AP Physics C: Electricity and Magnetism Conductors, Capacitors, Dielectrics — Worked Answer Explanations

Unit 2 · 12 questions explained

1. Question 1 · Easy

   A parallel-plate capacitor has plate area $A$ and plate separation $d$. The capacitance is:

   • A
     $C=\frac{{ϵ}_{0}A}{d}$Correct
   • B
     $C=\frac{{ϵ}_{0}d}{A}$

     Why not B: The fraction is inverted. Larger plate area increases capacitance; larger separation decreases it.
   • C
     $C=\frac{A}{4\pi {ϵ}_{0}d}$

     Why not C: The $4\pi$ factor appears in Coulomb's law for a sphere, not in the parallel-plate formula. Plate geometry yields ${ϵ}_{0}A/d$.
   • D
     $C=\frac{{ϵ}_{0}A}{2d}$

     Why not D: There is no factor of 2 in the parallel-plate formula; this might arise from confusing it with the field of a single plate ($\sigma /2{ϵ}_0$).
   Explanation

   The electric field between the plates: $E=\sigma /{ϵ}_0=Q/({ϵ}_{0}A)$.

   Voltage: $V=Ed=Qd/({ϵ}_{0}A)$.

   Capacitance $C=Q/V$:
   $$C=\frac{Q}{Qd/({ϵ}_{0}A)}=\frac{{ϵ}_{0}A}{d}$$
   Larger $A$ stores more charge at the same voltage; larger $d$ stores less.

   Key takeaway

   Parallel-plate: $C = \epsilon_0 A/d$. Capacitance increases with area and decreases with separation.

2. Question 2 · Easy

   A conductor in electrostatic equilibrium has which of the following properties?

   • A
     The electric field just outside the surface is $\sigma /{ϵ}_0$, directed perpendicular to the surface.Correct
   • B
     The electric field inside the conductor is $\sigma /{ϵ}_0$.

     Why not B: In electrostatic equilibrium the electric field inside a conductor is exactly zero — free charges rearrange to cancel any internal field.
   • C
     The electric potential varies parabolically throughout the conductor's volume.

     Why not C: The potential is constant (uniform) throughout the volume and on the surface of a conductor in equilibrium, since $E=-\nabla V=0$ inside.
   • D
     Excess charge distributes uniformly throughout the conductor's volume.

     Why not D: All excess charge resides on the surface, not the interior. By Gauss's law, the electric field inside is zero only if there is no net charge enclosed, so interior charge density must be zero.
   Explanation

   For a conductor in electrostatic equilibrium:

   1. $E=0$ inside.
   2. The surface is an equipotential.
   3. Excess charge resides entirely on the surface.
   4. The field just outside the surface is perpendicular to the surface with magnitude $\sigma /{ϵ}_0$ (from a Gaussian pillbox with one face inside the conductor where $E=0$: $EA=\sigma A/{ϵ}_0$).

   Note the field just outside is $\sigma /{ϵ}_0$ (twice the single-sheet value $\sigma /(2{ϵ}_0)$) because the conductor enforces $E=0$ on one side.

   Key takeaway

   Conductor equilibrium: $\vec{E} = 0$ inside, all charge on surface, $E_{\text{outside}} = \sigma/\epsilon_0$ perpendicular to surface.

3. Question 3 · Easy

   A spherical capacitor consists of an inner shell of radius $a$ and a concentric outer shell of radius $b$ ($b>a$). The capacitance of this spherical capacitor is:

   • A
     $C=4\pi {ϵ}_{0}ab/(b-a)$Correct
   • B
     $C=4\pi {ϵ}_0(b-a)/(ab)$

     Why not B: The fraction is inverted. A larger gap $(b-a)$ means a smaller capacitance, not larger.
   • C
     $C=4\pi {ϵ}_{0}a$

     Why not C: This is the capacitance of an isolated sphere of radius $a$ (taking $b\to \infty$). With a finite outer shell, the capacitance is larger.
   • D
     $C={ϵ}_0\cdot 4\pi a^2/(b-a)$

     Why not D: This substitutes the inner sphere's surface area $4\pi a^2$ for the product $ab$ — the correct geometry integral yields $4\pi {ϵ}_{0}ab/(b-a)$, not $4\pi {ϵ}_0{a}^2/(b-a)$.
   Explanation

   Place charge $+Q$ on inner shell, $-Q$ on outer shell. By Gauss's law, field in the gap ($a<r<b$):
   $$E=\frac{Q}{4\pi {ϵ}_0{r}^2}$$
   Potential difference:
   $$V={\int }_a^{b}Edr=\frac{Q}{4\pi {ϵ}_0}{\int }_a^b\frac{dr}{r^2}=\frac{Q}{4\pi {ϵ}_0}\left(\frac{1}{a}-\frac{1}{b}\right)=\frac{Q(b-a)}{4\pi {ϵ}_{0}ab}$$
   Capacitance:
   $$C=\frac{Q}{V}=\frac{4\pi {ϵ}_{0}ab}{b-a}$$

   Key takeaway

   Spherical capacitor: $C = 4\pi\epsilon_0 ab/(b-a)$. Derived by integrating $1/r^2$ field in the gap.

4. Question 4 · Easy

   A coaxial cylindrical capacitor has inner radius $a$, outer radius $b$, and length $L$. The capacitance per unit length is:

   • A
     $C/L=\frac{2\pi {ϵ}_0}{\ln (b/a)}$Correct
   • B
     $C/L=\frac{\pi {ϵ}_0}{\ln (b/a)}$

     Why not B: Missing a factor of 2. Integrating the radial field $E=\lambda /(2\pi {ϵ}_{0}r)$ gives $V=(\lambda /2\pi {ϵ}_0)\ln (b/a)$, yielding $C/L=2\pi {ϵ}_0/\ln (b/a)$.
   • C
     $C/L=\frac{2\pi {ϵ}_0(b-a)}{ab\ln (b/a)}$

     Why not C: This incorrectly mixes the spherical capacitor formula with the cylindrical geometry. For a cylinder the integral yields a simple logarithm without the $(b-a)/(ab)$ factor.
   • D
     $C/L=\frac{2\pi {ϵ}_0}{b-a}$

     Why not D: This is the flat-plate approximation (treating $b-a$ as a gap width), valid only when $b-a\ll a$. The exact result involves $\ln (b/a)$.
   Explanation

   Field between the cylinders (from Gauss's law): $E=\lambda /(2\pi {ϵ}_{0}r)$.

   Potential difference:
   $$V={\int }_a^b\frac{\lambda }{2\pi {ϵ}_{0}r}dr=\frac{\lambda }{2\pi {ϵ}_0}\ln \frac{b}{a}$$
   Capacitance: $C=Q/V=\lambda L/V$
   $$\frac{C}{L}=\frac{\lambda }{V/L}=\frac{2\pi {ϵ}_0}{\ln (b/a)}$$
   As $b-a\ll a$: $\ln (b/a)\approx (b-a)/a$, so $C/L\approx 2\pi {ϵ}_{0}a/(b-a)$ — approaches the flat-plate limit.

   Key takeaway

   Coaxial capacitor: $C/L = 2\pi\epsilon_0/\ln(b/a)$. The $\ln$ appears from integrating $1/r$ field.

5. Question 5 · Medium

   A capacitor of capacitance $C$ is fully charged to voltage $V_0$ and then disconnected from the battery. A dielectric of dielectric constant $\kappa$ is then inserted, filling the gap completely. What happens to the stored energy?

   • A
     The energy decreases by a factor of $\kappa$.Correct
   • B
     The energy increases by a factor of $\kappa$.

     Why not B: After disconnecting, charge $Q=C{V}_0$ is fixed. Inserting the dielectric increases capacitance to $\kappa C$, so voltage drops to $V_0/\kappa$. Energy $=Q^2/(2\kappa C)$ — this decreases by $\kappa$, not increases.
   • C
     The energy remains the same, because charge is conserved.

     Why not C: Charge is conserved, but energy depends on $Q^2/(2C)$. Increasing $C$ to $\kappa C$ reduces the energy even as $Q$ stays constant.
   • D
     The energy increases by a factor of ${\kappa }^2$.

     Why not D: This would apply if voltage were held fixed (battery connected), not charge. Here charge is fixed, so energy $=Q^2/(2C)$ scales as $1/C$, not $C$.
   Explanation

   After disconnecting, charge is fixed: $Q=C{V}_0$.

   Initial energy: $U_i=\frac{Q^2}{2C}=\frac{1}{2}C{V}_0^2$.

   After inserting dielectric: $C^{\prime }=\kappa C$.
   $$U_f=\frac{Q^2}{2C^{\prime }}=\frac{Q^2}{2\kappa C}=\frac{U_i}{\kappa }$$

   The energy decreases by factor $\kappa$ — the dielectric is pulled in, doing work, which reduces electrical potential energy.

   Contrast: if battery stays connected ($V$ fixed), then $Q^{\prime }=\kappa C{V}_0$ and $U_f=\frac{1}{2}\kappa C{V}_0^2=\kappa U_i$ — energy increases.

   Key takeaway

   Disconnected (constant $Q$): inserting dielectric reduces energy by $\kappa$. Battery connected (constant $V$): energy increases by $\kappa$.

6. Question 6 · Medium

   Three capacitors $C_1=6\mu \text{F}$, $C_2=3\mu \text{F}$, and $C_3=2\mu \text{F}$ are connected in series across a $12\text{V}$ battery. The total energy stored in the combination is:

   • A
     $72\mu \text{J}$Correct
   • B
     $648\mu \text{J}$

     Why not B: This uses $C_{\text{total}}=11\mu \text{F}$ (the parallel sum), not the series combination. Series capacitance is smaller than any individual value.
   • C
     $144\mu \text{J}$

     Why not C: This uses the smallest individual capacitor $C_3=2\mu \text{F}$ alone in $U=\frac{1}{2}C{V}^2$ — but the series formula must be used for the combination.
   • D
     $36\mu \text{J}$

     Why not D: This uses $V=6\text{V}$ (half the battery voltage) rather than the correct $V=12\text{V}$, perhaps from misidentifying the voltage across the combination.
   Explanation

   Series equivalent:
   $$\frac{1}{C_s}=\frac{1}{6}+\frac{1}{3}+\frac{1}{2}=\frac{1+2+3}{6}=1\mu {\text{F}}^{-1}$$
   $$C_s=1\mu \text{F}$$

   Energy stored:
   $$U=\frac{1}{2}{C}_s{V}^2=\frac{1}{2}(1\times 10^{-6})(12{)}^2=\frac{1}{2}(10^{-6})(144)=72\mu \text{J}$$

   Note: individual capacitor voltages: $V_1=Q/C_1$, where $Q=C_s\cdot V=12\mu \text{C}$.

   Key takeaway

   Series capacitors: $1/C_s = \sum 1/C_i$; energy $= \frac{1}{2}C_s V^2$.

7. Question 7 · Medium

   A parallel-plate capacitor with plate area $A=0.10{\text{m}}^2$ and gap $d=2\text{mm}$ is filled with a dielectric of $\kappa =5$. A voltage $V=100\text{V}$ is applied. The surface charge density ${\sigma }_f$ on the capacitor plates (free charge) is:

   • A
     ${\sigma }_f=\kappa {ϵ}_{0}V/d\approx 2.21\mu {\text{C/m}}^2$Correct
   • B
     ${\sigma }_f={ϵ}_{0}V/d\approx 0.44\mu {\text{C/m}}^2$

     Why not B: This is the free charge density without the dielectric. The dielectric increases capacitance by $\kappa$, so $D=\kappa {ϵ}_{0}E=\kappa {ϵ}_{0}V/d$ is the free surface charge density.
   • C
     ${\sigma }_f={ϵ}_{0}V/(\kappa d)\approx 0.089\mu {\text{C/m}}^2$

     Why not C: This divides by $\kappa$ rather than multiplying. The dielectric increases the charge that can be stored at a given voltage, not decreases it.
   • D
     ${\sigma }_f={\kappa }^2{ϵ}_{0}V/d\approx 11.1\mu {\text{C/m}}^2$

     Why not D: Applying $\kappa$ twice is incorrect. The free charge density is $D={ϵ}_{0}E+P=\kappa {ϵ}_{0}E$, giving one factor of $\kappa$.
   Explanation

   With a dielectric, the capacitance is $C=\kappa {ϵ}_{0}A/d$.

   Free charge on plates: $Q=CV=\kappa {ϵ}_{0}AV/d$.

   Free surface charge density:
   $${\sigma }_f=Q/A=\kappa {ϵ}_{0}V/d=5\cdot (8.85\times 10^{-12})\cdot \frac{100}{2\times 10^{-3}}$$
   $$=5\cdot 8.85\times 10^{-12}\cdot 5\times 10^4=5\cdot 4.425\times 10^{-7}\approx 2.21\mu {\text{C/m}}^2$$

   The dielectric reduces the electric field inside ($E=V/d$ is unchanged), but the plates must carry more free charge to maintain the same $E$ in the presence of the polarization field.

   Key takeaway

   $\sigma_f = \kappa\epsilon_0 V/d$; dielectric multiplies free charge at fixed voltage by factor $\kappa$.

8. Question 8 · Medium

   A charged isolated conductor has a sharp point (small radius of curvature $r$) and a flat region (large radius of curvature). Which statement correctly describes the charge distribution?

   • A
     Surface charge density and electric field are highest at the sharp point.Correct
   • B
     Surface charge density is uniform across the entire conductor surface.

     Why not B: Uniform surface charge density occurs only on a sphere. For an irregular conductor, charge concentrates where the surface curves most sharply.
   • C
     The electric field is stronger over flat regions because there is more area for charge to spread.

     Why not C: More area means charge spreads out (lower density). Field is proportional to $\sigma$, so it is weaker over flat regions.
   • D
     Electric field inside the conductor is largest near the sharp point.

     Why not D: The field inside any conductor in electrostatic equilibrium is identically zero everywhere, regardless of the external geometry.
   Explanation

   The surface of a conductor is an equipotential. Near a sharp point, the potential drops steeply (field lines converge), requiring a high $\sigma$ to maintain the equipotential condition.

   More rigorously, for a conductor shaped like an ellipsoid, $\sigma \propto 1/(\text{local radius of curvature})$. A smaller radius → higher $\sigma$ → higher $E=\sigma /{ϵ}_0$ just outside.

   This is the lightning-rod effect — sharp points have the highest field, which can trigger corona discharge and lightning strikes.

   Key takeaway

   Charge concentrates at sharp points (small radius of curvature); field is highest there. $E = 0$ inside always.

9. Question 9 · Medium

   A capacitor $C=10\mu \text{F}$ is charged by a battery $V=100\text{V}$ and then disconnected. The plates are then pulled apart from separation $d$ to separation $2d$. Which of the following correctly describes the new voltage across the capacitor?

   • A
     The voltage doubles to $200\text{V}$.Correct
   • B
     The voltage stays at $100\text{V}$, because charge is conserved.

     Why not B: Charge is conserved, but $V=Q/C$. Doubling $d$ halves $C$ (since $C={ϵ}_{0}A/d$), so $V=Q/C^{\prime }=Q/(C/2)=2Q/C=200\text{V}$.
   • C
     The voltage halves to $50\text{V}$.

     Why not C: Voltage halves only if capacitance doubles (e.g., plates moved closer together). Moving plates apart reduces $C$, which increases $V$ at fixed $Q$.
   • D
     The voltage stays at $100\text{V}$, because the electric field $E=V/d$ is unchanged when $d$ doubles.

     Why not D: If $V$ stayed the same and $d$ doubled, then $E=V/d$ would halve. But charge conservation requires $Q=\text{const}$, so $E=\sigma /{ϵ}_0=Q/({ϵ}_{0}A)$ is also constant — meaning $V=Ed$ must double.
   Explanation

   After disconnecting, charge $Q=C{V}_0=(10\mu \text{F})(100\text{V})=1000\mu \text{C}$ is fixed.

   Doubling $d$ halves the capacitance: $C^{\prime }={ϵ}_{0}A/(2d)=C/2=5\mu \text{F}$.

   New voltage:
   $$V^{\prime }=\frac{Q}{C^{\prime }}=\frac{1000\mu \text{C}}{5\mu \text{F}}=200\text{V}$$

   Energy also doubles: $U^{\prime }=Q^2/(2C^{\prime })=2U_0$ — the extra energy comes from the work done in pulling the plates apart against the attractive electric force.

   Key takeaway

   Disconnected capacitor at fixed $Q$: increasing $d$ halves $C$, doubling $V$ and doubling stored energy.

10. Question 10 · Hard

    Two capacitors $C_1=4\mu \text{F}$ (initially charged to $V_1=12\text{V}$) and $C_2=8\mu \text{F}$ (initially uncharged) are connected in parallel by closing a switch. The final common voltage and the energy dissipated in the connecting wires are:

    • A
      $V_f=4\text{V}$; $\Delta U_{\text{dissipated}}=192\mu \text{J}$Correct
    • B
      $V_f=6\text{V}$; $\Delta U_{\text{dissipated}}=144\mu \text{J}$

      Why not B: $V_f=6\text{V}$ is an arithmetic mean of voltages, not the charge-conserving result. Conservation of charge gives $V_f=Q_i/(C_1+C_2)=48/12=4\text{V}$.
    • C
      $V_f=4\text{V}$; $\Delta U_{\text{dissipated}}=0$

      Why not C: Energy is not conserved when capacitors share charge; the difference $U_i-U_f$ is always dissipated as heat or radiation, regardless of wire resistance.
    • D
      $V_f=12\text{V}$; $\Delta U_{\text{dissipated}}=576\mu \text{J}$

      Why not D: If no charge flows to $C_2$, voltage stays at $12\text{V}$ — but charge does flow until equilibrium is reached. The final voltage must be below $12\text{V}$.
    Explanation

    Charge conservation (isolated system after switch closes):
    $$Q_i=C_1{V}_1=4\times 12=48\mu \text{C}$$
    $$V_f=\frac{Q_i}{C_1+C_2}=\frac{48}{4+8}=4\text{V}$$

    Energy before:
    $$U_i=\frac{1}{2}{C}_1{V}_1^2=\frac{1}{2}(4\times 10^{-6})(144)=288\mu \text{J}$$
    Energy after:
    $$U_f=\frac{1}{2}(C_1+C_2)V_f^2=\frac{1}{2}(12\times 10^{-6})(16)=96\mu \text{J}$$
    Dissipated:
    $$\Delta U=288-96=192\mu \text{J}$$

    This energy goes to resistive heating of the connecting wires (even an ideal wire dissipates this energy via a transient current).

    Key takeaway

    Charge sharing: conserve charge to find $V_f = Q_i/(C_1+C_2)$; energy loss $= U_i - U_f = 192\,\mu\text{J}$ is always dissipated.

11. Question 11 · Hard

    The energy density stored in the electric field of a parallel-plate capacitor filled with dielectric $\kappa$ and field $E$ is:

    • A
      $u=\frac{1}{2}\kappa {ϵ}_0{E}^2$Correct
    • B
      $u=\frac{1}{2}{ϵ}_0{E}^2$

      Why not B: This is the vacuum result. With a dielectric, the effective permittivity is $\kappa {ϵ}_0$, increasing the stored energy density by $\kappa$.
    • C
      $u=\frac{1}{2\kappa {ϵ}_0{E}^2}$

      Why not C: Dividing by $E^2$ and $\kappa {ϵ}_0$ gives units of ${\text{m}}^4/\text{J}$, not ${\text{J/m}}^3$. The energy density must increase with $E^2$.
    • D
      $u=\kappa {ϵ}_0{E}^2$

      Why not D: Missing the factor of $\frac{1}{2}$. Energy density always has a $\frac{1}{2}$ from integrating $qdV$ or $CdV$.
    Explanation

    From $U=\frac{1}{2}C{V}^2$ for a parallel-plate capacitor with dielectric:
    $$U=\frac{1}{2}(\kappa {ϵ}_{0}A/d)(Ed{)}^2=\frac{1}{2}\kappa {ϵ}_0{E}^2(Ad)$$

    Divide by volume $Ad$:
    $$u=\frac{1}{2}\kappa {ϵ}_0{E}^2$$

    Alternatively, in terms of displacement field $D=\kappa {ϵ}_{0}E$:
    $$u=\frac{1}{2}D\cdot E=\frac{1}{2}\kappa {ϵ}_0{E}^2$$

    Key takeaway

    Energy density with dielectric: $u = \frac{1}{2}\kappa\epsilon_0 E^2$. Dielectric increases stored energy at fixed $E$.

12. Question 12 · Hard

    A parallel-plate capacitor (area $A$, separation $d$, no dielectric) is connected to a constant voltage source $V$. A dielectric slab of thickness $t<d$ and dielectric constant $\kappa$ is inserted so that it fills the cross-section but only part of the gap. The capacitance of the resulting configuration is:

    • A
      $C=\frac{{ϵ}_{0}A}{d-t+t/\kappa }$Correct
    • B
      $C=\frac{\kappa {ϵ}_{0}A}{d}$

      Why not B: This would apply if the dielectric filled the entire gap ($t=d$). With partial filling, the remaining vacuum gap reduces the total capacitance.
    • C
      $C=\frac{{ϵ}_{0}A}{d+t(\kappa -1)}$

      Why not C: This adds $t(\kappa -1)$ to $d$ rather than subtracting. Adding the dielectric should increase capacitance (reduce the effective gap), not decrease it.
    • D
      $C=\frac{\kappa {ϵ}_{0}A}{d-t}$

      Why not D: This ignores the vacuum gap and treats only the dielectric portion. The two regions (vacuum and dielectric) are in series, so both gaps must contribute.
    Explanation

    Model as two capacitors in series: the vacuum gap (thickness $d-t$) and the dielectric slab (thickness $t$):
    $$C_1=\frac{{ϵ}_{0}A}{d-t}{C}_2=\frac{\kappa {ϵ}_{0}A}{t}$$
    $$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{d-t}{{ϵ}_{0}A}+\frac{t}{\kappa {ϵ}_{0}A}=\frac{1}{{ϵ}_{0}A}\left(d-t+\frac{t}{\kappa }\right)$$
    $$C=\frac{{ϵ}_{0}A}{d-t+t/\kappa }$$

    Check limits: $t\to 0$: $C\to {ϵ}_{0}A/d$ (air capacitor). $t\to d$: $C\to \kappa {ϵ}_{0}A/d$ (full dielectric). Both limiting cases confirm the result.

    Key takeaway

    Partial dielectric fill = two capacitors in series (vacuum + dielectric); $C = \epsilon_0 A/(d - t + t/\kappa)$.