The Timpson Lyceum · University Physics II (2026) – Electricity & Magnetism

Chapter 24 — Capacitance & Dielectrics

Problems with worked solutions · 14 points · 3 extra credit

Chapter 24 — Capacitance & Dielectrics

Core: 14 points · Extra credit: 3 points. Show your reasoning; partial credit is given. Use $\varepsilon_0=8.85\times10^{-12}\,\text{C}^2/(\text{N}\cdot\text{m}^2)$ and $k=8.99\times10^{9}\,\text{N}\cdot\text{m}^2/\text{C}^2$.

1. 1. (2 pts ••) A parallel-plate capacitor has plate area $A=0.012\,\text{m}^2$ and separation $d=1.5\,\text{mm}$, connected to a $12\,\text{V}$ battery.

   (a) Find the capacitance.

   (b) Find the charge on each plate.

   (c) Find the energy stored.

   (d) Find the electric field between the plates.

   Solution

   1.

   (a) $C=\dfrac{\varepsilon_0 A}{d}=\dfrac{(8.85\times10^{-12})(0.012)}{1.5\times10^{-3}}\approx 7.1\times10^{-11}\,\text{F}=71\,\text{pF}$.

   (b) $Q=CV=(7.08\times10^{-11})(12)\approx 8.5\times10^{-10}\,\text{C}$.

   (c) $U=\tfrac12 CV^2=\tfrac12(7.08\times10^{-11})(12)^2\approx 5.1\times10^{-9}\,\text{J}$.

   (d) $E=\dfrac{V}{d}=\dfrac{12}{1.5\times10^{-3}}=8.0\times10^{3}\,\text{V/m}$.

2. 2. (2 pts ••) Three capacitors have capacitances $6.0\,\mu\text{F}$, $3.0\,\mu\text{F}$, and $2.0\,\mu\text{F}$.

   (a) Find the equivalent capacitance if all three are in series.

   (b) Find the equivalent capacitance if all three are in parallel.

   Solution

   2.

   (a) $\dfrac{1}{C_s}=\dfrac16+\dfrac13+\dfrac12=1\,\mu\text{F}^{-1}\Rightarrow C_s=1.0\,\mu\text{F}$.

   (b) $C_p=6.0+3.0+2.0=11\,\mu\text{F}$.

3. 3. (3 pts •••) A $6.0\,\mu\text{F}$ capacitor ($C_1$) is connected in series with the parallel combination of two $3.0\,\mu\text{F}$ capacitors ($C_2$ and $C_3$). This combination is connected across a $24\,\text{V}$ battery.

   (a) Find the equivalent capacitance of the network.

   (b) Find the total charge delivered by the battery.

   (c) Find the voltage across, and the charge on, each of the three capacitors.

   Solution

   3.

   (a) The parallel pair: $C_2+C_3=6.0\,\mu\text{F}$. In series with $C_1=6.0\,\mu\text{F}$: $C_\text{eq}=\dfrac{(6)(6)}{6+6}=3.0\,\mu\text{F}$.

   (b) $Q_\text{tot}=C_\text{eq}V=(3.0\,\mu\text{F})(24\,\text{V})=72\,\mu\text{C}$.

   (c) $C_1$ (in series) carries the full $72\,\mu\text{C}$, so $V_1=Q/C_1=72/6=12\,\text{V}$. The remaining $24-12=12\,\text{V}$ is across the parallel pair, so $V_2=V_3=12\,\text{V}$ and $Q_2=Q_3=(3.0)(12)=36\,\mu\text{C}$ each (they sum to $72\,\mu\text{C}$, as required).

4. 4. (2 pts ••) A $4.0\,\mu\text{F}$ capacitor is charged to $50\,\text{V}$ and then disconnected from the battery. Its plate separation is then doubled.

   (a) Find the new capacitance.

   (b) Find the new charge on the plates.

   (c) Find the new voltage.

   (d) Find the change in stored energy, and explain where that energy came from.

   Solution

   4.

   (a) $C=\varepsilon_0 A/d$, and $d$ doubles, so $C$ halves: $C_\text{new}=2.0\,\mu\text{F}$.

   (b) Disconnected, so $Q$ is fixed: $Q=C V=(4.0\,\mu\text{F})(50)=2.0\times10^{-4}\,\text{C}$ (unchanged).

   (c) $V_\text{new}=Q/C_\text{new}=\dfrac{2.0\times10^{-4}}{2.0\times10^{-6}}=100\,\text{V}$.

   (d) $U_i=\tfrac12 CV^2=5.0\times10^{-3}\,\text{J}$; $U_f=\tfrac12 C_\text{new}V_\text{new}^2=1.0\times10^{-2}\,\text{J}$, so $\Delta U=+5.0\times10^{-3}\,\text{J}$. The energy came from the work you did pulling the plates apart against their mutual attraction.

5. 5. (2 pts ••) An air-filled capacitor with $C_0=100\,\text{pF}$ is connected to a $12\,\text{V}$ battery. While it remains connected, a dielectric of constant $K=3.0$ is inserted, completely filling the gap.

   (a) Find the new capacitance.

   (b) Find the new charge on the plates.

   (c) Does the stored energy increase or decrease, and where does the difference come from?

   Solution

   5.

   (a) $C=KC_0=3.0(100\,\text{pF})=300\,\text{pF}$.

   (b) Still at $12\,\text{V}$: $Q=CV=(300\times10^{-12})(12)=3.6\times10^{-9}\,\text{C}$ (three times the original $1.2\,\text{nC}$).

   (c) The energy increases: $U=\tfrac12 CV^2$ triples to $2.16\times10^{-8}\,\text{J}$. Because the capacitor stays connected at fixed $V$, the battery pushes additional charge onto the plates, and that extra energy comes from the battery.

6. 6. (1 pt •) A capacitor stores $4.5\times10^{-4}\,\text{J}$ of energy when charged to $30\,\text{V}$. Find its capacitance.

   Solution

   6. $U=\tfrac12 CV^2\Rightarrow C=\dfrac{2U}{V^2}=\dfrac{2(4.5\times10^{-4})}{(30)^2}=1.0\times10^{-6}\,\text{F}=1.0\,\mu\text{F}$.

7. 7. (2 pts ••) A spherical capacitor has an inner sphere of radius $a=1.0\,\text{cm}$ and a concentric outer shell of radius $b=1.2\,\text{cm}$ (its capacitance is $C=4\pi\varepsilon_0\dfrac{ab}{b-a}$).

   (a) Find the capacitance.

   (b) Find the energy stored when it is charged to $500\,\text{V}$.

   Solution

   7.

   (a) $C=4\pi\varepsilon_0\dfrac{ab}{b-a}=\dfrac{1}{k}\cdot\dfrac{(0.010)(0.012)}{0.002}=\dfrac{0.060}{8.99\times10^{9}}\approx 6.7\times10^{-12}\,\text{F}=6.7\,\text{pF}$.

   (b) $U=\tfrac12 CV^2=\tfrac12(6.7\times10^{-12})(500)^2\approx 8.3\times10^{-7}\,\text{J}$.

8. Extra Credit (3 pts •••)

   A cylindrical capacitor consists of an inner conductor of radius $a$ and a coaxial outer conductor of radius $b$, both of length $L\gg b$, carrying charges $\pm Q$.

   (a) Use Gauss's law to find the electric field between the conductors.

   (b) Integrate the field to find the potential difference between them, and hence show that $C=\dfrac{2\pi\varepsilon_0 L}{\ln(b/a)}$.

   (c) Show that when the gap $d=b-a$ is very small compared with $a$, this reduces to the parallel-plate result $C=\varepsilon_0 A/d$.

   Solution

   Extra Credit

   (a) A coaxial Gaussian cylinder of radius $r$ ($a<r<b$) and length $\ell$ encloses charge $Q\ell/L$, so $E(2\pi r\ell)=\dfrac{Q\ell/L}{\varepsilon_0}$, giving $E=\dfrac{Q}{2\pi\varepsilon_0 L r}$ (radial).

   (b) $V=\displaystyle\int_a^b E\,dr=\frac{Q}{2\pi\varepsilon_0 L}\int_a^b \frac{dr}{r}=\frac{Q}{2\pi\varepsilon_0 L}\ln\!\frac{b}{a}$, so $C=\dfrac{Q}{V}=\dfrac{2\pi\varepsilon_0 L}{\ln(b/a)}$.

   (c) Write $b=a+d$ with $d\ll a$: $\ln(b/a)=\ln\!\left(1+\dfrac{d}{a}\right)\approx\dfrac{d}{a}$, so $C\approx\dfrac{2\pi\varepsilon_0 L a}{d}=\dfrac{\varepsilon_0(2\pi a L)}{d}=\dfrac{\varepsilon_0 A}{d}$, with $A=2\pi a L$ the cylinder's surface area — the parallel-plate formula.