor, and an uncharged 20 μF capacitor are in a single series loop. Find the current at t = 0, the final charge on the capacitor, and the time constant.
Show answer
\(I_0 = 12/100 = 0.12\ \text{A}\); at steady state \(I = 0\) and \(V_C = 12\ \text{V}\), so \(Q = (20\times10^{-6})(12) = 240\ \mu\text{C}\);
\(\tau = RC = (100)(20\times10^{-6}) = 2.0\times10^{-3}\ \text{s}\).
Lesson 11.5 · Unit 11 · CED topic 11.7
EMF, internal resistance, and real batteries
An ideal battery holds its voltage no matter what you connect to it.
A real one droops as you draw more current, because the chemistry
inside has resistance of its own. Modeling that as a small resistor
in series with an ideal source explains dimming headlights, warm
batteries, and the data from every battery lab you'll ever do.
Formula
A real battery is an ideal source of emf \(\varepsilon\) in
series with an internal resistance \(r\). When it delivers
current \(I\), the terminal voltage, what a voltmeter across
its terminals reads, is
\[V_{\text{terminal}} = \varepsilon - Ir.\]
On a graph of terminal voltage versus current, the vertical intercept
is \(\varepsilon\) (the open-circuit voltage) and the slope is \(-r\).
Power lost as heat inside the battery is \(I^2 r\).
Worked example · Reading ε and r from a V–I graph
A student varies the load on a battery and records the terminal voltage.
I (A)
0.50
1.0
1.5
2.0
V (V)
11.5
11.0
10.5
10.0
The points are linear. Slope: \(\dfrac{10.0 - 11.5}{2.0 - 0.50} = -1.0\ \text{V/A}\), so
\(r = 1.0\ \Omega\). Extend to \(I = 0\): \(\varepsilon = 11.5 + (0.50)(1.0) = 12.0\ \text{V}\).
On the exam, draw the best-fit line and use two far-apart points on the
line, not two data points: that's the graded technique.
Worked example · Battery driving a load
A battery with \(\varepsilon = 12\ \text{V}\) and \(r = 0.50\ \Omega\) is connected to a 5.5 Ω resistor. Find the current, terminal voltage, power to the load, and power wasted in the battery.
The whole loop is a series circuit: \(I = \dfrac{\varepsilon}{R + r} = \dfrac{12}{5.5 + 0.50} = 2.0\ \text{A}\).
\(V_{\text{terminal}} = \varepsilon - Ir = 12 - (2.0)(0.50) = 11.0\ \text{V}\) (equal to \(IR\), as it must be).
\(P_{\text{load}} = I^2R = (2.0)^2(5.5) = 22\ \text{W}\); \(P_{\text{internal}} = I^2 r = 2.0\ \text{W}\).
The battery converts 24 W of chemical energy per second and delivers 22 W of it: about 92% efficient.
In words: when you crank a car's starter, it draws a very large current
from the battery. The \(Ir\) drop inside the battery becomes large, so the
terminal voltage falls, and the headlights sharing those terminals
dim until the engine catches and the current drops back.
Try it
A battery with \(\varepsilon = 9.0\ \text{V}\) and \(r = 0.30\ \Omega\) shows a terminal voltage of 8.4 V while powering a resistor. Find the current and the resistance.
Ten problems covering the whole unit, in roughly exam order. Work each one on paper
before revealing the answer. Constant: \(e = 1.60\times10^{-19}\ \text{C}\). Treat batteries
as ideal unless the problem gives an internal resistance. Calculator allowed.
A lamp draws 0.75 A for 2.0 minutes. How much charge passes through it, and how many electrons is that? In which direction do the electrons actually move relative to the conventional current?
Show answer
\(\Delta Q = I\,\Delta t = (0.75\ \text{A})(120\ \text{s}) = 90\ \text{C}\). \(n = \dfrac{90}{1.60\times10^{-19}} \approx 5.6\times10^{20}\) electrons. Electrons drift opposite to conventional current, from the − terminal toward the +.
(a) A nichrome wire (\(\rho = 1.10\times10^{-6}\ \Omega\text{·m}\)) is 1.5 m long with diameter 0.40 mm. Find its resistance. (b) A different wire of resistance 3.0 Ω is stretched to twice its length without changing its volume. Find its new resistance and explain the factor.
Two identical bulbs, 1 and 2, are connected in series to an ideal battery. A third identical bulb is then added in parallel with bulb 2. Bulb 1 becomes
Adding a parallel branch lowers the resistance of the bulb-2 section from \(R\) to \(R/2\), so the total resistance drops from \(2R\) to \(1.5R\) and the battery current rises. All of that current passes through bulb 1, so its power \(I^2R\) rises and it brightens, while bulb 2 dims because it now shares the current and has less voltage across it.
Bulb 1 would dim only if the total current fell, which requires the total resistance to rise. Adding a resistor in parallel always lowers the resistance of that section, so the current goes up, not down.
Bulb 1 would be unchanged only if the battery were a constant-current source. An ideal battery holds the voltage fixed, so any change in total resistance changes the current, and all of it flows through bulb 1.
Bulb 1 sits in the only path from the battery, so it cannot go dark: every bit of current still flows through it. Only a short circuit across bulb 1 itself would darken it.
A 24 V battery is connected to \(R_1 = 6.0\ \Omega\), which then leads to \(R_2 = 12\ \Omega\) and \(R_3 = 4.0\ \Omega\) wired in parallel, returning to the battery. Find the battery current, the current through each resistor, and the power dissipated in \(R_3\).
Resistors of resistance \(R\), \(2R\), and \(3R\) are connected to the same ideal battery, first all in parallel and then all in series. For each arrangement, rank the resistors by power dissipated, greatest to least, and justify in one sentence each.
Show answer
Parallel: same \(\Delta V\) across each, so \(P = (\Delta V)^2/R\) is largest for the smallest resistance: \(R \gt 2R \gt 3R\). Series: same \(I\) through each, so \(P = I^2R\) is largest for the largest resistance: \(3R \gt 2R \gt R\). Pick the power formula whose shared quantity is the one held in common.
Three branches join a top node to a bottom node. Left branch: a 12 V battery (+ up) in series with 3.0 Ω. Middle branch: 6.0 Ω. Right branch: a 6.0 V battery (+ up) in series with 3.0 Ω. Find the three currents, and interpret any negative result.
Show answer
Guess \(I_1\) up the left, \(I_3\) up the right, \(I_2\) down the middle; junction: \(I_1 + I_3 = I_2\). Left loop: \(12 - 3.0I_1 - 6.0I_2 = 0\). Right loop: \(6.0 - 3.0I_3 - 6.0I_2 = 0\). So \(I_1 = 4 - 2I_2\) and \(I_3 = 2 - 2I_2\); the junction rule gives \(6 - 4I_2 = I_2\), \(I_2 = 1.2\ \text{A}\). Then \(I_1 = 1.6\ \text{A}\) and \(I_3 = -0.40\ \text{A}\). The minus sign means 0.40 A actually flows down the right branch: the 12 V battery is pushing current backward through the 6 V battery, charging it. Check left loop: \(12 - 4.8 - 7.2 = 0\). ✓
A 4.0 μF and a 12 μF capacitor are connected in series across 24 V. Find the charge on each and the voltage across each, and the total energy stored. Then find the total charge drawn from the battery if the same two capacitors are placed in parallel across 24 V.
A 9.0 V battery is in series with \(R_1 = 3.0\ \Omega\), which leads to \(R_2 = 6.0\ \Omega\) and an uncharged 10 μF capacitor connected in parallel with each other. The switch closes at \(t = 0\). Find the battery current at \(t = 0\) and at steady state, and the final charge on the capacitor. Explain in words why the capacitor behaves differently at the two moments.
Show answer
At \(t = 0\) the capacitor is uncharged, so there is no voltage across it: it acts like a wire and shorts out \(R_2\). \(I_0 = \dfrac{9.0}{3.0} = 3.0\ \text{A}\). At steady state no more charge flows onto it, so it acts like an open switch and the current runs through \(R_1\) and \(R_2\) in series: \(I_\infty = \dfrac{9.0}{9.0} = 1.0\ \text{A}\). The capacitor sits across \(R_2\): \(V_C = (1.0)(6.0) = 6.0\ \text{V}\), \(Q = CV_C = (10\times10^{-6})(6.0) = 60\ \mu\text{C}\). In between, charge builds up on the plates, its voltage grows, and the current through \(R_1\) falls from 3.0 A to 1.0 A.
A battery's terminal voltage is 5.7 V when it delivers 1.0 A and 5.1 V when it delivers 3.0 A. (a) Find its emf and internal resistance. (b) It is then connected to a 1.2 Ω resistor. Find the current, the terminal voltage, the power delivered to the resistor, and the power wasted inside the battery.
Show answer
(a) \(V = \varepsilon - Ir\). Slope of \(V\) versus \(I\): \(\dfrac{5.1 - 5.7}{3.0 - 1.0} = -0.30\ \text{V/A}\), so \(r = 0.30\ \Omega\); \(\varepsilon = 5.7 + (1.0)(0.30) = 6.0\ \text{V}\). (b) \(I = \dfrac{\varepsilon}{R + r} = \dfrac{6.0}{1.5} = 4.0\ \text{A}\); \(V_{\text{terminal}} = 6.0 - (4.0)(0.30) = 4.8\ \text{V}\); \(P_{\text{load}} = I^2R = (16)(1.2) = 19\ \text{W}\); \(P_{\text{internal}} = I^2r = (16)(0.30) = 4.8\ \text{W}\). The battery supplies 24 W and delivers 80% of it.
Lesson 12.1 · Unit 12 · CED topics 12.1–12.2
Magnetic fields and the force on a moving charge
Magnetic field lines leave a north pole, enter a south pole, and always
close on themselves: there are no isolated poles. Unlike an electric
field, a magnetic field ignores a charge at rest. It pushes only on a
charge that is moving, and it pushes sideways.
Formula
The magnetic force on a charge \(q\) moving with velocity \(\vec{v}\) in a field \(\vec{B}\) is
\[F = |q|\,v\,B\sin\theta,\]
where \(\theta\) is the angle between \(\vec{v}\) and \(\vec{B}\). The force is
perpendicular to both. Unit: tesla, \(1\ \mathrm{T} = 1\ \mathrm{N/(A\cdot m)}\).
Method · Right-hand rule
Point the fingers of your right hand along \(\vec{v}\).
Curl them toward \(\vec{B}\) through the smaller angle.
Your thumb points along the force on a positive charge.
For a negative charge, reverse it.
Because the force is always perpendicular to the velocity, it does no work: it can
turn the charge but never change its speed. A charge moving perpendicular to a uniform
field therefore travels in a circle. Setting \(qvB = \dfrac{mv^2}{r}\) gives
\[r = \frac{mv}{qB}.\]
Worked example · Force and radius
A proton (\(m = 1.67\times10^{-27}\ \mathrm{kg}\), \(q = e = 1.60\times10^{-19}\ \mathrm{C}\))
moves east at \(3.0\times10^{6}\ \mathrm{m/s}\) through a \(0.50\ \mathrm{T}\) field pointing north. Find the force and the radius of its path.
Picture a map: east to the right, north up the page. Fingers east, curl north: the thumb
comes out of the page, so the initial force is upward, out of the map. With \(\theta = 90^\circ\),
\[F = qvB = (1.60\times10^{-19}\ \mathrm{C})(3.0\times10^{6}\ \mathrm{m/s})(0.50\ \mathrm{T}) = 2.4\times10^{-13}\ \mathrm{N}.\]
\[r = \frac{mv}{qB} = \frac{(1.67\times10^{-27}\ \mathrm{kg})(3.0\times10^{6}\ \mathrm{m/s})}{(1.60\times10^{-19}\ \mathrm{C})(0.50\ \mathrm{T})} = 6.3\times10^{-2}\ \mathrm{m},\]
about 6.3 cm, in the vertical plane containing east and up.
Worked example · Direction only
An electron moves to the right across the page in a field pointing into the page. Which way is it deflected?
Fingers right, curl into the page: thumb up the page. That is the force on a positive charge,
so the electron is pushed down the page. Write "negative charge, so reverse" explicitly: graders look for it.
Try it
An electron (\(m = 9.11\times10^{-31}\ \mathrm{kg}\)) moves at \(2.0\times10^{7}\ \mathrm{m/s}\)
perpendicular to a \(1.5\times10^{-3}\ \mathrm{T}\) field. Find the radius of its path.
Show answer
\(r = \dfrac{mv}{qB} = \dfrac{(9.11\times10^{-31})(2.0\times10^{7})}{(1.60\times10^{-19})(1.5\times10^{-3})}
= 7.6\times10^{-2}\ \mathrm{m}\), about 7.6 cm.
Lesson 12.2 · Unit 12 · CED topic 12.3
Force on a current-carrying wire and the magnetic fields of currents
A current is a stream of moving charges, so a wire in a magnetic field feels
a force. The reverse is also true: moving charges are the source of
magnetic fields: every magnetic field, a bar magnet's included, comes from moving charge.
Formulas
Force on a straight wire of length \(L\) carrying current \(I\) at angle \(\theta\) to the field:
\[F = B\,I\,L\sin\theta.\]
Field a distance \(r\) from a long straight wire:
\[B = \frac{\mu_0 I}{2\pi r}, \qquad \mu_0 = 4\pi\times10^{-7}\ \mathrm{T\cdot m/A}.\]
The field lines are circles around the wire, and the field falls off as \(1/r\), not \(1/r^2\).
Method · Two right-hand rules
Force on a wire: fingers along the current, curl toward \(\vec{B}\); the thumb gives the force. Same rule as for a positive charge.
Field of a wire: grip the wire with your thumb along the current. Your fingers curl in the direction of the field lines.
Worked example · Force on a wire
A horizontal wire 0.40 m long carries 6.0 A to the right in a 0.25 T field pointing into the page. Find the force.
\[F = BIL\sin 90^\circ = (0.25\ \mathrm{T})(6.0\ \mathrm{A})(0.40\ \mathrm{m}) = 0.60\ \mathrm{N}.\]
Fingers right, curl into the page: thumb up the page. The force is 0.60 N upward.
Worked example · Force between parallel wires
Two long parallel wires 2.0 cm apart each carry 15 A in the same direction. Find the field one wire produces at the other, and the force per meter between them.
\[B_1 = \frac{\mu_0 I_1}{2\pi r} = \frac{(4\pi\times10^{-7}\ \mathrm{T\cdot m/A})(15\ \mathrm{A})}{2\pi(0.020\ \mathrm{m})} = 1.5\times10^{-4}\ \mathrm{T}.\]
Wire 2 sits in that field:
\[\frac{F}{L} = B_1 I_2 = (1.5\times10^{-4}\ \mathrm{T})(15\ \mathrm{A}) = 2.3\times10^{-3}\ \mathrm{N/m}.\]
Direction: put both currents up the page, wire 1 on the left. Gripping wire 1 with the thumb up, the fingers point into the page at wire 2. At wire 2, fingers up, curl into the page: thumb points left, toward wire 1. Parallel currents attract; antiparallel currents repel.
Combined, this is the equation-sheet form \(F/L = \mu_0 I_1 I_2 /(2\pi d)\). By Newton's third law the forces on the two wires are equal and opposite even when the currents differ.
Try it
Two parallel wires 5.0 cm apart carry 12 A and 8.0 A in opposite directions. Find the force on a 2.0 m length of either wire. Attract or repel?
A steady field does nothing to a loop of wire at rest. But change the
amount of field threading the loop and a voltage appears around it. That is
electromagnetic induction: the way nearly all electricity is generated.
Definition
The magnetic flux through a flat loop of area \(A\) is
\[\Phi_B = B\,A\cos\theta,\]
where \(\theta\) is the angle between \(\vec{B}\) and the normal to the loop:
largest when the field passes straight through, zero when it skims the plane. Unit: \(1\ \mathrm{Wb} = 1\ \mathrm{T\cdot m^2}\).
Faraday's law and Lenz's law
The emf induced in a coil of \(N\) turns is
\[\varepsilon = -N\,\frac{\Delta\Phi_B}{\Delta t}.\]
Flux changes when \(B\) changes, \(A\) changes, or the loop rotates.
The minus sign is Lenz's law: the induced current's own field opposes the change in flux.
Worked example · A flux-change problem
A 200-turn circular coil of radius 5.0 cm lies flat on the page. A uniform field into the page rises steadily from 0.10 T to 0.50 T in 0.25 s. The coil's resistance is 4.0 Ω. Find the induced emf, the current, and its direction.
\(A = \pi r^2 = \pi(0.050\ \mathrm{m})^2 = 7.85\times10^{-3}\ \mathrm{m^2}\), and \(\cos\theta = 1\):
\[\Delta\Phi_B = \Delta B\,A = (0.40\ \mathrm{T})(7.85\times10^{-3}\ \mathrm{m^2}) = 3.14\times10^{-3}\ \mathrm{Wb}.\]
\[|\varepsilon| = N\frac{\Delta\Phi_B}{\Delta t} = 200\cdot\frac{3.14\times10^{-3}\ \mathrm{Wb}}{0.25\ \mathrm{s}} = 2.5\ \mathrm{V},
\qquad I = \frac{\varepsilon}{R} = \frac{2.51\ \mathrm{V}}{4.0\ \Omega} = 0.63\ \mathrm{A}.\]
The into-the-page flux is growing, so the induced current must make a field out of the page inside the loop. Point your right thumb out of the page (the field the loop must produce at its center); your fingers curl counterclockwise, and so does the current.
Worked example · Explain in words
A bar magnet is dropped, north pole first, through a coil. Explain why the induced current must repel the magnet rather than attract it.
Suppose it attracted. The magnet would speed up, the flux would change faster, the current would grow, and the magnet would accelerate further: kinetic and electrical energy both appearing from nothing. That violates conservation of energy. So the current must oppose the motion, making the coil's near end a north pole and slowing the magnet; the kinetic energy the magnet loses is exactly the electrical energy the coil gains. Lenz's law is energy conservation applied to induction.
Try it
A single square loop of side 0.20 m sits perpendicular to a 0.80 T field. It is turned a quarter turn in 0.10 s, so its plane ends up parallel to the field. Find the average induced emf.
Induction in practice: motional emf, generators, and transformers
Faraday's law becomes hardware in three ways: a conductor sliding through a
field, a coil spinning in a field, and two coils sharing a changing flux.
Each is a flux-change problem in disguise.
Formulas
Motional emf: a rod of length \(L\) moving at speed \(v\) perpendicular to \(B\) has \(\varepsilon = B L v\), because it sweeps out area \(Lv\,\Delta t\) in time \(\Delta t\).
Generator: a coil rotating steadily has flux \(BA\cos\theta\) with \(\theta\) growing steadily, so the emf is sinusoidal: alternating current. Spinning faster raises both the frequency and the peak emf.
Ideal transformer: \(\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}\) and \(I_p V_p = I_s V_s\). It needs a changing (AC) primary current.
Worked example · Motional emf and energy
A 0.50 m rod slides at 4.0 m/s along frictionless rails, perpendicular to a 0.30 T field into the page. The rails are joined by a 2.0 Ω resistor. Find the emf, the current, the force needed to keep the rod moving, and compare mechanical and electrical power.
\[\varepsilon = BLv = (0.30\ \mathrm{T})(0.50\ \mathrm{m})(4.0\ \mathrm{m/s}) = 0.60\ \mathrm{V}, \qquad I = \frac{0.60\ \mathrm{V}}{2.0\ \Omega} = 0.30\ \mathrm{A}.\]
The current-carrying rod sits in the field, and by Lenz's law the force on it opposes the motion:
\[F = BIL = (0.30\ \mathrm{T})(0.30\ \mathrm{A})(0.50\ \mathrm{m}) = 0.045\ \mathrm{N}.\]
Someone must pull with 0.045 N. Mechanical power in: \(Fv = (0.045\ \mathrm{N})(4.0\ \mathrm{m/s}) = 0.18\ \mathrm{W}\).
Electrical power out: \(I^2R = (0.30\ \mathrm{A})^2(2.0\ \Omega) = 0.18\ \mathrm{W}\). They match: the pull pays for the heat in the resistor.
Worked example · Transformer
A charger steps 120 V down to 6.0 V with a 500-turn primary. How many secondary turns? If the secondary delivers 2.0 A, what current does the primary draw?
\[N_s = N_p\frac{V_s}{V_p} = 500\cdot\frac{6.0\ \mathrm{V}}{120\ \mathrm{V}} = 25\ \text{turns}, \qquad
I_p = \frac{I_s V_s}{V_p} = \frac{(2.0\ \mathrm{A})(6.0\ \mathrm{V})}{120\ \mathrm{V}} = 0.10\ \mathrm{A}.\]
Voltage down by 20, current up by 20; the 12 W is the same on both sides.
Eddy currents. Move a solid metal sheet through a field and the changing flux induces swirling currents
throughout it, which by Lenz's law brake the sheet and warm it. Induction stovetops and train brakes use this on purpose; transformer cores are laminated to prevent it.
Try it
A transformer has 40 primary turns and 800 secondary turns, with 12 V AC on the primary. Find the secondary voltage and, if the secondary supplies 0.50 A, the primary current.
Show answer
\(V_s = 12\ \mathrm{V}\cdot\dfrac{800}{40} = 240\ \mathrm{V}\).
Power is conserved: \(I_p = \dfrac{(0.50\ \mathrm{A})(240\ \mathrm{V})}{12\ \mathrm{V}} = 10\ \mathrm{A}\).
Unit 12 practice · 10 problems
Unit 12 practice: Magnetism and Electromagnetism
Ten problems covering the whole unit, in roughly exam order. Work each one on paper
before revealing the answer. Constants: \(e = 1.60\times10^{-19}\ \text{C}\),
\(m_e = 9.11\times10^{-31}\ \text{kg}\), \(\mu_0 = 4\pi\times10^{-7}\ \text{T·m/A}\), g = 9.8 m/s².
Calculator allowed.
A proton moves at 2.0 × 10⁶ m/s at 30° to a uniform 0.80 T magnetic field. Find the magnitude of the magnetic force on it. Does the force change the proton's speed? Explain.
Show answer
\(F = qvB\sin\theta = (1.60\times10^{-19})(2.0\times10^{6})(0.80)\sin 30^\circ = 1.3\times10^{-13}\ \text{N}\). No: the force is always perpendicular to the velocity, so it does no work and changes only the direction of motion, never the speed.
An electron moves at 3.0 × 10⁶ m/s perpendicular to a 2.0 × 10⁻³ T field. Find the radius of its circular path and the time for one revolution.
A positive charge moves due north through a region where the magnetic field points straight up, away from the ground. The magnetic force on it points
Point your fingers along \(\vec{v}\) (north) and curl them toward \(\vec{B}\) (up): the thumb points east, the direction of \(q\vec{v}\times\vec{B}\) for a positive charge.
West is the answer for a negative charge such as an electron, whose force is opposite to \(\vec{v}\times\vec{B}\). It also results from curling from \(\vec{B}\) toward \(\vec{v}\) instead of from \(\vec{v}\) toward \(\vec{B}\).
The magnetic force is always perpendicular to the velocity, so it can never point along or against the direction of motion. That is why a magnetic field changes a charge's direction but never its speed.
The magnetic force is also perpendicular to \(\vec{B}\), so it cannot point up or down when the field is vertical. A force along the field line is what an electric field produces, not a magnetic one.
A horizontal wire 0.50 m long carries 8.0 A perpendicular to a 0.30 T field. (a) Find the force on it. (b) The wire has mass 0.020 kg; what current would make the magnetic force just balance its weight?
Show answer
(a) \(F = BIL\sin 90^\circ = (0.30)(8.0)(0.50) = 1.2\ \text{N}\). (b) Set \(BIL = mg\): \(I = \dfrac{mg}{BL} = \dfrac{(0.020)(9.8)}{(0.30)(0.50)} = 1.3\ \text{A}\), with the current directed so the right-hand rule gives an upward force.
A long straight wire carries 20 A. (a) Find the magnetic field 4.0 cm from it. (b) A second long wire 4.0 cm away carries 10 A in the opposite direction. Find the force per meter between the wires and state whether they attract or repel.
Show answer
(a) \(B = \dfrac{\mu_0 I}{2\pi r} = \dfrac{(4\pi\times10^{-7})(20)}{2\pi(0.040)} = 1.0\times10^{-4}\ \text{T}\), circling the wire. (b) The second wire sits in that field: \(\dfrac{F}{L} = BI_2 = (1.0\times10^{-4})(10) = 1.0\times10^{-3}\ \text{N/m}\). Antiparallel currents repel.
Rank the magnetic flux through these flat loops, greatest to least. Loop A: area 0.10 m², B = 0.50 T, field perpendicular to the loop. Loop B: area 0.20 m², B = 0.50 T, field parallel to the plane of the loop. Loop C: area 0.20 m², B = 0.30 T, field at 60° to the normal. Loop D: area 0.050 m², B = 1.2 T, field perpendicular to the loop.
Show answer
\(\Phi = BA\cos\theta\) with \(\theta\) measured from the normal. A: \((0.50)(0.10)(1) = 0.050\ \text{Wb}\). B: field skims the plane, \(\theta = 90^\circ\), \(\Phi = 0\). C: \((0.30)(0.20)\cos 60^\circ = 0.030\ \text{Wb}\). D: \((1.2)(0.050)(1) = 0.060\ \text{Wb}\). Ranking: D > A > C > B.
A 50-turn circular coil of radius 0.10 m and resistance 2.5 Ω lies flat on the page in a uniform field directed into the page. The field falls steadily from 0.60 T to zero in 0.20 s. Find the induced emf and current, and give the current's direction with a Lenz's-law justification.
Show answer
\(A = \pi(0.10)^2 = 3.14\times10^{-2}\ \text{m}^2\); \(\Delta\Phi = \Delta B\,A = (0.60)(3.14\times10^{-2}) = 1.88\times10^{-2}\ \text{Wb}\). \(|\varepsilon| = N\dfrac{\Delta\Phi}{\Delta t} = 50\cdot\dfrac{1.88\times10^{-2}}{0.20} = 4.7\ \text{V}\); \(I = \dfrac{4.71}{2.5} = 1.9\ \text{A}\). The into-the-page flux is decreasing, so the induced current makes its own field into the page inside the loop to oppose the loss: thumb into the page, fingers curl clockwise.
A strong magnet dropped down a vertical copper pipe falls far more slowly than in free fall, even though copper is not magnetic. Explain in words what is happening, and account for the energy.
Show answer
As the magnet falls, the flux through each ring of the pipe changes: increasing below the magnet, decreasing above it. By Faraday's law, currents are induced around the pipe, and by Lenz's law they flow so as to oppose the change: the ring below repels the approaching magnet and the ring above attracts the departing one. Both forces point up, so the magnet reaches a slow terminal speed. The gravitational potential energy it loses is not becoming kinetic energy; it is dissipated as heat by the induced currents in the copper. If the currents helped the motion instead, energy would be created from nothing.
A 0.40 m conducting rod slides at 5.0 m/s along frictionless rails, perpendicular to a 0.50 T field. The rails are joined through a 4.0 Ω resistor. Find the emf, the current, the force needed to keep the rod moving at constant speed, and show that the mechanical power in equals the electrical power out.
Show answer
\(\varepsilon = BLv = (0.50)(0.40)(5.0) = 1.0\ \text{V}\); \(I = \dfrac{1.0}{4.0} = 0.25\ \text{A}\). The current-carrying rod feels \(F = BIL = (0.50)(0.25)(0.40) = 0.050\ \text{N}\) opposing its motion (Lenz), so the same 0.050 N must pull it. Mechanical power \(Fv = (0.050)(5.0) = 0.25\ \text{W}\); electrical power \(I^2R = (0.25)^2(4.0) = 0.25\ \text{W}\). Equal, as energy conservation demands.
An ideal transformer steps 120 V down to 9.0 V. The primary has 2400 turns. (a) How many turns does the secondary have? (b) If the secondary delivers 1.5 A, what current does the primary draw? (c) Explain in one or two sentences why the transformer does nothing if the primary is connected to a steady DC source.
Show answer
(a) \(\dfrac{N_s}{N_p} = \dfrac{V_s}{V_p}\), so \(N_s = 2400\cdot\dfrac{9.0}{120} = 180\) turns. (b) Power is conserved: \(I_p = \dfrac{I_sV_s}{V_p} = \dfrac{(1.5)(9.0)}{120} = 0.11\ \text{A}\). (c) A steady current makes a steady flux through the core, and only a changing flux induces an emf in the secondary. With DC, nothing changes after the first instant, so the secondary voltage is zero.
Lesson 13.1 · Unit 13 · CED topic 13.1
Reflection and plane mirrors
Geometric optics treats light as rays: straight lines that bend only at a
surface. A flat mirror is the cleanest place to learn what an "image" actually is.
Law of reflection
The angle of incidence equals the angle of reflection, \(\theta_i = \theta_r\), both measured
from the normal (the perpendicular to the surface at the point of contact).
A rough surface reflects diffusely: the law still holds ray by ray, but the normals point every which way.
Plane-mirror image
The image is virtual, upright, the same size as the object, and as far behind the mirror as the object is in front:
\(d_i = -d_o\), \(m = +1\). Virtual means no light actually passes through the image point: your eye traces the reflected rays backward to where they seem to come from.
Worked example · Ray diagram in words
Locate the image of a candle flame 40 cm in front of a plane mirror.
Draw the mirror as a vertical line with the flame 40 cm to its left. One ray from the flame hits the mirror head-on and reflects straight back. A second hits higher up at an angle and reflects with equal angles about the normal. The reflected rays diverge and never meet, so extend them backward (dashed) behind the mirror. They cross 40 cm behind it, directly opposite the flame: \(d_i = -40\ \mathrm{cm}\), the minus sign meaning "behind the mirror, virtual."
Worked example · Shortest full-length mirror
A person 1.80 m tall has eyes 1.70 m above the floor. What is the shortest vertical mirror in which they can see their whole body, and where should it hang?
The ray from toes to eyes reflects halfway between them, at height \(\tfrac{1.70}{2} = 0.85\ \mathrm{m}\). The ray from the top of the head reflects halfway between head and eyes, at \(\tfrac{1.80 + 1.70}{2} = 1.75\ \mathrm{m}\). The mirror runs from 0.85 m to 1.75 m: length \(0.90\ \mathrm{m}\), half the person's height, regardless of how far away they stand.
Try it
An object 40 cm in front of a plane mirror is moved back to 60 cm. How far does its image move, and how far apart are object and image at the end?
Show answer
The image goes from 40 cm behind the mirror to 60 cm behind it: it moves 20 cm.
Object–image separation: \(60 + 60 = 120\ \mathrm{cm}\).
Lesson 13.2 · Unit 13 · CED topic 13.2
Spherical mirrors: ray diagrams and the mirror equation
Curve the mirror and images can be magnified, shrunk, or flipped. A
concave mirror (reflecting inside of the sphere) brings parallel rays to a real
focus; a convex mirror (reflecting outside) spreads them out as if from a focus behind it.
Formula and sign conventions
\[\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}, \qquad f = \frac{R}{2}, \qquad m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}.\]
\(d_o \gt 0\) for a real object in front of the mirror.
\(d_i \gt 0\): image in front of the mirror, real (can be caught on a screen). \(d_i \lt 0\): behind the mirror, virtual.
From the tip of the object: (1) a ray parallel to the axis reflects through the focal point F (for convex, as if from F behind the mirror); (2) a ray through F (or aimed at F, for convex) reflects parallel to the axis; (3) a ray to the center of the mirror reflects symmetrically about the axis. Where the reflected rays cross is the real image tip; where their backward extensions cross is the virtual image tip. Any two rays suffice: the third is a check.
Worked example · Concave, real image
A 5.0 cm tall object stands 30 cm in front of a concave mirror of radius 40 cm. Locate and describe the image.
\(f = R/2 = 20\ \mathrm{cm}\). The object is between F and the center of curvature, so the diagram shows the reflected rays converging beyond C.
\[\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{20\ \mathrm{cm}} - \frac{1}{30\ \mathrm{cm}} = \frac{3 - 2}{60\ \mathrm{cm}} = \frac{1}{60\ \mathrm{cm}}, \qquad d_i = 60\ \mathrm{cm}.\]
\[m = -\frac{d_i}{d_o} = -\frac{60}{30} = -2.0, \qquad h_i = m h_o = -10\ \mathrm{cm}.\]
Real (\(d_i \gt 0\)), inverted (\(m \lt 0\)), and twice the size: a 10 cm tall inverted image 60 cm in front of the mirror.
Worked example · Convex, virtual image
The same object is placed 30 cm in front of a convex mirror with \(f = -15\ \mathrm{cm}\).
\[\frac{1}{d_i} = \frac{1}{-15\ \mathrm{cm}} - \frac{1}{30\ \mathrm{cm}} = -\frac{3}{30\ \mathrm{cm}}, \qquad d_i = -10\ \mathrm{cm}, \qquad m = -\frac{-10}{30} = +0.33.\]
Virtual, upright, one-third size, 10 cm behind the mirror. A convex mirror gives this kind of image for every real object, which is why car side mirrors show a wide, shrunken view.
Try it
An object is 8.0 cm in front of a concave mirror with \(f = 12\ \mathrm{cm}\). Find \(d_i\) and \(m\), and describe the image.
Show answer
\(\dfrac{1}{d_i} = \dfrac{1}{12} - \dfrac{1}{8} = \dfrac{2 - 3}{24} = -\dfrac{1}{24}\), so \(d_i = -24\ \mathrm{cm}\) and \(m = -\dfrac{-24}{8} = +3.0\).
Virtual, upright, three times larger: a makeup mirror. An object inside F of a concave mirror always gives this.
Lesson 13.3 · Unit 13 · CED topic 13.3
Refraction and total internal reflection
Light slows down inside glass or water, and a ray crossing the boundary at an
angle bends. The frequency is set by the source and cannot change at the
boundary, so it is the wavelength that shrinks along with the speed.
Formulas
Index of refraction: \(n = \dfrac{c}{v}\) (so \(n \ge 1\); \(\lambda_n = \lambda/n\)). At a boundary,
Snell's law:
\[n_1\sin\theta_1 = n_2\sin\theta_2,\]
angles measured from the normal. Entering a higher \(n\), the ray bends toward the normal; entering a lower \(n\), away from it.
Going from high \(n_1\) to low \(n_2\), the refracted angle reaches 90° at the critical angle
\[\sin\theta_c = \frac{n_2}{n_1}.\]
Beyond \(\theta_c\) nothing refracts: total internal reflection.
Worked example · Snell's law
Light in air (\(n = 1.00\)) strikes a glass block (\(n = 1.52\)) at 40° from the normal. Find the refracted angle and the speed of light in the glass. Take \(c = 3.00\times10^{8}\ \mathrm{m/s}\).
\[\sin\theta_2 = \frac{n_1\sin\theta_1}{n_2} = \frac{(1.00)\sin 40^\circ}{1.52} = \frac{0.643}{1.52} = 0.423, \qquad \theta_2 = 25.0^\circ.\]
The ray bends toward the normal, as expected for slower light.
\[v = \frac{c}{n} = \frac{3.00\times10^{8}\ \mathrm{m/s}}{1.52} = 1.97\times10^{8}\ \mathrm{m/s}.\]
Exiting the far side of a parallel-faced block, the ray bends back to 40°, offset sideways but parallel to its original direction.
Worked example · Critical angle and fiber optics
(a) Find the critical angle for light inside water (\(n = 1.33\)) meeting air. (b) An optical fiber has a core of \(n = 1.50\) clad in glass of \(n = 1.46\). Find its critical angle.
(a) \(\sin\theta_c = \dfrac{1.00}{1.33} = 0.752\), so \(\theta_c = 48.8^\circ\). A diver looking up sees the whole sky squeezed into a cone of that half-angle; outside it the surface acts as a mirror.
(b) \(\sin\theta_c = \dfrac{1.46}{1.50} = 0.973\), so \(\theta_c = 76.7^\circ\). Light traveling nearly along the fiber strikes the core–cladding boundary at grazing angles larger than this, so it is totally reflected at every bounce and none leaks out, even around gentle bends. A common exam error: forgetting that TIR needs \(n_1 \gt n_2\). Light cannot be trapped going from air into glass.
Try it
Light inside glass (\(n = 1.60\)) reaches a boundary with air. (a) What is the critical angle? (b) What happens to a ray incident at 45°? (c) At 30°?
Show answer
(a) \(\sin\theta_c = 1/1.60 = 0.625\), \(\theta_c = 38.7^\circ\). (b) \(45^\circ \gt 38.7^\circ\): totally internally reflected at 45°.
(c) \(\sin\theta_2 = 1.60\sin 30^\circ = 0.800\), \(\theta_2 = 53.1^\circ\); it refracts into the air, bending away from the normal.
Lesson 13.4 · Unit 13 · CED topics 13.4–13.5
Thin lenses: ray diagrams, the thin-lens equation, and real vs. virtual images
A converging lens (thicker in the middle) bends parallel rays to a real focal
point on the far side; a diverging lens (thinner in the middle) spreads them
as if from a focal point on the near side.
\(d_i \gt 0\): image on the far side from the object: real, inverted, can be projected. \(d_i \lt 0\): same side as the object; virtual, upright.
\(m\) works as for mirrors: negative is inverted, \(|m| \gt 1\) is enlarged.
Method · Three principal rays
From the object's tip: (1) a ray parallel to the axis refracts through the far focal point (diverging: emerges as if from the near one); (2) a ray through the center of the lens goes straight; (3) a ray through the near focal point (diverging: aimed at the far one) emerges parallel to the axis. Real image where the rays cross; virtual image where their backward extensions cross.
Worked example · Camera (real image)
A camera lens has \(f = 50\ \mathrm{mm}\). A 1.8 m tall person stands 2.0 m away. Where must the sensor be, and how tall is the image?
\[\frac{1}{d_i} = \frac{1}{0.050\ \mathrm{m}} - \frac{1}{2.0\ \mathrm{m}} = 20.0 - 0.50 = 19.5\ \mathrm{m^{-1}}, \qquad d_i = 0.0513\ \mathrm{m} = 51.3\ \mathrm{mm}.\]
\[m = -\frac{0.0513}{2.0} = -0.0256, \qquad h_i = (-0.0256)(1.8\ \mathrm{m}) = -0.046\ \mathrm{m}.\]
The sensor sits 51.3 mm behind the lens (just past F, as for any distant object) and records a real, inverted image 4.6 cm tall.
Worked example · Magnifying glass (virtual image)
A converging lens with \(f = 10\ \mathrm{cm}\) is held 6.0 cm from a stamp.
\[\frac{1}{d_i} = \frac{1}{10\ \mathrm{cm}} - \frac{1}{6.0\ \mathrm{cm}} = \frac{3 - 5}{30\ \mathrm{cm}} = -\frac{1}{15\ \mathrm{cm}}, \qquad d_i = -15\ \mathrm{cm}, \qquad m = -\frac{-15}{6.0} = +2.5.\]
Virtual, upright, 2.5 times larger, 15 cm from the lens on the stamp's side. Any object inside F of a converging lens behaves this way.
Worked example · Explain in words
Why can't the magnifying glass's image be projected onto a screen?
After the lens, the rays from any point on the stamp are still diverging; they never reconverge. Your eye traces them backward to where they appear to originate, and that is where you see the image. A screen lights up only where rays actually meet, and on the far side of the lens they are spread out, so a screen shows a blur. Only a real image can be caught on a screen.
Try it
An object is 30 cm from a diverging lens with \(f = -20\ \mathrm{cm}\). Find the image position and magnification, and describe the image.
Show answer
\(\dfrac{1}{d_i} = \dfrac{1}{-20} - \dfrac{1}{30} = -\dfrac{5}{60} = -\dfrac{1}{12}\), so \(d_i = -12\ \mathrm{cm}\) and \(m = -\dfrac{-12}{30} = +0.40\).
Virtual, upright, reduced to 40%, 12 cm from the lens on the object's side.
Unit 13 practice · 10 problems
Unit 13 practice: Geometric Optics
Ten problems covering the whole unit, in roughly exam order. Work each one on paper
before revealing the answer. Sign conventions as in the lessons: \(d_i \gt 0\) is real,
\(d_i \lt 0\) is virtual; \(f \gt 0\) for concave mirrors and converging lenses.
Constant: \(c = 3.00\times10^{8}\ \text{m/s}\). Calculator allowed.
A person 1.70 m tall stands 2.0 m in front of a vertical plane mirror. Where is the image, how far is the person from the image, and what is the shortest mirror in which they can see their entire body?
Show answer
The image is virtual, upright, the same size, and 2.0 m behind the mirror (\(d_i = -2.0\ \text{m}\)); person-to-image distance is 4.0 m. The ray from any body point reflects at the midpoint height between that point and the eyes, so the mirror need only span half the height: \(0.85\ \text{m}\), independent of distance.
An object stands 45 cm in front of a concave mirror whose radius of curvature is 30 cm. Locate the image, find the magnification, and describe the image.
Show answer
\(f = R/2 = 15\ \text{cm}\). \(\dfrac{1}{d_i} = \dfrac{1}{f} - \dfrac{1}{d_o} = \dfrac{1}{15} - \dfrac{1}{45} = \dfrac{2}{45}\), so \(d_i = 22.5\ \text{cm}\). \(m = -\dfrac{d_i}{d_o} = -\dfrac{22.5}{45} = -0.50\). Real (\(d_i \gt 0\)), inverted, half size, 22.5 cm in front of the mirror, between F and C, as expected for an object beyond C.
The same object is placed 40 cm in front of a convex mirror with \(f = -20\ \text{cm}\). Locate and describe the image.
Show answer
\(\dfrac{1}{d_i} = \dfrac{1}{-20} - \dfrac{1}{40} = -\dfrac{3}{40}\), so \(d_i = -13\ \text{cm}\); \(m = -\dfrac{-13.3}{40} = +0.33\). Virtual, upright, one-third size, 13 cm behind the mirror. A convex mirror gives this kind of image for every real object.
An object is placed between a concave mirror and its focal point. The image is
A real, inverted, enlarged image is what a concave mirror gives when the object sits between \(f\) and \(2f\), not inside \(f\). Inside the focal point the reflected rays diverge and never cross in front of the mirror.
A real, inverted, reduced image requires the object to be beyond \(2f\), farther from the mirror than the center of curvature: the opposite placement from the one described.
With \(d_o \lt f\), \(\dfrac{1}{d_i} = \dfrac{1}{f} - \dfrac{1}{d_o} \lt 0\), so the image is virtual and behind the mirror, and \(|d_i| \gt d_o\) makes \(|m| \gt 1\): upright and enlarged. This is the makeup-mirror case.
A virtual, upright, reduced image is what a convex mirror (or a diverging lens) produces for any object position. A concave mirror's virtual image is always magnified.
Light of wavelength 600 nm in air strikes a water surface (\(n = 1.33\)) at 50° from the normal. Find the angle of refraction, the speed of light in the water, and the wavelength in the water. What happens to the frequency?
Show answer
\(\sin\theta_2 = \dfrac{n_1\sin\theta_1}{n_2} = \dfrac{\sin 50^\circ}{1.33} = \dfrac{0.766}{1.33} = 0.576\), so \(\theta_2 = 35^\circ\), bent toward the normal. \(v = \dfrac{c}{n} = \dfrac{3.00\times10^{8}}{1.33} = 2.26\times10^{8}\ \text{m/s}\); \(\lambda_n = \dfrac{\lambda}{n} = \dfrac{600}{1.33} = 451\ \text{nm}\). The frequency is unchanged: it is set by the source, and \(v = f\lambda\) makes \(\lambda\) shrink with \(v\).
(a) Find the critical angle for light inside diamond (\(n = 2.42\)) meeting air, and explain in one sentence why a cut diamond sparkles more than a glass imitation (\(n = 1.50\)). (b) Rank the critical angles for light leaving diamond, glass, and water (\(n = 1.33\)) into air, smallest to largest.
Show answer
(a) \(\sin\theta_c = \dfrac{n_2}{n_1} = \dfrac{1.00}{2.42} = 0.413\), so \(\theta_c = 24^\circ\). Any ray inside striking a facet at more than 24° is totally reflected, so light bounces around inside and exits only through the facets it was cut for; glass, with \(\theta_c = 42^\circ\), lets far more light leak out the back. (b) \(\theta_c\) grows as \(n_1\) shrinks: diamond (24°) < glass (42°) < water (49°).
A 2.0 cm tall object stands 15 cm from a converging lens of focal length 10 cm. Describe in words how you would draw the three principal rays from the tip of the object and where they meet, then confirm with the thin-lens equation and find the image height.
Show answer
Draw the axis, the lens as a vertical line, focal points 10 cm on each side, and the arrow 15 cm to the left. From its tip: (1) a ray parallel to the axis refracts through the far focal point; (2) a ray through the center of the lens continues straight; (3) a ray through the near focal point emerges parallel to the axis. All three cross on the far side, below the axis, at the tip of an inverted, enlarged real image. Equation: \(\dfrac{1}{d_i} = \dfrac{1}{10} - \dfrac{1}{15} = \dfrac{1}{30}\), so \(d_i = 30\ \text{cm}\); \(m = -\dfrac{30}{15} = -2.0\); \(h_i = (-2.0)(2.0) = -4.0\ \text{cm}\), 4.0 cm tall and inverted, 30 cm beyond the lens.
A magnifying glass with \(f = 8.0\ \text{cm}\) is held 5.0 cm above a page. Locate the image, find the magnification, and explain why you must look through the lens to see it.
Show answer
\(\dfrac{1}{d_i} = \dfrac{1}{8.0} - \dfrac{1}{5.0} = \dfrac{5 - 8}{40} = -\dfrac{3}{40}\), so \(d_i = -13\ \text{cm}\); \(m = -\dfrac{-13.3}{5.0} = +2.7\). Virtual, upright, 2.7 times larger, 13 cm below the lens on the page's side. The rays leaving the lens are still diverging; they never meet, so no screen can catch the image. Your eye traces them back to where they appear to come from.
An object is 30 cm from a diverging lens with \(f = -15\ \text{cm}\). Locate and describe the image. Could a diverging lens alone ever form a real image of a real object? Explain.
Show answer
\(\dfrac{1}{d_i} = \dfrac{1}{-15} - \dfrac{1}{30} = -\dfrac{3}{30} = -\dfrac{1}{10}\), so \(d_i = -10\ \text{cm}\); \(m = -\dfrac{-10}{30} = +0.33\). Virtual, upright, one-third size, 10 cm from the lens on the object's side. No: with \(f \lt 0\) and \(d_o \gt 0\), \(\dfrac{1}{d_i} = \dfrac{1}{f} - \dfrac{1}{d_o}\) is always negative, so the image is always virtual.
A camera lens has \(f = 50\ \text{mm}\). A subject at 1.0 m is in sharp focus. The subject then steps closer, to 0.50 m. Find the lens-to-sensor distance in each case and explain in words which way, and how far, the lens must move to refocus.
Show answer
At 1.0 m: \(\dfrac{1}{d_i} = \dfrac{1}{0.050} - \dfrac{1}{1.0} = 19.0\ \text{m}^{-1}\), \(d_i = 52.6\ \text{mm}\). At 0.50 m: \(\dfrac{1}{d_i} = 20 - 2.0 = 18.0\ \text{m}^{-1}\), \(d_i = 55.6\ \text{mm}\). A closer object makes \(1/d_o\) larger, so \(1/d_i\) is smaller and \(d_i\) is larger: the lens must move away from the sensor by about 2.9 mm. That is exactly what a focusing ring does.
Lesson 14.1 · Unit 14 · CED topics 14.1–14.2
Wave properties: amplitude, wavelength, frequency, and speed
A wave carries energy from one place to another without carrying the
medium along. Shake one end of a rope and the hump travels; the rope itself
just bobs. Everything in this unit (sound, light, interference) rests on
a handful of definitions and one equation.
Definitions
Transverse wave: the medium oscillates perpendicular to the direction of travel (rope, light). Longitudinal: parallel to it (sound; compressions and rarefactions).
Amplitude \(A\): maximum displacement from equilibrium. Wavelength \(\lambda\): distance from crest to crest. Period \(T\): time for one full cycle. Frequency \(f = 1/T\), in hertz.
Energy carried by a wave is proportional to \(A^2\): double the amplitude, four times the energy.
Formula
\[v = f\lambda = \frac{\lambda}{T}.\]
The speed is set by the medium (tension and mass per length for a string, temperature for air), not by the source. The frequency is set by the source. So when a wave enters a new medium, \(f\) stays fixed and \(\lambda\) adjusts.
Worked example · Reading a wave
On a rope, adjacent crests are 1.5 m apart and 8 crests pass a fixed point in 4.0 s. Find the frequency, period, and wave speed.
\[f = \frac{8\ \text{cycles}}{4.0\ \mathrm{s}} = 2.0\ \mathrm{Hz}, \qquad T = \frac{1}{f} = 0.50\ \mathrm{s}, \qquad v = f\lambda = (2.0\ \mathrm{Hz})(1.5\ \mathrm{m}) = 3.0\ \mathrm{m/s}.\]
Shaking the rope faster would shorten the wavelength but leave \(v\) at 3.0 m/s: only tightening the rope changes the speed.
Worked example · Crossing into a new medium
A 680 Hz sound wave travels through air at 340 m/s and then into water, where sound travels at 1480 m/s. Find its wavelength in each medium.
\[\lambda_{\text{air}} = \frac{v}{f} = \frac{340\ \mathrm{m/s}}{680\ \mathrm{Hz}} = 0.50\ \mathrm{m}, \qquad
\lambda_{\text{water}} = \frac{1480\ \mathrm{m/s}}{680\ \mathrm{Hz}} = 2.2\ \mathrm{m}.\]
The frequency is unchanged, the water surface is driven at 680 Hz by the air and pushes the water at 680 Hz, so the wavelength stretches by the same factor as the speed.
Exam tip: a graph of displacement versus position shows \(\lambda\); a graph versus time shows \(T\). Read the axis label before you read the distance between peaks.
Try it
A displacement-vs-position graph shows a crest and the next trough 0.30 m apart; a displacement-vs-time graph of the same wave shows a period of 0.25 s. Find the wave speed.
Show answer
Crest to trough is half a wavelength, so \(\lambda = 0.60\ \mathrm{m}\). \(v = \lambda/T = 0.60\ \mathrm{m}/0.25\ \mathrm{s} = 2.4\ \mathrm{m/s}\).
Lesson 14.2 · Unit 14 · CED topics 14.3–14.4
Superposition and standing waves
Two waves in the same place don't collide; they add. Where crests meet crests the
medium moves more, where crests meet troughs it moves less, and afterward each wave
continues as if the other had never been there. That simple rule produces standing
waves, beats, and every interference pattern in this unit.
Principle of superposition
The displacement of the medium is the sum of the displacements of the individual waves.
Same-sign displacements give constructive interference; opposite-sign give
destructive interference. Two identical waves traveling in opposite directions
make a standing wave: points that never move are nodes,
points of maximum motion are antinodes, and adjacent nodes are \(\lambda/2\) apart.
Formula · String fixed at both ends
Both ends must be nodes, so a whole number of half-wavelengths must fit in the length \(L\):
\[\lambda_n = \frac{2L}{n}, \qquad f_n = \frac{n v}{2L} = n f_1, \qquad n = 1, 2, 3, \dots\]
\(f_1\) is the fundamental; \(f_n\) is the \(n\)th harmonic. The \(n\)th harmonic has \(n\) loops (antinodes) and \(n + 1\) nodes counting the ends.
Worked example · Harmonics of a string
Waves travel at 240 m/s on a 0.60 m string fixed at both ends. Find the first three harmonic frequencies, and describe the third-harmonic pattern.
\[f_1 = \frac{v}{2L} = \frac{240\ \mathrm{m/s}}{2(0.60\ \mathrm{m})} = 200\ \mathrm{Hz}, \qquad f_2 = 400\ \mathrm{Hz}, \qquad f_3 = 600\ \mathrm{Hz}.\]
The third harmonic has \(\lambda_3 = 2L/3 = 0.40\ \mathrm{m}\): three loops, with nodes at the two ends and at 0.20 m and 0.40 m from one end.
Worked example · From a picture to the speed
A photo of a 1.2 m string driven at 90 Hz shows three loops. Find the wave speed and the fundamental frequency.
Why does a guitar string produce only certain frequencies?
The ends are clamped, so they cannot move: both must be nodes. A standing wave has nodes every half-wavelength, so the string can hold only wavelengths for which a whole number of half-wavelengths fits between the clamps. Since the speed is fixed by tension and thickness, each allowed wavelength means one allowed frequency. Any other frequency sends waves that reflect off the ends out of step with themselves and cancel. Pressing a fret shortens \(L\), raising every allowed frequency together.
Try it
A 0.90 m string vibrates in its second harmonic at 220 Hz. Find the wavelength, the wave speed, and the fundamental frequency.
Sound, resonance in pipes, beats, and the Doppler effect
Sound is a longitudinal pressure wave; in air at room temperature it travels
at about 340 m/s. Pitch is frequency, loudness is amplitude. Three sound
phenomena show up on every exam: pipes, beats, and moving sources.
Formulas · Pipes
An open end is a displacement antinode; a closed end is a node.
Two tones of nearly equal frequency produce a loudness that pulses at \(f_{\text{beat}} = |f_1 - f_2|\).
A source and observer moving toward each other raise the heard frequency; moving apart lowers it:
\[f' = f\,\frac{v \pm v_o}{v \mp v_s},\]
with \(v\) the speed of sound. Choose the upper signs for approach (numerator up, denominator down make \(f'\) bigger).
Worked example · A closed pipe
A pipe 0.25 m long is closed at one end. With \(v = 340\ \mathrm{m/s}\), find its three lowest resonant frequencies. What changes if the closed end is opened?
\[f_1 = \frac{v}{4L} = \frac{340\ \mathrm{m/s}}{4(0.25\ \mathrm{m})} = 340\ \mathrm{Hz}, \qquad f_3 = 3f_1 = 1020\ \mathrm{Hz}, \qquad f_5 = 5f_1 = 1700\ \mathrm{Hz}.\]
The fundamental is a quarter wavelength: node at the closed end, antinode at the open end. Opening the closed end makes the pipe open–open with \(f_1 = v/(2L) = 680\ \mathrm{Hz}\): an octave higher, and now every harmonic is present.
Worked example · Tuning with beats
A guitar string sounded with a 440 Hz tuning fork gives 3 beats per second. Tightening the string slightly increases the beat rate. What was the string's frequency?
The string is at \(440 \pm 3\), so 437 Hz or 443 Hz. Tightening raises its frequency; if the beats got faster, the string moved away from 440 Hz, so it was already above: 443 Hz. Exam tip: beats alone give two candidates; the direction of change picks one.
Worked example · Doppler shift
An ambulance siren emits 700 Hz while driving at 30 m/s toward a stationary listener. What frequency is heard as it approaches, and after it passes?
The observer is still (\(v_o = 0\)), the source moves. Approaching, the denominator shrinks:
\[f' = f\,\frac{v}{v - v_s} = (700\ \mathrm{Hz})\frac{340}{340 - 30} = (700)\frac{340}{310} = 768\ \mathrm{Hz}.\]
Receding: \(f' = (700)\dfrac{340}{340 + 30} = 643\ \mathrm{Hz}\). The pitch drops abruptly as it passes: the classic siren swoop. Physically, the moving source crowds its wavefronts together ahead of it, shortening the wavelength the listener receives.
Try it
An open–open pipe has a fundamental of 256 Hz when \(v = 340\ \mathrm{m/s}\). Find its length. If one end is then closed, what is the new fundamental?
Show answer
\(L = \dfrac{v}{2f_1} = \dfrac{340}{2(256)} = 0.66\ \mathrm{m}\). Closed at one end: \(f_1 = \dfrac{v}{4L} = 128\ \mathrm{Hz}\), half the original.
Lesson 14.4 · Unit 14 · CED topic 14.7
Electromagnetic waves and polarization
Light needs no medium. An electromagnetic wave is an oscillating electric field
and an oscillating magnetic field, perpendicular to each other and to the
direction of travel, each regenerating the other as they move. In vacuum every
EM wave, from radio to gamma rays, travels at the same speed.
Formula and the spectrum
\[c = f\lambda, \qquad c = 3.00\times10^{8}\ \mathrm{m/s}.\]
In order of increasing frequency (decreasing wavelength): radio, microwave, infrared, visible (about 700 nm red to 400 nm violet), ultraviolet, X-ray, gamma. All are transverse waves of the same kind; only \(f\) and \(\lambda\) differ.
Polarization
A wave is polarized when its electric field oscillates along one fixed direction.
Ordinary light is unpolarized (all directions mixed). A polarizing filter transmits the
component of \(\vec{E}\) along its axis. Unpolarized light through one filter emerges polarized at half the intensity;
polarized light of intensity \(I_0\) through a filter at angle \(\theta\) to its polarization emerges with
\(I = I_0\cos^2\theta\) (Malus's law). Light reflecting off water or glass also comes out partly polarized horizontally, which is why polarized sunglasses cut glare.
Worked example · Wavelength and frequency
An FM station broadcasts at 101.1 MHz. Green light has \(\lambda = 550\ \mathrm{nm}\). Find the radio wavelength and the light's frequency.
\[\lambda = \frac{c}{f} = \frac{3.00\times10^{8}\ \mathrm{m/s}}{1.011\times10^{8}\ \mathrm{Hz}} = 2.97\ \mathrm{m}, \qquad
f = \frac{c}{\lambda} = \frac{3.00\times10^{8}\ \mathrm{m/s}}{5.50\times10^{-7}\ \mathrm{m}} = 5.45\times10^{14}\ \mathrm{Hz}.\]
Radio antennas are meters long and light interacts with atoms: the wavelength tells you the scale a wave "sees."
Worked example · Two polarizers
Unpolarized light of intensity \(I_0\) passes through a polarizer, then through a second one whose axis is rotated 60° from the first. What fraction gets through?
First filter: \(I_1 = \tfrac{1}{2}I_0\), now polarized along the first axis. Second filter:
\[I_2 = I_1\cos^2 60^\circ = \tfrac{1}{2}I_0\,(0.50)^2 = 0.125\,I_0,\]
or 12.5%. At 90° (crossed polarizers) nothing passes. Explain-in-words version: the second filter keeps only the component of the field along its axis, \(E\cos\theta\), and intensity goes as the field squared.
Polarization is the proof that light is transverse. A longitudinal wave like sound oscillates along its direction of travel, so there is no sideways direction to filter: sound cannot be polarized.
Try it
Unpolarized light of intensity \(I_0\) passes through two polarizers whose axes differ by 30°. Find the transmitted intensity.
Show answer
\(I = \tfrac{1}{2}I_0\cos^2 30^\circ = \tfrac{1}{2}I_0\,(0.866)^2 = 0.375\,I_0\), or 37.5% of the original.
Lesson 14.5 · Unit 14 · CED topics 14.8–14.9
Interference and diffraction: double slits, single slits, and thin films
Shine light through two narrow slits and the screen shows bright and dark
bands, not two bright stripes. That pattern is superposition at work: light
from the two slits travels different distances, and where the difference is
a whole number of wavelengths the waves arrive in step.
Formulas
Double slit, separation \(d\): bright fringes where \(d\sin\theta = m\lambda\), dark where \(d\sin\theta = (m + \tfrac{1}{2})\lambda\), \(m = 0, 1, 2, \dots\)
For a screen at distance \(L\) and small angles, the \(m\)th bright fringe is at \(y_m = \dfrac{m\lambda L}{d}\), so adjacent fringes are \(\Delta y = \dfrac{\lambda L}{d}\) apart.
Single slit, width \(a\): dark fringes where \(a\sin\theta = m\lambda\), \(m = 1, 2, \dots\); the central bright band is twice as wide as the others. Narrower slit, wider spread.
Thin film of index \(n\) and thickness \(t\): the wavelength inside is \(\lambda/n\), and a reflection from a higher-index medium flips the wave by half a wavelength; from a lower-index medium it does not.
Worked example · Fringe spacing
Light of wavelength 600 nm passes through two slits 0.25 mm apart onto a screen 2.0 m away. Find the fringe spacing and the position of the third bright fringe from center.
\[\Delta y = \frac{\lambda L}{d} = \frac{(6.00\times10^{-7}\ \mathrm{m})(2.0\ \mathrm{m})}{2.5\times10^{-4}\ \mathrm{m}} = 4.8\times10^{-3}\ \mathrm{m} = 4.8\ \mathrm{mm}.\]
The third bright fringe (\(m = 3\)) is at \(y_3 = 3\,\Delta y = 14\ \mathrm{mm}\). Red light or a farther screen spreads the fringes; wider slit separation squeezes them together. Sanity check on the small-angle assumption: \(\sin\theta \approx 0.0144/2.0 = 0.0072\), tiny.
Worked example · Thin-film reasoning
A soap film (\(n = 1.33\)) in air is lit by 500 nm light. What minimum thickness makes the reflected light strongly constructive?
Track the two reflected waves. The one off the top surface goes air → soap (low to high \(n\)): it flips by half a wavelength. The one off the bottom goes soap → air (high to low): no flip. So the reflections start half a wavelength apart, and the extra path \(2t\) inside the film must supply another half wavelength (in the film) to bring them back in step:
\[2t = \left(m + \tfrac{1}{2}\right)\frac{\lambda}{n}, \qquad t_{\min} = \frac{\lambda}{4n} = \frac{500\ \mathrm{nm}}{4(1.33)} = 94\ \mathrm{nm}.\]
A film much thinner than this looks dark, the two reflections cancel, which is exactly what you see at the top of a draining soap film just before it pops.
Try it
In a double-slit experiment the bright fringes are 3.2 mm apart on a screen 1.6 m away, with slits 0.30 mm apart. Find the wavelength.
Unit 14 practice: Waves, Sound, and Physical Optics
Ten problems covering the whole unit, in roughly exam order. Work each one on paper
before revealing the answer. Take the speed of sound in air as 340 m/s and
\(c = 3.00\times10^{8}\ \text{m/s}\). Calculator allowed.
Twelve crests of a wave on a rope pass a fixed point in 3.0 s, and adjacent crests are 0.50 m apart. Find the frequency, period, and wave speed. If the amplitude is doubled, what happens to the energy the wave carries and to its speed?
Show answer
\(f = \dfrac{12}{3.0} = 4.0\ \text{Hz}\); \(T = \dfrac{1}{f} = 0.25\ \text{s}\); \(v = f\lambda = (4.0)(0.50) = 2.0\ \text{m/s}\). Energy \(\propto A^2\), so it quadruples; the speed is set by the rope, not the amplitude, and stays 2.0 m/s.
A 1700 Hz sound travels through air (340 m/s) into a steel rail where sound travels at 5100 m/s. Find the wavelength in each medium and explain why the frequency does not change.
Show answer
\(\lambda_{\text{air}} = \dfrac{340}{1700} = 0.20\ \text{m}\); \(\lambda_{\text{steel}} = \dfrac{5100}{1700} = 3.0\ \text{m}\). The steel surface is driven by the air at 1700 oscillations per second, so it vibrates at 1700 Hz too; only the wavelength adjusts, by the same factor of 15 as the speed.
Waves travel at 270 m/s on a 0.90 m string fixed at both ends. Find the fundamental frequency and the frequency and wavelength of the third harmonic, and give the positions of the nodes in the third-harmonic pattern.
Show answer
\(f_1 = \dfrac{v}{2L} = \dfrac{270}{2(0.90)} = 150\ \text{Hz}\); \(f_3 = 3f_1 = 450\ \text{Hz}\); \(\lambda_3 = \dfrac{2L}{3} = 0.60\ \text{m}\). Nodes every half wavelength (0.30 m): at 0, 0.30 m, 0.60 m, and 0.90 m: four nodes and three loops.
A guitar string's tension is increased so that the wave speed on it doubles, with its length unchanged. Its fundamental frequency
Halving would happen if the wavelength doubled at fixed speed, or if \(f\) were inversely related to \(v\). But \(f = v/\lambda\) with \(\lambda\) fixed, so a faster wave means a higher frequency.
The frequency would be unchanged only if the wavelength changed in step with the speed, but the fundamental wavelength \(\lambda_1 = 2L\) is set by the clamped ends and does not depend on tension. Only the speed changes, so the frequency must change too.
The fundamental wavelength \(\lambda_1 = 2L\) is fixed by the clamped ends, so \(f_1 = v/(2L)\) scales directly with \(v\). Doubling the speed doubles the pitch, one octave up.
Quadrupling would require a factor-of-4 change in speed. Doubling the wave speed does require quadrupling the tension (since \(v = \sqrt{F_T/\mu}\)), but the frequency tracks the speed, not the tension, so it only doubles.
(a) A pipe 0.85 m long is closed at one end. Find its three lowest resonant frequencies. (b) Rank the fundamental frequencies of these pipes, greatest to least: (i) open–open, length \(L\); (ii) open–closed, length \(L\); (iii) open–open, length \(2L\); (iv) open–closed, length \(L/2\).
Show answer
(a) Closed pipes hold odd harmonics only: \(f_1 = \dfrac{v}{4L} = \dfrac{340}{4(0.85)} = 100\ \text{Hz}\), \(f_3 = 300\ \text{Hz}\), \(f_5 = 500\ \text{Hz}\). (b) (i) \(\dfrac{v}{2L}\); (ii) \(\dfrac{v}{4L}\); (iii) \(\dfrac{v}{4L}\); (iv) \(\dfrac{v}{4(L/2)} = \dfrac{v}{2L}\). Ranking: (i) = (iv) > (ii) = (iii). Closing one end or doubling the length each halve the fundamental.
A string sounded together with a 512 Hz tuning fork produces 4 beats per second. When the string is loosened slightly, the beat rate decreases. What was the string's original frequency? Justify.
Show answer
\(f_{\text{beat}} = |f_1 - f_2| = 4\ \text{Hz}\), so the string was at 508 Hz or 516 Hz. Loosening lowers its frequency; since the beats slowed, it moved toward 512 Hz, so it started above: 516 Hz.
A train horn emits 500 Hz while the train approaches a stationary listener at 25 m/s. Find the frequency heard as the train approaches and after it passes, and explain in words why the pitch is higher on approach.
Show answer
Approaching (source moving, observer still): \(f' = f\dfrac{v}{v - v_s} = 500\cdot\dfrac{340}{340 - 25} = 540\ \text{Hz}\). Receding: \(f' = 500\cdot\dfrac{340}{340 + 25} = 466\ \text{Hz}\). Each wavefront is emitted from a point closer to the listener than the one before, so the fronts ahead of the train are crowded together: shorter wavelength, more fronts per second arriving, higher frequency. The wave speed in the air is unchanged.
(a) A microwave oven operates at 2.45 GHz. Find the wavelength. (b) Unpolarized light of intensity \(I_0\) passes through two polarizers whose axes differ by 45°. Find the transmitted intensity. (c) A third polarizer is inserted between two crossed (90°) polarizers, at 45° to each. Explain in words why light now gets through, and find how much.
Show answer
(a) \(\lambda = \dfrac{c}{f} = \dfrac{3.00\times10^{8}}{2.45\times10^{9}} = 0.122\ \text{m}\), about 12 cm. (b) The first filter halves the intensity and polarizes the light; the second passes \(\cos^2 45^\circ = 0.50\) of that: \(I = \tfrac{1}{2}I_0(0.50) = 0.25\,I_0\). (c) Crossed filters alone pass nothing because the light leaving the first has no component along the second's axis. The middle filter rotates the polarization to 45°, and that light does have a component along the last axis: \(I = \tfrac{1}{2}I_0\cos^2 45^\circ\cos^2 45^\circ = 0.125\,I_0\).
Light of wavelength 500 nm passes through two slits 0.20 mm apart onto a screen 3.0 m away. Find the spacing of the bright fringes, the distance from the center to the second dark fringe, and the angle of the second-order bright fringe.
Show answer
\(\Delta y = \dfrac{\lambda L}{d} = \dfrac{(5.00\times10^{-7})(3.0)}{2.0\times10^{-4}} = 7.5\times10^{-3}\ \text{m} = 7.5\ \text{mm}\). Dark fringes sit at \((m + \tfrac{1}{2})\Delta y\): the second (\(m = 1\)) is at \(1.5(7.5) = 11\ \text{mm}\). Second-order bright: \(d\sin\theta = 2\lambda\), \(\sin\theta = \dfrac{2(5.00\times10^{-7})}{2.0\times10^{-4}} = 5.0\times10^{-3}\), \(\theta = 0.29^\circ\): small-angle approximation fully justified.
A thin film of oil (\(n = 1.45\)) floats on water (\(n = 1.33\)) and is viewed in air under 580 nm light. Track the phase change at each reflecting surface, then find the minimum oil thickness for which the reflected light is strongly constructive.
Show answer
Top surface, air → oil (low to high \(n\)): the reflection flips by half a wavelength. Bottom surface, oil → water (high to low \(n\)): no flip. The two reflections start half a wavelength apart, so the round trip inside the film must add another half wavelength (in the oil) for constructive interference: \(2t = \left(m + \tfrac{1}{2}\right)\dfrac{\lambda}{n}\). Minimum (\(m = 0\)): \(t = \dfrac{\lambda}{4n} = \dfrac{580}{4(1.45)} = 100\ \text{nm}\).
Lesson 15.1 · Unit 15 · CED topics 15.1–15.2
Radioactive decay and nuclear reactions
A nucleus is written \({}^{A}_{Z}\mathrm{X}\): \(Z\) protons (the element), \(A\) nucleons
(protons plus neutrons). Some combinations are unstable and rearrange themselves
spontaneously, spitting out a particle. Which particle, and how fast, are the two
things this lesson pins down.
Rules · Three decays and two conservation laws
Alpha: emits \({}^{4}_{2}\mathrm{He}\). \(A\) drops by 4, \(Z\) by 2.
Beta-minus: a neutron becomes a proton, emitting an electron \({}^{\ 0}_{-1}\mathrm{e}\) and an antineutrino \(\bar{\nu}\). \(A\) unchanged, \(Z\) up by 1.
Gamma: a photon \({}^{0}_{0}\gamma\) carries away energy; \(A\) and \(Z\) unchanged.
In every nuclear equation the total \(A\) (nucleon number) and the total \(Z\) (charge) are the same on both sides. Balance both: that is the whole method.
Formula · Half-life
Each nucleus decays at random, but a large sample halves in a fixed time \(T_{1/2}\):
\[N = N_0\left(\tfrac{1}{2}\right)^{t/T_{1/2}}.\]
The same law holds for mass, and for activity (decays per second), since both are proportional to \(N\).
Worked example · Balancing nuclear equations
Write the equation for the alpha decay of uranium-238, and for the beta-minus decay of carbon-14.
Alpha: the daughter has \(A = 238 - 4 = 234\) and \(Z = 92 - 2 = 90\), which is thorium:
\[{}^{238}_{92}\mathrm{U} \to {}^{234}_{90}\mathrm{Th} + {}^{4}_{2}\mathrm{He}.\]
Check: \(234 + 4 = 238\) and \(90 + 2 = 92\).
Beta: \(A\) stays 14, \(Z\) goes from 6 to 7 (nitrogen), and the electron's \(-1\) balances the charge:
\[{}^{14}_{6}\mathrm{C} \to {}^{14}_{7}\mathrm{N} + {}^{\ 0}_{-1}\mathrm{e} + \bar{\nu}.\]
Check: \(14 = 14 + 0\) and \(6 = 7 + (-1)\).
Worked example · Half-life
Iodine-131 has a half-life of 8.0 days. A hospital receives 80 mg. How much remains after 24 days? After 20 days? When will 5.0 mg remain?
24 days is 3 half-lives: \(80 \to 40 \to 20 \to 10\ \mathrm{mg}\).
20 days is 2.5 half-lives:
\[m = (80\ \mathrm{mg})\left(\tfrac{1}{2}\right)^{20/8} = (80\ \mathrm{mg})(0.177) = 14\ \mathrm{mg}.\]
5.0 mg is \(80/16\), and \(16 = 2^4\), so 4 half-lives: \(t = 4(8.0\ \mathrm{d}) = 32\ \mathrm{days}\).
Exam tip: count halvings first; use the exponent form only when the time isn't a whole number of half-lives.
Try it
(a) Radium-226 (\(Z = 88\)) alpha-decays. Write the daughter nucleus with its \(A\) and \(Z\) (the element with \(Z = 86\) is radon, Rn).
(b) A sample's activity falls to one-eighth of its initial value in 45 minutes. Find the half-life.
Show answer
(a) \({}^{226}_{88}\mathrm{Ra} \to {}^{222}_{86}\mathrm{Rn} + {}^{4}_{2}\mathrm{He}\).
(b) One-eighth is \((1/2)^3\), three half-lives, so \(T_{1/2} = 45/3 = 15\ \mathrm{min}\).
Lesson 15.2 · Unit 15 · CED topic 15.3
Mass–energy equivalence, mass defect, and binding energy
Weigh a helium nucleus and it comes out lighter than two protons plus two
neutrons weighed separately. The missing mass didn't vanish: it left as energy
when the nucleus formed. Einstein's relation is the exchange rate.
Formulas
\[E = mc^2, \qquad \Delta E = (\Delta m)c^2.\]
The mass defect of a nucleus is
\(\Delta m = Z m_p + (A - Z) m_n - m_{\text{nucleus}}\), and its binding energy is \(\Delta m\, c^2\): the energy needed to pull it completely apart.
Nuclear masses are given in atomic mass units, and the handy conversion is
\[1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^2, \qquad\text{so}\qquad E\ (\mathrm{MeV}) = \Delta m\ (\mathrm{u}) \times 931.5.\]
(\(1\ \mathrm{eV} = 1.60\times10^{-19}\ \mathrm{J}\); \(m_p = 1.007276\ \mathrm{u}\), \(m_n = 1.008665\ \mathrm{u}\).)
Worked example · Binding energy of helium-4
The helium-4 nucleus has mass 4.001506 u. Find its mass defect and binding energy, total and per nucleon.
\[\Delta m = 2(1.007276\ \mathrm{u}) + 2(1.008665\ \mathrm{u}) - 4.001506\ \mathrm{u} = 4.031882 - 4.001506 = 0.030376\ \mathrm{u}.\]
\[E_B = (0.030376\ \mathrm{u})(931.5\ \mathrm{MeV/u}) = 28.3\ \mathrm{MeV}, \qquad \frac{E_B}{A} = \frac{28.3}{4} = 7.07\ \mathrm{MeV/nucleon}.\]
Compare a few eV for a chemical bond: nuclear energies are about a million times larger, which is the whole reason a gram of fuel matters.
Worked example · Energy released in fusion
In the deuterium–tritium reaction \({}^{2}_{1}\mathrm{H} + {}^{3}_{1}\mathrm{H} \to {}^{4}_{2}\mathrm{He} + {}^{1}_{0}\mathrm{n}\), the nuclear masses are 2.013553 u, 3.015501 u, 4.001506 u, and 1.008665 u. Find the energy released.
\[\Delta m = (2.013553 + 3.015501) - (4.001506 + 1.008665) = 5.029054 - 5.010171 = 0.018883\ \mathrm{u}.\]
\[E = (0.018883\ \mathrm{u})(931.5\ \mathrm{MeV/u}) = 17.6\ \mathrm{MeV}.\]
Products lighter than reactants means energy out. The same bookkeeping works for fission: the fragments of a split uranium nucleus weigh less than the original, by roughly 0.2 u, or about 200 MeV per fission.
Worked example · Explain in words
Why does a stable nucleus weigh less than the sum of its parts?
The nucleons attract one another through the strong force, so pulling them apart takes work: you must add energy to separate them. That added energy ends up as mass in the separated pieces, because energy and mass are the same quantity in different units. Run it backward: when nucleons bind, they release that energy (as gamma rays and kinetic energy), and the bound system is left lighter by exactly \(E_B/c^2\). A bigger mass defect per nucleon means a more tightly bound, more stable nucleus.
Try it
In one fission of uranium-235 the products are 0.190 u lighter than the reactants. Find the energy released in MeV and in joules.
Shine light on a metal and electrons pop out, but only if the light's
frequency is high enough, no matter how bright it is. Wave theory can't
explain that. Einstein could: light arrives in packets, and one packet
kicks out one electron.
Formulas
A photon of frequency \(f\) carries energy
\[E = hf = \frac{hc}{\lambda}, \qquad h = 6.63\times10^{-34}\ \mathrm{J\cdot s}.\]
With \(c = 3.00\times10^{8}\ \mathrm{m/s}\) and \(1\ \mathrm{eV} = 1.60\times10^{-19}\ \mathrm{J}\), the product \(hc \approx 1.24\times10^{3}\ \mathrm{eV\cdot nm}\), so \(E\ (\mathrm{eV}) = 1240/\lambda\ (\mathrm{nm})\).
An electron absorbing one photon leaves the metal with at most
\[K_{\max} = hf - \phi,\]
where \(\phi\) is the work function, the minimum energy to escape. The stopping potential \(V_s\) is the reverse voltage that just halts the fastest electrons: \(K_{\max} = eV_s\). The threshold frequency is \(f_0 = \phi/h\); below it, no electrons at all.
Worked example · Stopping potential
Sodium has \(\phi = 2.28\ \mathrm{eV}\). It is lit with 400 nm light. Find the photon energy, \(K_{\max}\), the stopping potential, and the longest wavelength that still ejects electrons.
\[E = \frac{hc}{\lambda} = \frac{1240\ \mathrm{eV\cdot nm}}{400\ \mathrm{nm}} = 3.10\ \mathrm{eV}, \qquad K_{\max} = 3.10 - 2.28 = 0.82\ \mathrm{eV} = 1.3\times10^{-19}\ \mathrm{J}.\]
\(K_{\max} = eV_s\), so \(V_s = 0.82\ \mathrm{V}\), in electron-volts the numbers match by construction.
Threshold: \(E = \phi\) when \(\lambda_0 = \dfrac{1240\ \mathrm{eV\cdot nm}}{2.28\ \mathrm{eV}} = 544\ \mathrm{nm}\). Green light works; red (650 nm) does nothing, however bright.
Worked example · Explain in words
The 400 nm lamp is replaced by one four times as intense. What changes, and what doesn't?
Intensity means photons per second, not energy per photon. Four times as many photons arrive, so four times as many electrons are ejected per second: the photocurrent quadruples. But each electron still absorbs a single 3.10 eV photon and still pays 2.28 eV to escape, so \(K_{\max}\) and the stopping potential are unchanged. To raise \(K_{\max}\) you must raise the frequency. Graders want the phrase "one photon, one electron."
The graph of \(K_{\max}\) versus \(f\) is a straight line with slope \(h\) and vertical intercept \(-\phi\); it crosses the axis at \(f_0\). Every metal gives the same slope: that universality was the evidence for Planck's constant.
Try it
A metal has \(\phi = 4.20\ \mathrm{eV}\). Find the stopping potential for 250 nm light. Does 300 nm light eject any electrons?
The photoelectric effect showed that light, a wave, also behaves as particles.
De Broglie asked the reverse question: do particles also behave as waves? They do.
Electrons fired at a crystal produce a diffraction pattern, just as X-rays of the
same wavelength do, and electrons sent one at a time through a double slit still
build up interference fringes.
Formula
A particle of momentum \(p = mv\) has a wavelength
\[\lambda = \frac{h}{p} = \frac{h}{mv}, \qquad h = 6.63\times10^{-34}\ \mathrm{J\cdot s}.\]
Wave behavior (diffraction, interference) shows up only when \(\lambda\) is comparable to the size of the opening or spacing the particle meets. Bigger momentum, shorter wavelength.
Worked example · An electron
An electron (\(m = 9.11\times10^{-31}\ \mathrm{kg}\)) is accelerated from rest through 100 V. Find its de Broglie wavelength.
Kinetic energy gained: \(K = qV = (1.60\times10^{-19}\ \mathrm{C})(100\ \mathrm{V}) = 1.60\times10^{-17}\ \mathrm{J}\).
\[v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2(1.60\times10^{-17}\ \mathrm{J})}{9.11\times10^{-31}\ \mathrm{kg}}} = 5.93\times10^{6}\ \mathrm{m/s}, \qquad
p = mv = 5.40\times10^{-24}\ \mathrm{kg\cdot m/s}.\]
\[\lambda = \frac{h}{p} = \frac{6.63\times10^{-34}\ \mathrm{J\cdot s}}{5.40\times10^{-24}\ \mathrm{kg\cdot m/s}} = 1.23\times10^{-10}\ \mathrm{m}.\]
That is 0.123 nm: about the spacing between atoms in a crystal. So a crystal acts as a diffraction grating for these electrons, which is exactly what Davisson and Germer observed.
Worked example · A baseball
A 0.145 kg baseball is pitched at 40 m/s. Find its wavelength and explain why it never diffracts.
\[p = mv = (0.145\ \mathrm{kg})(40\ \mathrm{m/s}) = 5.8\ \mathrm{kg\cdot m/s}, \qquad
\lambda = \frac{6.63\times10^{-34}}{5.8} = 1.1\times10^{-34}\ \mathrm{m}.\]
No opening in the universe is anywhere near \(10^{-34}\) m wide: an atomic nucleus is \(10^{-15}\) m. With \(\lambda\) unimaginably smaller than any slit, the diffraction angle is effectively zero and the ball travels in a straight line. The wave nature of matter is real for everything, but observable only for tiny momenta.
Exam tip: to compare wavelengths, compare momenta. A proton and an electron with the same speed have wavelengths in the ratio of their masses (the electron's is about 1800 times longer); with the same kinetic energy, use \(p = \sqrt{2mK}\).
Try it
Find the de Broglie wavelength of a proton (\(m = 1.67\times10^{-27}\ \mathrm{kg}\)) moving at \(1.0\times10^{5}\ \mathrm{m/s}\).
Heat hydrogen gas and it glows in a few sharp colors, not a rainbow. The
reason is that an electron bound to an atom can hold only certain energies.
It jumps between them by emitting or absorbing a photon whose energy equals
the gap, no more, no less.
Formulas · Hydrogen levels and transitions
\[E_n = -\frac{13.6\ \mathrm{eV}}{n^2}, \qquad n = 1, 2, 3, \dots\]
Energies are negative because the electron is bound; \(E = 0\) means free (ionized). Levels crowd together as \(n\) grows.
A transition between levels \(n_i\) and \(n_f\) involves a photon with
\[E_{\text{photon}} = |E_i - E_f| = hf = \frac{hc}{\lambda}.\]
Emission: electron drops, photon leaves. Absorption: photon of exactly the right energy is taken in, electron rises. Use \(hc = 1240\ \mathrm{eV\cdot nm}\).
Worked example · An emission line
A hydrogen electron drops from \(n = 3\) to \(n = 2\). Find the wavelength and frequency of the emitted photon.
\[E_3 = -\frac{13.6}{9} = -1.51\ \mathrm{eV}, \qquad E_2 = -\frac{13.6}{4} = -3.40\ \mathrm{eV}, \qquad E_{\text{photon}} = -1.51 - (-3.40) = 1.89\ \mathrm{eV}.\]
\[\lambda = \frac{hc}{E} = \frac{1240\ \mathrm{eV\cdot nm}}{1.89\ \mathrm{eV}} = 656\ \mathrm{nm}, \qquad
f = \frac{E}{h} = \frac{(1.89)(1.60\times10^{-19}\ \mathrm{J})}{6.63\times10^{-34}\ \mathrm{J\cdot s}} = 4.56\times10^{14}\ \mathrm{Hz}.\]
That is the red line of hydrogen, the brightest in its visible spectrum. Picture the energy-level diagram: horizontal rungs at \(-13.6\), \(-3.40\), \(-1.51\), \(-0.85\) eV, getting closer together toward zero, with an arrow drawn downward from the third rung to the second for emission.
Worked example · Absorption and ionization
(a) A beam of 2.0 eV photons passes through cool hydrogen in the ground state. Are they absorbed? (b) What is the longest photon wavelength that can ionize hydrogen from \(n = 2\)?
(a) No. From \(n = 1\) the smallest possible jump is to \(n = 2\), needing \(-3.40 - (-13.6) = 10.2\ \mathrm{eV}\). A 2.0 eV photon matches no gap, so it passes straight through; an atom can't absorb part of a photon.
(b) Ionizing from \(n = 2\) takes the electron from \(-3.40\ \mathrm{eV}\) to 0: \(E = 3.40\ \mathrm{eV}\), so
\(\lambda = \dfrac{1240}{3.40} = 365\ \mathrm{nm}\) (ultraviolet). Any shorter wavelength also works: the excess becomes the freed electron's kinetic energy, which is why absorption above the ionization energy is continuous rather than a line.
Dark absorption lines in a star's spectrum sit at exactly the wavelengths of the gas's emission lines: the same gaps, crossed in the opposite direction.
Try it
Find the wavelength of the photon emitted when a hydrogen electron drops from \(n = 4\) to \(n = 2\). Is it visible?
Ten problems covering the whole unit, in roughly exam order. Work each one on paper
before revealing the answer. Constants: \(h = 6.63\times10^{-34}\ \text{J·s}\),
\(hc = 1240\ \text{eV·nm}\), \(1\ \text{eV} = 1.60\times10^{-19}\ \text{J}\),
\(1\ \text{u} = 931.5\ \text{MeV}/c^2\), \(m_p = 1.007276\ \text{u}\), \(m_n = 1.008665\ \text{u}\),
\(m_e = 9.11\times10^{-31}\ \text{kg}\). Calculator allowed.
Complete and balance each nuclear equation, identifying the unknown nucleus by its \(A\) and \(Z\). (a) Polonium-210 (\(Z = 84\)) undergoes alpha decay. (b) Cobalt-60 (\(Z = 27\)) undergoes beta-minus decay. (c) \({}^{235}_{92}\mathrm{U} + {}^{1}_{0}\mathrm{n} \to {}^{141}_{56}\mathrm{Ba} + \mathrm{X} + 3\,{}^{1}_{0}\mathrm{n}\). (Lead is \(Z = 82\), nickel \(Z = 28\), krypton \(Z = 36\).)
Show answer
Conserve nucleon number and charge. (a) \({}^{210}_{84}\mathrm{Po} \to {}^{206}_{82}\mathrm{Pb} + {}^{4}_{2}\mathrm{He}\): \(210 = 206 + 4\), \(84 = 82 + 2\). (b) \({}^{60}_{27}\mathrm{Co} \to {}^{60}_{28}\mathrm{Ni} + {}^{\ 0}_{-1}\mathrm{e} + \bar{\nu}\): \(A\) unchanged, \(27 = 28 - 1\). (c) \(A\): \(235 + 1 = 141 + A + 3\), so \(A = 92\); \(Z\): \(92 = 56 + Z\), so \(Z = 36\). X is \({}^{92}_{36}\mathrm{Kr}\).
A radioactive sample of 160 mg has a half-life of 6.0 h. How much remains after 24 h? After 15 h? When will 5.0 mg remain?
Show answer
24 h is 4 half-lives: \(160 \to 80 \to 40 \to 20 \to 10\ \text{mg}\). 15 h is 2.5 half-lives: \(m = 160\left(\tfrac{1}{2}\right)^{15/6} = 160(0.177) = 28\ \text{mg}\). \(5.0 = 160/32\) and \(32 = 2^5\): five half-lives, \(t = 30\ \text{h}\).
In beta-minus decay, which of the following is unchanged?
In beta-minus decay a neutron becomes a proton, so the proton count rises by one. That is what moves the daughter nucleus one place to the right in the periodic table.
A neutron turns into a proton plus an electron and an antineutrino: neutrons drop by one, protons rise by one, and the total nucleon count \(A = Z + N\) stays the same. Only alpha decay changes \(A\).
The neutron count falls by one, since one neutron is converted into a proton. Beta-minus decay happens in nuclei with too many neutrons and is how they shed one.
\(Z\) is the number of protons, so it increases by one: the daughter is a different element. Picking this usually means confusing \(Z\) with the mass number \(A\).
The carbon-12 nucleus has mass 11.996706 u. Find its mass defect, total binding energy in MeV, and binding energy per nucleon.
Show answer
\(\Delta m = 6m_p + 6m_n - m_{\text{nuc}} = 6(1.007276) + 6(1.008665) - 11.996706 = 12.095646 - 11.996706 = 0.098940\ \text{u}\). \(E_B = (0.098940)(931.5) = 92.2\ \text{MeV}\); per nucleon \(\dfrac{92.2}{12} = 7.68\ \text{MeV}\). Carbon-12 is bound more tightly per nucleon than helium-4 (7.07 MeV).
In the fusion reaction \({}^{2}_{1}\mathrm{H} + {}^{2}_{1}\mathrm{H} \to {}^{3}_{2}\mathrm{He} + {}^{1}_{0}\mathrm{n}\) the nuclear masses are 2.013553 u (each deuteron), 3.014932 u, and 1.008665 u. Find the energy released in MeV and in joules, and explain in one sentence where it comes from.
Show answer
\(\Delta m = 2(2.013553) - (3.014932 + 1.008665) = 4.027106 - 4.023597 = 0.003509\ \text{u}\). \(E = (0.003509)(931.5) = 3.27\ \text{MeV} = (3.27\times10^{6})(1.60\times10^{-19}) = 5.2\times10^{-13}\ \text{J}\). The products are lighter than the reactants; the missing mass appears as kinetic energy of the helium nucleus and neutron, by \(E = \Delta m\,c^2\).
Potassium has a work function of 2.30 eV. It is lit with 350 nm light. Find the photon energy, the maximum kinetic energy of the ejected electrons, the stopping potential, and the longest wavelength that ejects any electrons.
A graph of \(K_{\max}\) versus light frequency for a metal is a straight line crossing the frequency axis at 5.5 × 10¹⁴ Hz. (a) What does the slope represent? (b) Find the work function in joules and eV. (c) Find \(K_{\max}\) for 8.0 × 10¹⁴ Hz light. (d) Explain in words why doubling the light's intensity leaves the graph unchanged.
Show answer
(a) \(K_{\max} = hf - \phi\), so the slope is Planck's constant \(h\), the same for every metal. (b) At the intercept \(K_{\max} = 0\): \(\phi = hf_0 = (6.63\times10^{-34})(5.5\times10^{14}) = 3.6\times10^{-19}\ \text{J} = 2.3\ \text{eV}\). (c) \(K_{\max} = h(f - f_0) = (6.63\times10^{-34})(2.5\times10^{14}) = 1.7\times10^{-19}\ \text{J} \approx 1.0\ \text{eV}\). (d) Intensity is photons per second; each electron absorbs one photon, whose energy depends only on frequency. More photons means more electrons per second, not faster ones.
(a) Find the de Broglie wavelength of an electron with kinetic energy 50 eV. (b) An electron, a proton, and an alpha particle move at the same speed; rank their de Broglie wavelengths, longest to shortest.
Show answer
(a) \(K = 50(1.60\times10^{-19}) = 8.0\times10^{-18}\ \text{J}\); \(p = \sqrt{2mK} = \sqrt{2(9.11\times10^{-31})(8.0\times10^{-18})} = 3.8\times10^{-24}\ \text{kg·m/s}\); \(\lambda = \dfrac{h}{p} = \dfrac{6.63\times10^{-34}}{3.8\times10^{-24}} = 1.7\times10^{-10}\ \text{m}\), about 0.17 nm: atomic spacing, so a crystal diffracts it. (b) \(\lambda = h/(mv)\) at equal \(v\) goes as \(1/m\): electron > proton > alpha, the electron's about 1800 times the proton's.
A hydrogen electron drops from \(n = 5\) to \(n = 2\). Find the photon's energy, wavelength, and frequency. Then find the wavelength of the photon a ground-state atom must absorb to reach \(n = 3\).
(a) Rank the wavelengths of the hydrogen photons emitted in the transitions \(2 \to 1\), \(3 \to 2\), and \(4 \to 3\), longest to shortest. (b) Explain in words why a hot gas emits a line spectrum rather than a continuous rainbow, and why the dark lines in a star's spectrum sit at exactly the same wavelengths.
Show answer
(a) Energies: \(2 \to 1\): \(10.2\ \text{eV}\); \(3 \to 2\): \(1.89\ \text{eV}\); \(4 \to 3\): \(0.66\ \text{eV}\). Longer wavelength means smaller energy: \(4 \to 3\) (1880 nm) > \(3 \to 2\) (656 nm) > \(2 \to 1\) (122 nm). The levels crowd together as \(n\) grows, so high transitions are gentle. (b) A bound electron can hold only certain energies, so a photon can carry away only the difference between two allowed levels: a discrete set of energies, hence discrete wavelengths. Cool gas in a star's atmosphere absorbs photons of exactly those same energies, lifting electrons across the same gaps in the other direction, and removes them from the continuous light behind it.
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