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AP Physics 2 Unit 10 practice questions

Try 10 multiple-choice questions from Unit 10 here. Pick an answer to see why it's right or wrong, then take the timed test above for a full AP-style set.

  1. Question 1

    Two point charges, +Q+Q and −Q-Q, are fixed on the x-axis at x=−dx = -d and x=+dx = +d, respectively. Which statement correctly describes the electric potential at the origin (x=0x = 0)?

    The potential is zero because the scalar contributions from the positive and negative charges cancel out

    The potential is zero because the electric field vectors from the two charges point in opposite directions

    The potential is nonzero because the scalar contributions from the two charges add up constructively

    The potential is nonzero because the electric field vectors from the two charges point in the same direction

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    Answer: (A)

    Applying the principle of superposition requires summing the scalar potentials from each charge (V=kQ/d+k(−Q)/d=0V = kQ/d + k(-Q)/d = 0). The correct answer recognizes that electric potential is a scalar quantity where equal positive and negative contributions at the same distance sum to zero.

  2. Question 2

    A conducting sphere of radius RR with an initial charge +Q+Q is connected by a thin wire to an uncharged conducting sphere of radius 3R3R. Once equilibrium is reached, what is the final charge on the larger sphere?

    The final charge is 3Q/43Q/4 because charge flows until the electric potentials of both spheres are equal

    The final charge is Q/2Q/2 because charge flows until the charge magnitudes on both spheres are equal

    The final charge is Q/4Q/4 because charge flows until the surface electric fields of both spheres are equal

    The final charge is Q/3Q/3 because charge flows until the electric potential energies of both spheres are equal

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    Answer: (A)

    Applying the condition for electrostatic equilibrium between conductors requires equating their potentials (V=kq/rV = kq/r). Since the larger sphere has 3 times the radius, it must hold 3 times the charge (qlarge=3qsmallq_{large} = 3q_{small}) to have the same potential, resulting in 3/43/4 of the total charge QQ.

  3. Question 3

    Four point charges, each with magnitude +Q+Q, are fixed at the corners of a square with side length dd. Which expression correctly represents the total electric potential at the center of the square?

    The potential is 42kQ/d4\sqrt{2}kQ/d because the scalar potentials from all four charges add together

    The potential is 00 because the electric field vectors from all four charges cancel out

    The potential is 4kQ/d4kQ/d because the distance to the center is treated as dd for each charge

    The potential is 22kQ/d2\sqrt{2}kQ/d because only diagonal pairs contribute to the total potential

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    Answer: (A)

    Calculating the potential requires summing the scalar contribution from each charge (V=∑kq/rV = \sum kq/r). The distance from each corner to the center is d/2d/\sqrt{2}, so the total potential is 4×(kQ/(d/2))=42kQ/d4 \times (kQ / (d/\sqrt{2})) = 4\sqrt{2}kQ/d.

  4. Question 4

    A positive point charge +Q+Q is fixed at the origin. A student measures the electric potential VV and electric field magnitude EE at a distance rr from the charge. If the student moves to a distance 2r2r, how do the new values compare?

    The new potential is V/2V/2 and the new field is E/4E/4 because V∝1/rV \propto 1/r and E∝1/r2E \propto 1/r^2

    The new potential is V/4V/4 and the new field is E/2E/2 because V∝1/r2V \propto 1/r^2 and E∝1/rE \propto 1/r

    The new potential is V/2V/2 and the new field is E/2E/2 because both VV and EE are proportional to 1/r1/r

    The new potential is V/4V/4 and the new field is E/4E/4 because both VV and EE are proportional to 1/r21/r^2

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    Answer: (A)

    Applying the theoretical definitions of potential and field for a point charge reveals their different dependencies on distance. Potential scales inversely with distance (1/r1/r), while field magnitude scales inversely with the square of distance (1/r21/r^2).

  5. Question 5

    A point charge +Q+Q is placed at x=0x=0 and a point charge −2Q-2Q is placed at x=Dx=D. At which position on the x-axis between the charges is the electric potential zero?

    At x=D/3x = D/3, because the potential from the positive charge cancels the potential from the negative charge

    At x=D/2x = D/2, because the zero potential point is always equidistant from two opposite point charges

    At x=2D/3x = 2D/3, because the potential is zero closer to the charge with the larger magnitude

    At x=D/4x = D/4, because the potential varies with the square of the distance from each charge

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    Answer: (A)

    Solving for the zero potential point requires setting the sum of potentials to zero: kQ/x+k(−2Q)/(D−x)=0kQ/x + k(-2Q)/(D-x) = 0. This simplifies to 1/x=2/(D−x)1/x = 2/(D-x), which yields x=D/3x = D/3, the point where the scalar contributions cancel.

  6. Question 6

    An alpha particle (charge +2e+2e) moves from a location where the electric potential is 100 V100 \text{ V} to a location where the potential is 120 V120 \text{ V} under the influence of only the electric force. Which statement correctly describes the change in the particle's kinetic energy?

    The kinetic energy decreases by 40 eV40 \text{ eV} because the particle moves to a higher potential

    The kinetic energy increases by 40 eV40 \text{ eV} because the electric field does positive work

    The kinetic energy decreases by 20 eV20 \text{ eV} because the potential difference is 20 V20 \text{ V}

    The kinetic energy increases by 20 eV20 \text{ eV} because potential energy is converted to kinetic

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    Answer: (A)

    Applying conservation of energy, the change in kinetic energy equals the negative change in potential energy (ΔK=−qΔV\Delta K = -q\Delta V). Moving a positive charge to a higher potential increases potential energy (+2e×20 V=+40 eV+2e \times 20\text{ V} = +40\text{ eV}), causing a corresponding decrease in kinetic energy.

  7. Question 7

    A solid conducting sphere of radius RR carries a net positive charge +Q+Q and is in electrostatic equilibrium. Which statement correctly describes the electric potential at a distance r=R/2r = R/2 from the center?

    It equals kQ/RkQ/R, because the potential is constant throughout a conductor in equilibrium

    It equals 2kQ/R2kQ/R, because the potential increases as the distance to the center decreases

    It equals 00, because the electric field is zero everywhere inside a conductor in equilibrium

    It equals kQ/(2R)kQ/(2R), because the potential decreases as the distance to the center decreases

    Show answer

    Answer: (A)

    Applying the properties of conductors requires recognizing that the entire conductor is an equipotential volume. The potential anywhere inside (r<Rr < R) is the same as the potential at the surface (kQ/RkQ/R).

  8. Question 8

    Three point charges are arranged at the vertices of an equilateral triangle of side ss. Two vertices have charge +q+q and the third has charge −q-q. What is the electric potential at the center of the triangle?

    The potential is 3kq/s\sqrt{3}kq/s because the scalar potentials from the three charges sum algebraically

    The potential is 00 because the contributions from the positive and negative charges cancel out

    The potential is 33kq/s3\sqrt{3}kq/s because the magnitudes of the potentials from all charges add up

    The potential is kq/skq/s because the effective charge at the center is the net charge +q+q

    Show answer

    Answer: (A)

    Calculating the potential requires summing scalars: V=k(q)/r+k(q)/r+k(−q)/r=kq/rV = k(q)/r + k(q)/r + k(-q)/r = kq/r. The distance rr from center to vertex is s/3s/\sqrt{3}, so V=kq/(s/3)=3kq/sV = kq / (s/\sqrt{3}) = \sqrt{3}kq/s.

  9. Question 9

    A battery is connected to a circuit. Which statement best describes the electric potential difference created by the battery in terms of energy and charge?

    It is the work done per unit charge by chemical processes to separate positive and negative charges

    It is the force exerted per unit charge by the chemical processes on the electrons in the terminals

    It is the rate of energy transfer per unit charge as current flows through the internal resistance

    It is the total potential energy stored in the battery divided by the mass of the chemical reactants

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    Answer: (A)

    Applying the definition of electric potential difference (ΔV=ΔUE/q\Delta V = \Delta U_E / q) in the context of a battery (LO 10.5.A.3.i) identifies it as the energy provided per unit charge to create charge separation.

  10. Question 10

    Two identical spheres are initially charged: Sphere A has charge +4μC+4 \mu C and Sphere B has charge −2μC-2 \mu C. They are brought into contact and then separated. Which statement correctly compares their final electric potentials?

    Both spheres have the same positive electric potential because charge redistributes until potentials equalize

    Both spheres have zero electric potential because the positive and negative charges neutralize each other

    Sphere A has a positive potential and Sphere B has a negative potential because they retain their original signs

    Sphere A has a higher potential than Sphere B because it started with a greater magnitude of charge

    Show answer

    Answer: (A)

    Applying conservation of charge and the equilibrium condition: total charge is +2μC+2 \mu C, which splits equally (+1μC+1 \mu C each) between identical spheres. Since both have positive charge and equal radius, they share the same positive potential.

What's on the AP Physics 2 Unit 10 practice test

The standard version takes about 88 minutes and follows the timing and directions of the AP exam, sized to one unit:

  • 20 Multiple Choice Questions (40 minutes)
  • 1 FRQ – FRQ 1 – Mathematical Routines (24 minutes)
  • 1 FRQ – FRQ 2 – Translation Between Representations (24 minutes)

Questions are drawn from 1424 multiple-choice questions and 35 FRQs written for Unit 10, Electric Force, Field, and Potential.

Practice AP Physics 2 Unit 10 FRQs

Answer the unit's free-response questions one at a time and get a score on every part.

Review Unit 10 before you start

Other AP Physics 2 unit tests