AP Physics 2 Electric Force, Field, and Potential — Worked Answer Explanations

Unit 2 · 16% of the AP exam · 12 questions explained

1. Question 1 · Easy

   Two point charges $+Q$ and $-Q$ are separated by distance $r$. According to Coulomb's law, the magnitude of the electric force between them is:

   • A
     $F=\frac{k{Q}^2}{r}$

     Why not A: The distance should be squared in the denominator; this omits the $r^2$ dependence.
   • B
     $F=\frac{k{Q}^2}{r^2}$Correct
   • C
     $F=\frac{kQ}{r^2}$

     Why not C: Coulomb's law depends on the product of both charges $q_1{q}_2=Q\cdot Q=Q^2$, not just one charge.
   • D
     $F=\frac{2k{Q}^2}{r^2}$

     Why not D: There is no factor of 2; Coulomb's law is $F=k{q}_1{q}_2/r^2$.
   Explanation

   Coulomb's law for the force between two point charges:
   $$F=k\frac{|q_1||q_2|}{r^2}$$
   where $k=9\times 10^9\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2$. For charges $+Q$ and $-Q$: $|q_1||q_2|=Q\cdot Q=Q^2$. The force is attractive (opposite signs) and has magnitude $k{Q}^2/r^2$. The force obeys an inverse-square law — doubling the distance quarters the force.

   Key takeaway

   Coulomb's law: $F = k|q_1||q_2|/r^2$; force depends on the product of both charge magnitudes and the inverse square of separation.

2. Question 2 · Easy

   Equipotential surfaces around a positive point charge are:

   • A
     Parallel planes perpendicular to the radial electric field lines.

     Why not A: Parallel planes are equipotentials for a uniform field, not for a point charge.
   • B
     Concentric spheres centered on the charge.Correct
   • C
     Lines radiating outward from the charge.

     Why not C: Radial lines describe the electric field direction, not equipotential surfaces.
   • D
     Ellipsoidal surfaces elongated along the direction of the field.

     Why not D: Ellipsoidal equipotentials arise near multiple charges or non-spherical charge distributions, not a single point charge.
   Explanation

   The electric potential due to a point charge $+Q$ is $V=kQ/r$. All points at the same distance $r$ from the charge have the same potential, so the equipotential surfaces are concentric spheres. The electric field $E=-\nabla V$ is always perpendicular to equipotential surfaces — for a point charge, the field points radially outward, which is indeed perpendicular to spherical shells.

   Key takeaway

   Equipotentials for a point charge are concentric spheres; electric field lines are always perpendicular to equipotential surfaces.

3. Question 3 · Easy

   A positive test charge $+q$ is placed in an electric field $E$. The electric force on the test charge is:

   • A
     $F=E/q$

     Why not A: This inverts the relationship. The field is defined as force per unit charge: $E=F/q$.
   • B
     $F=qE$, directed opposite to $E$ since the charge is positive.

     Why not B: A positive charge experiences force in the same direction as $E$, not opposite.
   • C
     $F=qE$, directed in the same direction as $E$.Correct
   • D
     $F=q^{2}E$

     Why not D: Force is linear in charge, not quadratic: $F=qE$.
   Explanation

   The electric field $E$ is defined by $E=F/q_0$, where $q_0$ is a positive test charge. Rearranging: $F=qE$.

   • For a positive charge ($q>0$): $F$ is in the same direction as $E$.
   • For a negative charge ($q<0$): $F$ is in the opposite direction from $E$.

   Electric field lines point in the direction of force on a positive test charge.

   Key takeaway

   Electric force $\vec{F} = q\vec{E}$; positive charges accelerate along field lines, negative charges accelerate opposite to field lines.

4. Question 4 · Easy

   A parallel-plate capacitor has plate area $A$ and plate separation $d$. If the separation is doubled while the charge on the plates remains constant, how does the voltage across the capacitor change?

   • A
     The voltage is halved.

     Why not A: Doubling $d$ decreases capacitance $C={\epsilon }_{0}A/d$ by half, but since $V=Q/C$, voltage doubles, not halves.
   • B
     The voltage is unchanged.

     Why not B: Voltage depends on both charge and capacitance; changing $d$ changes $C$, which changes $V$ at fixed $Q$.
   • C
     The voltage doubles.Correct
   • D
     The voltage quadruples.

     Why not D: Capacitance varies linearly with $d$, so voltage also changes linearly, not as $d^2$.
   Explanation

   Capacitance of a parallel-plate capacitor: $C={\epsilon }_{0}A/d$.

   Doubling $d$: $C^{\prime }={\epsilon }_{0}A/(2d)=C/2$

   Voltage with fixed charge $Q$:
   $$V=\frac{Q}{C}\Rightarrow V^{\prime }=\frac{Q}{C/2}=2V$$

   The voltage doubles. Physically, the same charge now spreads over a larger gap, weakening the field slightly — actually the surface charge density $\sigma$ stays the same, but the field $E=\sigma /{\epsilon }_0$ is unchanged, so $V=Ed$ doubles because $d$ doubled.

   Key takeaway

   At fixed $Q$, $V = Q/C$; doubling plate separation halves $C$, which doubles $V$.

5. Question 5 · Medium

   A charge $q=+2\mu \mathrm{C}$ is moved from point A to point B in an electric field. The potential at A is $V_A=100\mathrm{V}$ and at B is $V_B=40\mathrm{V}$. What is the work done by the electric field on the charge?

   • A
     $W=+1.2\times 10^{-4}\mathrm{J}$Correct
   • B
     $W=-1.2\times 10^{-4}\mathrm{J}$

     Why not B: This has the wrong sign. Moving a positive charge to lower potential means the field does positive work.
   • C
     $W=+2.8\times 10^{-4}\mathrm{J}$

     Why not C: This uses $V_A+V_B$ instead of $V_A-V_B$ for the potential difference.
   • D
     $W=+8.0\times 10^{-5}\mathrm{J}$

     Why not D: This uses $V_B=40\mathrm{V}$ alone rather than the potential difference $\Delta V=V_A-V_B=60\mathrm{V}$.
   Explanation

   Work done by the electric field on charge $q$ moving from A to B:
   $$W_{\text{field}}=q(V_A-V_B)$$

   The field does positive work when a positive charge moves from high to low potential (analogous to gravity doing positive work when an object falls).

   $$W=(2\times 10^{-6}\mathrm{C})(100-40\mathrm{V})=(2\times 10^{-6})(60)=1.2\times 10^{-4}\mathrm{J}$$

   Note: Work done by an external agent would be $W_{\text{ext}}=q(V_B-V_A)=-1.2\times 10^{-4}\mathrm{J}$.

   Key takeaway

   Work by electric field: $W = q(V_A - V_B) = -q\Delta V$; positive for a positive charge moving to lower potential.

6. Question 6 · Medium

   Three point charges are placed along a line: $+Q$ at $x=0$, $-Q$ at $x=d$, and $+Q$ at $x=2d$. At the point $x=d$ (the location of the $-Q$ charge), the net electric field due to the other two charges points in which direction?

   • A
     In the $+x$ direction

     Why not A: The left $+Q$ pushes the field rightward at $x=d$, but the right $+Q$ also pushes the field leftward there — they cancel, not add.
   • B
     In the $-x$ direction

     Why not B: By symmetry, both $+Q$ charges are equidistant from the midpoint and produce equal and opposite fields there; they cancel.
   • C
     The net electric field at $x=d$ from the two $+Q$ charges is zero.Correct
   • D
     Perpendicular to the line (in the $y$-direction)

     Why not D: All charges are collinear; components perpendicular to the line are zero for point charges on the line.
   Explanation

   The $-Q$ charge at $x=d$ is equidistant ($d$) from the $+Q$ at $x=0$ and the $+Q$ at $x=2d$.

   Field from left $+Q$ (at $x=0$): points in the $+x$ direction at $x=d$, magnitude $E_1=kQ/d^2$.

   Field from right $+Q$ (at $x=2d$): points in the $-x$ direction at $x=d$, magnitude $E_2=kQ/d^2$.

   Since $E_1=E_2$ and they point in opposite directions, the net field is zero. This is the expected result by symmetry — the midpoint of two equal positive charges is a field zero.

   Key takeaway

   Equal charges equidistant on opposite sides produce equal and opposite fields at the midpoint; the net field is zero there.

7. Question 7 · Medium

   A dielectric material with dielectric constant $\kappa =3$ is inserted between the plates of a parallel-plate capacitor connected to a battery (constant voltage). How does the energy stored in the capacitor change?

   • A
     The stored energy decreases by a factor of 3.

     Why not A: At constant voltage, adding a dielectric increases capacitance and therefore increases stored energy.
   • B
     The stored energy increases by a factor of 3.Correct
   • C
     The stored energy is unchanged because the voltage is constant.

     Why not C: Although $V$ is constant, capacitance $C$ increases with $\kappa$; since $U=\frac{1}{2}C{V}^2$, the energy changes.
   • D
     The stored energy increases by a factor of 9.

     Why not D: The factor of $\kappa$ appears linearly in $U=\frac{1}{2}C{V}^2$, not as ${\kappa }^2$.
   Explanation

   Inserting a dielectric increases capacitance: $C^{\prime }=\kappa C_0=3C_0$.

   At constant voltage $V$ (battery connected):
   $$U=\frac{1}{2}C{V}^2\Rightarrow U^{\prime }=\frac{1}{2}(3C_0)V^2=3U_0$$

   The stored energy increases by a factor of 3. The battery does extra work to push additional charge onto the plates (the dielectric allows more charge per volt). This contrasts with the constant charge case (battery disconnected), where $U=Q^2/(2C)$ and inserting a dielectric decreases the energy.

   Key takeaway

   At constant $V$: $U = \frac{1}{2}CV^2$ increases by factor $\kappa$ when dielectric is inserted. At constant $Q$: $U = Q^2/2C$ decreases by factor $\kappa$.

8. Question 8 · Medium

   An electron (charge $-e$) is released from rest at a point where the electric potential is $V=-50\mathrm{V}$ and moves to a point where $V=+50\mathrm{V}$. By how much does the electron's kinetic energy increase?

   • A
     $\Delta KE=-1.6\times 10^{-17}\mathrm{J}$ (electron loses energy)

     Why not A: Negative KE change means the electron slows down; but moving to higher potential with negative charge means the field does positive work on it, so it gains KE.
   • B
     $\Delta KE=+8.0\times 10^{-18}\mathrm{J}$

     Why not B: This uses $|q|\cdot |\Delta V|/2$, incorrectly halving the potential difference; $\Delta V=50-(-50)=100\mathrm{V}$, not $50\mathrm{V}$.
   • C
     $\Delta KE=+1.6\times 10^{-17}\mathrm{J}$Correct
   • D
     $\Delta KE=0$ (electron starts from rest so gains no energy)

     Why not D: Starting from rest means initial KE = 0, but the electric force does work on the electron as it moves, giving it kinetic energy.
   Explanation

   Work done by the electric field on the electron:
   $$W=q(V_i-V_f)=(-e)((-50)-(+50))=(-1.6\times 10^{-19})(-100)=+1.6\times 10^{-17}\mathrm{J}$$

   By the work-energy theorem, $\Delta KE=W=+1.6\times 10^{-17}\mathrm{J}$.

   Physically: electrons naturally move from low-potential to high-potential regions (opposite to positive charges), so the electric force does positive work on the electron as it moves from $-50\mathrm{V}$ to $+50\mathrm{V}$, increasing its kinetic energy.

   Key takeaway

   For an electron (negative charge), moving to higher potential means the field does positive work; $\Delta KE = q(V_i - V_f)$.

9. Question 9 · Hard

   Two capacitors, $C_1=4\mu \mathrm{F}$ and $C_2=12\mu \mathrm{F}$, are connected in series across a $24\mathrm{V}$ battery. What is the charge on $C_1$?

   • A
     $Q_1=96\mu \mathrm{C}$

     Why not A: This uses $C_1\times V=4\times 24$, treating $C_1$ as if it were directly across $24\mathrm{V}$ (parallel, not series).
   • B
     $Q_1=288\mu \mathrm{C}$

     Why not B: This uses $C_2\times V=12\times 24$, again treating $C_2$ as across the full $24\mathrm{V}$.
   • C
     $Q_1=72\mu \mathrm{C}$Correct
   • D
     $Q_1=18\mu \mathrm{C}$

     Why not D: This adds the capacitances ($4+12=16$) instead of combining in series; series combination gives a smaller equivalent capacitance.
   Explanation

   For series capacitors, the equivalent capacitance is:
   $$\frac{1}{C_{\text{eq}}}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{4}+\frac{1}{12}=\frac{3}{12}+\frac{1}{12}=\frac{4}{12}\Rightarrow C_{\text{eq}}=3\mu \mathrm{F}$$

   In series, both capacitors store the same charge equal to $Q=C_{\text{eq}}\times V$:
   $$Q=3\mu \mathrm{F}\times 24\mathrm{V}=72\mu \mathrm{C}$$

   So $Q_1=Q_2=72\mu \mathrm{C}$. The voltages differ: $V_1=Q/C_1=18\mathrm{V}$, $V_2=Q/C_2=6\mathrm{V}$, and $V_1+V_2=24\mathrm{V}$ ✓

   Key takeaway

   Series capacitors share the same charge $Q = C_{\text{eq}}V$; find $C_{\text{eq}}$ from $1/C_{\text{eq}} = 1/C_1 + 1/C_2$ first.

10. Question 10 · Hard

    Point P is located midway between two point charges: $+3Q$ at $x=-d$ and $-Q$ at $x=+d$. What is the direction of the net electric field at point P ($x=0$)?

    • A
      The field points in the $+x$ direction.Correct
    • B
      The field points in the $-x$ direction.

      Why not B: The $+3Q$ charge contributes a field pointing right (away from it) that is larger than the $-Q$ field pointing right (toward it). Net field is rightward.
    • C
      The net field is zero because positive and negative charges cancel.

      Why not C: Fields from charges of opposite signs do not automatically cancel; their magnitudes depend on the charge values, and here $3Q\ne Q$.
    • D
      The field points in the $+y$ direction.

      Why not D: All charges lie on the $x$-axis; by symmetry, there is no $y$-component of the electric field at any point on the $x$-axis.
    Explanation

    At point P ($x=0$), both charges are a distance $d$ away.

    Field from $+3Q$ (at $x=-d$): points away from $+3Q$, i.e., in the $+x$ direction:
    $$E_1=\frac{k(3Q)}{d^2}$$

    Field from $-Q$ (at $x=+d$): points toward $-Q$, i.e., in the $+x$ direction:
    $$E_2=\frac{kQ}{d^2}$$

    Both fields point in the $+x$ direction at P! Net field:
    $$E_{\text{net}}=E_1+E_2=\frac{4kQ}{d^2}(\text{in }+x\text{ direction})$$

    Key takeaway

    Always find the direction each charge's field contributes at the point of interest; don't assume opposite-sign charges create opposing fields without checking directions.

11. Question 11 · Hard

    A parallel-plate capacitor with plate separation $d$ is connected to a battery maintaining voltage $V_0$. A conducting slab of thickness $d/2$ is inserted between the plates (not touching either plate). How does the capacitance change?

    • A
      The capacitance is unchanged.

      Why not A: Inserting a conductor effectively reduces the gap; capacitance increases.
    • B
      The capacitance doubles.Correct
    • C
      The capacitance is halved.

      Why not C: This confuses the effect: reducing the effective gap increases, not decreases, capacitance.
    • D
      The capacitance increases by a factor of 4.

      Why not D: The factor is the ratio $d/d_{\text{eff}}$; here $d_{\text{eff}}=d/2$, giving a factor of 2, not 4.
    Explanation

    A conducting slab inside a capacitor has zero electric field inside it. Effectively, the conducting slab of thickness $d/2$ removes $d/2$ from the effective plate separation.

    Effective gap = $d-d/2=d/2$

    $$C^{\prime }=\frac{{\epsilon }_{0}A}{d/2}=\frac{2{\epsilon }_{0}A}{d}=2C_0$$

    The capacitance doubles. This is equivalent to two capacitors in series, each with gap $d/4$, giving combined capacitance:
    $$\frac{1}{C_{\text{series}}}=\frac{d/4}{{\epsilon }_{0}A}+\frac{d/4}{{\epsilon }_{0}A}\Rightarrow C=2C_0$$

    Key takeaway

    A conducting slab of thickness $t$ reduces effective gap to $d - t$; capacitance becomes $C = \varepsilon_0 A/(d-t)$.

12. Question 12 · Hard

    An electron is fired horizontally into a uniform electric field directed vertically downward with magnitude $E=500N/C$. The electron has initial speed $v_0=2.0\times 10^{6}m/s$. What is the magnitude of the electron's acceleration? ($m_e=9.11\times 10^{-31}\mathrm{kg}$, $e=1.6\times 10^{-19}\mathrm{C}$)

    • A
      $a\approx 8.8\times 10^{13}m/{\mathrm{s}}^2$Correct
    • B
      $a\approx 4.4\times 10^{13}m/{\mathrm{s}}^2$

      Why not B: This is half the correct value, possibly from using $E/2$ or dividing force by $2m_e$ incorrectly.
    • C
      $a\approx 8.8\times 10^{10}m/{\mathrm{s}}^2$

      Why not C: This is off by $10^3$, likely from a unit error (using charge in $\mu \mathrm{C}$ instead of $\mathrm{C}$).
    • D
      $a\approx 5.5\times 10^{14}m/{\mathrm{s}}^2$

      Why not D: This results from forgetting to divide force by mass (i.e., just computing $F=qE$ without $a=F/m$).
    Explanation

    The electric force on the electron (charge $-e$, field $E$ downward):
    $$F=eE=(1.6\times 10^{-19}\mathrm{C})(500N/C)=8.0\times 10^{-17}\mathrm{N}$$

    Note: the field points down, so force on negative charge points up (magnitude only asked).

    Acceleration by Newton's second law:
    $$a=\frac{F}{m_e}=\frac{8.0\times 10^{-17}}{9.11\times 10^{-31}}\approx 8.8\times 10^{13}m/{\mathrm{s}}^2$$

    The horizontal motion (at $v_0$) is unaffected by this vertical acceleration — the electron follows a parabolic path, analogous to projectile motion but with electric force instead of gravity. Note the enormous acceleration compared to gravity ($g\approx 10m/{\mathrm{s}}^2$) — electromagnetic forces dominate for electrons.

    Key takeaway

    Use $F = qE$ to find force, then $a = F/m$; for electrons the tiny mass gives enormous accelerations even in modest fields.