AP Physics C: Electricity and Magnetism Electrostatics — Worked Answer Explanations

Unit 1 · 12 questions explained

1. Question 1 · Easy

   Two point charges $q_1=+3\mu \text{C}$ and $q_2=-3\mu \text{C}$ are separated by $r=0.30\text{m}$. Using $k=9\times 10^9{\text{N⋅m}}^2/{\text{C}}^2$, the magnitude of the electric force between them is:

   • A
     $0.90\text{N}$Correct
   • B
     $0.30\text{N}$

     Why not B: This results from using $r$ in the denominator rather than $r^2$, neglecting the inverse-square dependence of Coulomb's law.
   • C
     $0.10\text{N}$

     Why not C: This would require $k{q}^2/r^3$ — an extra power of $r$ that has no physical basis in Coulomb's law.
   • D
     $8.1\times 10^{-3}\text{N}$

     Why not D: This uses $(3\times 10^{-6}{)}^2/(0.30{)}^2=10^{-10}$ but divides by $k$ instead of multiplying, inverting the formula.
   Explanation

   $$F=k\frac{|q_1||q_2|}{r^2}=(9\times 10^9)\frac{(3\times 10^{-6}{)}^2}{(0.30{)}^2}$$
   $$=(9\times 10^9)\frac{9\times 10^{-12}}{0.09}=(9\times 10^9)(10^{-10})=0.90\text{N}$$
   The force is attractive (opposite signs).

   Key takeaway

   Coulomb's law: $F = k|q_1||q_2|/r^2$. Always square $r$ in the denominator.

2. Question 2 · Easy

   A uniformly charged thin ring of radius $R$ and total charge $Q$ lies in the $xy$-plane centered at the origin. The electric field on the axis of the ring at distance $z$ from the center points:

   • A
     Radially outward from the ring's center, perpendicular to the axis.

     Why not A: By symmetry, radial components from opposite ring elements cancel exactly; only the axial component survives.
   • B
     Along the $z$-axis (axial direction) for $z\ne 0$.Correct
   • C
     In the plane of the ring, toward the nearest point of the ring.

     Why not C: The symmetry of the uniform ring cancels all in-plane components; the field is purely axial on axis.
   • D
     Zero everywhere on the axis.

     Why not D: The field is zero only at the center ($z=0$) by symmetry, not for $z\ne 0$.
   Explanation

   Each element $dq$ of the ring contributes a field $dE$ pointing from $dq$ toward the field point. The component perpendicular to the $z$-axis from element $dq$ is canceled by the diametrically opposite element. The axial components add constructively:
   $$E_z=\frac{kQz}{(z^2+R^2{)}^{3/2}}$$
   At $z=0$, $E_z=0$ (center of ring). The direction is $+\hat{z}$ for $Q>0,z>0$.

   Key takeaway

   Symmetry cancels transverse components on the ring's axis; only the axial component survives and equals $kQz/(z^2+R^2)^{3/2}$.

3. Question 3 · Easy

   An infinitely long line charge has linear charge density $\lambda >0$. Using Gauss's law with a coaxial cylindrical Gaussian surface of radius $r$ and length $L$, the electric field at distance $r$ from the line is:

   • A
     $E=\frac{\lambda }{2\pi {ϵ}_0{r}^2}$

     Why not A: This has an extra factor of $1/r$. The curved surface area of the Gaussian cylinder is $2\pi rL$, giving $E\propto 1/r$, not $1/r^2$.
   • B
     $E=\frac{\lambda }{2\pi {ϵ}_{0}r}$Correct
   • C
     $E=\frac{\lambda L}{2\pi {ϵ}_{0}r}$

     Why not C: The length $L$ cancels when $Q_{enc}=\lambda L$ is divided by the surface area $2\pi rL$; the result is independent of $L$.
   • D
     $E=\frac{\lambda }{4\pi {ϵ}_{0}r}$

     Why not D: The factor $4\pi {ϵ}_0$ appears in Coulomb's law for a point charge, not for a line charge. The cylinder geometry yields a $2\pi {ϵ}_0$ denominator.
   Explanation

   Apply Gauss's law: $\oint E\cdot dA=Q_{\text{enc}}/{ϵ}_0$.

   For a cylindrical Gaussian surface of radius $r$, length $L$:

   • End caps contribute zero flux (field is radial, perpendicular to caps).
   • Curved surface: $E(2\pi rL)=\lambda L/{ϵ}_0$

   $$\boxed{E=\frac{\lambda }{2\pi {ϵ}_{0}r}}$$

   The field falls off as $1/r$ (not $1/r^2$) because the source is one-dimensional.

   Key takeaway

   Infinite line charge: $E = \lambda/(2\pi\epsilon_0 r)$, derived via cylindrical Gauss surface; field falls as $1/r$.

4. Question 4 · Easy

   A solid insulating sphere of radius $R$ carries uniform volume charge density $\rho$. For a point inside the sphere at radius $r<R$, Gauss's law gives the electric field magnitude:

   • A
     $E=\frac{\rho R^3}{3{ϵ}_0{r}^2}$

     Why not A: This is the field outside the sphere ($r>R$). Inside, only the charge within radius $r$ contributes.
   • B
     $E=\frac{\rho r}{3{ϵ}_0}$Correct
   • C
     $E=\frac{\rho R}{3{ϵ}_0}$

     Why not C: This is the field at the surface ($r=R$), not at a general interior point $r<R$.
   • D
     $E=0$

     Why not D: The field is zero inside a conducting shell, not inside a uniformly charged insulating sphere. The enclosed charge grows as $r^3$.
   Explanation

   Inside the sphere, apply Gauss's law with a spherical surface of radius $r<R$:
   $$Q_{\text{enc}}=\rho \cdot \frac{4}{3}\pi r^3$$
   $$E(4\pi r^2)=\frac{\rho \cdot \frac{4}{3}\pi r^3}{{ϵ}_0}$$
   $$E=\frac{\rho r}{3{ϵ}_0}$$

   The field increases linearly with $r$ inside, then falls as $1/r^2$ outside. At $r=R$: $E=\rho R/(3{ϵ}_0)$ — the two expressions match at the boundary.

   Key takeaway

   Inside a uniform sphere: $E = \rho r/(3\epsilon_0)$ — linear in $r$. Outside: $E = \rho R^3/(3\epsilon_0 r^2)$ — inverse square.

5. Question 5 · Medium

   A finite line segment of length $2L$ lies along the $x$-axis, centered at the origin, with uniform linear charge density $\lambda$. What is the electric potential $V$ at a point $P$ on the perpendicular bisector (the $y$-axis) at distance $d$ from the origin?

   • A
     $V=\frac{\lambda }{2\pi {ϵ}_0}\ln \left(\frac{L+\sqrt{L^2+d^2}}{d}\right)$Correct
   • B
     $V=\frac{\lambda L}{2\pi {ϵ}_{0}d}$

     Why not B: This treats the line as a point charge $Q=\lambda (2L)$ at distance $d$. The integral over the distributed source yields a logarithm, not a simple $1/d$ form.
   • C
     $V=\frac{\lambda }{4\pi {ϵ}_0}\ln \left(\frac{L+d}{L-d}\right)$

     Why not C: This is a ratio of collinear distances, not the result of the correct integration. The denominator should involve $d$ (the perpendicular distance), not $L-d$.
   • D
     $V=\frac{\lambda }{2\pi {ϵ}_0}\ln \left(\frac{d}{L+\sqrt{L^2+d^2}}\right)$

     Why not D: The argument of the logarithm is inverted relative to the correct answer, giving a negative (unphysical) potential for $\lambda >0$.
   Explanation

   Set up the integral with $x$ ranging from $-L$ to $L$. The distance from element $dx$ at position $x$ to the point $P$ at $(0,d)$ is $\sqrt{x^2+d^2}$:
   $$V={\int }_{-L}^L\frac{k\lambda dx}{\sqrt{x^2+d^2}}=2k\lambda {\int }_0^L\frac{dx}{\sqrt{x^2+d^2}}$$
   Using $\int dx/\sqrt{x^2+d^2}=\ln (x+\sqrt{x^2+d^2})+C$:
   $$V=2k\lambda [\ln (x+\sqrt{x^2+d^2}){]}_0^L=2k\lambda \ln \left(\frac{L+\sqrt{L^2+d^2}}{d}\right)$$
   Substituting $k=1/(4\pi {ϵ}_0)$:
   $$V=\frac{\lambda }{2\pi {ϵ}_0}\ln \left(\frac{L+\sqrt{L^2+d^2}}{d}\right)$$

   Key takeaway

   Potential from a line segment is found by integration; result is logarithmic. Potential (scalar) is easier to integrate than field (vector).

6. Question 6 · Medium

   An infinite plane of charge has surface charge density $\sigma$. Using Gauss's law with a pillbox Gaussian surface that straddles the sheet, the electric field magnitude on each side of the sheet is:

   • A
     $E=\frac{\sigma }{{ϵ}_0}$

     Why not A: This omits the factor of 2 from the two faces of the pillbox that both contribute flux. Each face has area $A$, giving total flux $2EA=\sigma A/{ϵ}_0$.
   • B
     $E=\frac{\sigma }{2{ϵ}_0}$Correct
   • C
     $E=\frac{2\sigma }{{ϵ}_0}$

     Why not C: This overcounts by a factor of 4. The correct result from two equal faces contributing to the flux integral yields $\sigma /(2{ϵ}_0)$.
   • D
     $E=\frac{\sigma }{4\pi {ϵ}_0}$

     Why not D: The factor $4\pi {ϵ}_0$ is the denominator in Coulomb's law for a point charge. Plane geometry gives $2{ϵ}_0$ in the denominator, not $4\pi {ϵ}_0$.
   Explanation

   Place a cylindrical "pillbox" of cross-section area $A$ symmetrically through the sheet:

   • Side wall: zero flux (field parallel to sheet)
   • Two flat faces (each area $A$): $\Phi =2EA$
     $$2EA=\frac{Q_{\text{enc}}}{{ϵ}_0}=\frac{\sigma A}{{ϵ}_0}$$
     $$E=\frac{\sigma }{2{ϵ}_0}$$
     The field points away from the sheet (for $\sigma >0$) on both sides. Note: Between two parallel conducting plates with surface charge $\sigma$, the fields from both plates add on the inside, giving $E=\sigma /{ϵ}_0$.
   Key takeaway

   Infinite sheet: $E = \sigma/(2\epsilon_0)$ from Gauss's law with pillbox. Conductor plates sandwich yields $\sigma/\epsilon_0$ between them.

7. Question 7 · Medium

   The electric potential in a region is $V(x,y)=3x^{2}y-y^3$ (in volts, with $x,y$ in meters). The $x$-component of the electric field at the point $(1,1)$ is:

   • A
     $E_x=-6\text{V/m}$Correct
   • B
     $E_x=+6\text{V/m}$

     Why not B: The correct sign is negative: $E=-\nabla V$. Forgetting the negative sign gives the wrong direction.
   • C
     $E_x=-3\text{V/m}$

     Why not C: This may result from taking the partial derivative with respect to $y$ rather than $x$, or from missing the chain-rule factor of 2 in differentiating $3x^{2}y$.
   • D
     $E_x=+3\text{V/m}$

     Why not D: Wrong both in magnitude (missing factor of 2$y$) and sign.
   Explanation

   The electric field is the negative gradient of potential:
   $$E=-\nabla V=-\frac{\partial V}{\partial x}\hat{x}-\frac{\partial V}{\partial y}\hat{y}$$
   $$\frac{\partial V}{\partial x}=6xy\Rightarrow E_x=-6xy$$
   At $(x,y)=(1,1)$:
   $$E_x=-6(1)(1)=-6\text{V/m}$$
   Similarly, $\partial V/\partial y=3x^2-3y^2$, so $E_y=-(3(1{)}^2-3(1{)}^2)=0$ at this point.

   Key takeaway

   $\vec{E} = -\nabla V$; always include the negative sign. Partial derivatives apply for multivariable potentials.

8. Question 8 · Medium

   A spherical conducting shell of inner radius $a$ and outer radius $b$ carries a net charge $+Q$. A point charge $+q$ is placed at the center. By Gauss's law, the surface charge density on the outer surface of the shell is:

   • A
     ${\sigma }_{\text{outer}}=\frac{Q}{4\pi b^2}$

     Why not A: This ignores the central charge $+q$. Charge induction requires $-q$ on the inner surface, so the outer surface must carry $Q+q$ to keep the shell neutral overall.
   • B
     ${\sigma }_{\text{outer}}=\frac{Q+q}{4\pi b^2}$Correct
   • C
     ${\sigma }_{\text{outer}}=\frac{Q-q}{4\pi b^2}$

     Why not C: The inner surface carries $-q$ (induced), so the outer surface carries $Q-(-q)=Q+q$, not $Q-q$.
   • D
     ${\sigma }_{\text{outer}}=\frac{q}{4\pi b^2}$

     Why not D: This only accounts for the induced redistribution and ignores the net charge $Q$ already on the shell.
   Explanation

   By Gauss's law, the electric field inside the conductor (between $a$ and $b$) must be zero. A Gaussian sphere in this region encloses the central charge $+q$ plus the inner surface charge, so the inner surface must carry $-q$.

   Charge conservation: the shell has net charge $+Q$, and $-q$ is on the inner surface, so:
   $$Q_{\text{inner}}+Q_{\text{outer}}=Q\Rightarrow Q_{\text{outer}}=Q-(-q)=Q+q$$
   $${\sigma }_{\text{outer}}=\frac{Q+q}{4\pi b^2}$$

   Key takeaway

   Central charge induces $-q$ on inner surface; outer surface carries $Q+q$ so the shell's net charge is preserved.

9. Question 9 · Medium

   The electric potential at a distance $r$ from the center of a uniformly charged insulating sphere of radius $R$ and total charge $Q$, for $r<R$, is:

   • A
     $V(r)=\frac{kQ}{r}$

     Why not A: This is the potential outside the sphere ($r>R$). The potential inside differs because only the enclosed charge contributes directly, and $V$ must be continuous at $r=R$.
   • B
     $V(r)=\frac{kQ}{2R}\left(3-\frac{r^2}{R^2}\right)$Correct
   • C
     $V(r)=\frac{kQ}{R}$

     Why not C: This is the constant potential at and outside the surface for a conductor. For an insulator with uniform $\rho$, the interior potential varies with $r$.
   • D
     $V(r)=\frac{kQ{r}^2}{R^3}$

     Why not D: The potential increases toward the center for $Q>0$, not decreases. Also this doesn't match the boundary condition $V(R)=kQ/R$.
   Explanation

   Inside ($r<R$), integrate $E$ from $r$ to $\infty$:
   $$V(r)=V(R)+{\int }_r^R{E}_{\text{inside}}d{r}^{\prime }=\frac{kQ}{R}+{\int }_r^R\frac{kQ{r}^{\prime }}{R^3}d{r}^{\prime }$$
   $$=\frac{kQ}{R}+\frac{kQ}{R^3}\cdot \frac{R^2-r^2}{2}=\frac{kQ}{2R}\left(3-\frac{r^2}{R^2}\right)$$
   At $r=0$: $V=3kQ/(2R)$ (maximum). At $r=R$: $V=kQ/R$ (matches exterior). $V$ is continuous but its derivative $dV/dr$ is discontinuous at $r=R$.

   Key takeaway

   Inside uniform sphere: $V(r) = \frac{kQ}{2R}(3 - r^2/R^2)$; parabolic in $r$. Maximum at center, matches $kQ/R$ at surface.

10. Question 10 · Hard

    A thin disk of radius $R$ has uniform surface charge density $\sigma$. Using integration over rings, the electric field at a point on the axis of the disk at distance $z$ from the center is:

    • A
      $E=\frac{\sigma }{2{ϵ}_0}\left(1-\frac{z}{\sqrt{z^2+R^2}}\right)$Correct
    • B
      $E=\frac{\sigma }{2{ϵ}_0}\left(1+\frac{z}{\sqrt{z^2+R^2}}\right)$

      Why not B: A sign error in the integration: after evaluating $[-z/\sqrt{z^2+r^2}{]}_0^R$, the result has a minus sign on the $z/\sqrt{z^2+R^2}$ term.
    • C
      $E=\frac{\sigma R}{2{ϵ}_0\sqrt{z^2+R^2}}$

      Why not C: This omits the lower limit of the integration and doesn't produce the correct infinite-sheet limit as $R\to \infty$.
    • D
      $E=\frac{\sigma z}{2{ϵ}_0(z^2+R^2)}$

      Why not D: This has incorrect dimensional form and doesn't reduce to $\sigma /(2{ϵ}_0)$ as $R\to \infty$.
    Explanation

    Divide the disk into rings of radius $r$ and width $dr$. Ring charge: $dq=\sigma (2\pi rdr)$. Field from ring on axis at distance $z$:
    $$dE=\frac{kzdq}{(z^2+r^2{)}^{3/2}}=\frac{\sigma zrdr}{2{ϵ}_0(z^2+r^2{)}^{3/2}}$$
    Integrate:
    $$E=\frac{\sigma z}{2{ϵ}_0}{\int }_0^R\frac{rdr}{(z^2+r^2{)}^{3/2}}=\frac{\sigma z}{2{ϵ}_0}{\left[-\frac{1}{\sqrt{z^2+r^2}}\right]}_0^R$$
    $$=\frac{\sigma z}{2{ϵ}_0}\left(\frac{1}{z}-\frac{1}{\sqrt{z^2+R^2}}\right)=\frac{\sigma }{2{ϵ}_0}\left(1-\frac{z}{\sqrt{z^2+R^2}}\right)$$
    As $R\to \infty$: $E\to \sigma /(2{ϵ}_0)$ — the infinite-sheet result.

    Key takeaway

    Disk field: build from rings; integrate $r\,dr/(z^2+r^2)^{3/2}$. Result reduces to infinite-sheet $\sigma/(2\epsilon_0)$ as $R\to\infty$.

11. Question 11 · Hard

    Two large parallel conducting plates separated by distance $d$ each carry surface charge density $+\sigma$ and $-\sigma$ respectively. A small conducting sphere of radius $a\ll d$ with net charge $q$ is held midway between the plates. Ignoring image charges, the electric force on the sphere is:

    • A
      $F=q\sigma /{ϵ}_0$, directed from the positive to the negative plate.Correct
    • B
      $F=q\sigma /(2{ϵ}_0)$, directed from the positive to the negative plate.

      Why not B: This uses the field of a single plate $\sigma /(2{ϵ}_0)$. Between oppositely charged plates the fields from both plates add, giving $E=\sigma /{ϵ}_0$.
    • C
      $F=0$, because the sphere is midway between the plates.

      Why not C: The midplane location makes the field uniform, not zero. The field $E=\sigma /{ϵ}_0$ is uniform everywhere between the plates.
    • D
      $F=2q\sigma /{ϵ}_0$, directed from the positive to the negative plate.

      Why not D: There is no extra factor of 2 beyond the plate-pair sum. The field between the plates is $\sigma /{ϵ}_0$, not $2\sigma /{ϵ}_0$.
    Explanation

    Between two infinite parallel plates with $+\sigma$ and $-\sigma$:

    • Field from the $+\sigma$ plate: $\sigma /(2{ϵ}_0)$, directed from $+$ to $-$.
    • Field from the $-\sigma$ plate: $\sigma /(2{ϵ}_0)$, also directed from $+$ to $-$ (field lines end on $-$).
    • Total: $E=\sigma /{ϵ}_0$, uniform between the plates.

    Force on charge $q$:
    $$F=qE=\frac{q\sigma }{{ϵ}_0}$$
    directed from the $+$ plate toward the $-$ plate (for $q>0$).

    Key takeaway

    Parallel plates with $\pm\sigma$: fields from both plates add between them, giving $E = \sigma/\epsilon_0$. Force $= qE$.

12. Question 12 · Hard

    A solid insulating cylinder of radius $R$ and infinite length has non-uniform volume charge density $\rho (r)={\rho }_{0}r/R$, where $r$ is the distance from the axis. Using Gauss's law, the electric field magnitude for $r<R$ is:

    • A
      $E=\frac{{\rho }_0{r}^2}{3{ϵ}_{0}R}$Correct
    • B
      $E=\frac{{\rho }_{0}r}{3{ϵ}_0}$

      Why not B: This is the result for uniform density ${\rho }_0$, not ${\rho }_{0}r/R$. The non-uniform density requires integration and introduces an extra factor of $r/R$.
    • C
      $E=\frac{{\rho }_0{r}^2}{2{ϵ}_{0}R}$

      Why not C: Integrating $\rho (r^{\prime })={\rho }_0{r}^{\prime }/R$ over the cylinder volume yields $Q_{\text{enc}}={\rho }_0\pi r^{3}L/(R\cdot ...)$; the coefficient comes out to $1/3$, not $1/2$.
    • D
      $E=\frac{{\rho }_{0}R}{3{ϵ}_0}$

      Why not D: This is the field at the surface $r=R$, not a general interior expression. The interior field must depend on $r$.
    Explanation

    Gaussian surface: coaxial cylinder, radius $r<R$, length $L$.
    $$Q_{\text{enc}}={\int }_0^r\rho (r^{\prime })\cdot 2\pi r^{\prime }Ld{r}^{\prime }={\int }_0^r\frac{{\rho }_0{r}^{\prime }}{R}\cdot 2\pi r^{\prime }Ld{r}^{\prime }=\frac{2\pi {\rho }_{0}L}{R}{\int }_0^r{r}^{\prime 2}d{r}^{\prime }=\frac{2\pi {\rho }_{0}L{r}^3}{3R}$$

    Gauss's law:
    $$E(2\pi rL)=\frac{Q_{\text{enc}}}{{ϵ}_0}=\frac{2\pi {\rho }_{0}L{r}^3}{3{ϵ}_{0}R}$$
    $$E=\frac{{\rho }_0{r}^2}{3{ϵ}_{0}R}$$

    At $r=R$: $E={\rho }_{0}R/(3{ϵ}_0)$, which should match the exterior solution evaluated at $r=R$ (verifiable with the total linear charge density $\lambda ={\rho }_0\pi R^2\cdot (2/3)/...$).

    Key takeaway

    Non-uniform $\rho(r)$: integrate over shells to find $Q_{\text{enc}}$, then apply Gauss's law. Here $E \propto r^2$ inside due to $\rho \propto r$.