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AP Physics C: E&M Unit 8 practice questions

Try 10 multiple-choice questions from Unit 8 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

    A conducting sphere of radius R carries an initial net charge +Q, while a second conducting sphere of radius 3R is initially neutral. The two spheres are connected by a long, thin conducting wire until equilibrium is reached. What is the final charge on the sphere of radius 3R?

    A final charge of +Q/4

    A final charge of +Q/3

    A final charge of +Q/2

    A final charge of +3Q/4

    Show answer

    Answer: (D)

    Equating the electric potentials (V = kQ/r) of the connected spheres reveals that charge distributes proportional to radius. The sphere with radius 3R holds three times the charge of the sphere with radius R, accounting for 3/4 of the total charge +Q.

  2. Question 2

    Two isolated conducting spheres of radii R and 4R are connected by a thin conducting wire and charged to a total potential V. Which expression correctly compares the surface charge density σ_small of the smaller sphere to the surface charge density σ_large of the larger sphere?

    The charge density σ_small is one-fourth the density σ_large.

    The charge density σ_small is equal to the density σ_large.

    The charge density σ_small is two times the density σ_large.

    The charge density σ_small is four times the density σ_large.

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    Answer: (D)

    Combining the potential formula (V = kQ/R) with the density formula (σ = Q/4πR²) shows that σ is proportional to V/R. Since V is constant for both connected spheres, the sphere with 1/4 the radius must have 4 times the surface charge density.

  3. Question 3

    A cylindrical wire of radius r delivers a constant current density J to charge an initially neutral object for time t. If the wire is replaced by one with radius 2r carrying the same current density J, what is the total charge delivered over the same time t?

    The total charge is two times the original charge.

    The total charge is four times the original charge.

    The total charge is eight times the original charge.

    The total charge is sixteen times the original charge.

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    Answer: (B)

    Current I is the product of current density J and cross-sectional area A (I = Jπr²). Doubling the radius increases the area by a factor of 4. Since Q = It, the total charge delivered also increases by a factor of 4.

  4. Question 4

    Two small spheres with charges q1q_1 and q2q_2 are separated by a distance rr and exert an electrostatic force of magnitude FF. If charge q1q_1 is doubled and the distance rr is tripled, what is the new magnitude of the force?

    29F\frac{2}{9}F

    23F\frac{2}{3}F

    49F\frac{4}{9}F

    6F6F

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    Answer: (A)

    Applying Coulomb's Law, the force is proportional to the product of charges and inversely proportional to the square of the distance. Doubling one charge multiplies the force by 2, and tripling the distance divides the force by 32=93^2=9, resulting in a factor of 2/92/9.

  5. Question 5

    Two identical conducting spheres carry charges +Q+Q and −3Q-3Q respectively and are separated by a distance dd. They attract each other with a force of magnitude F0F_0. The spheres are brought into contact, allowing charge to equilibrate, and then returned to the same distance dd. What is the new magnitude of the force between them?

    F0/3F_0/3

    F0/9F_0/9

    3F03F_0

    9F09F_0

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    Answer: (A)

    Calculating the initial force proportional to ∣(+Q)(−3Q)∣=3Q2|(+Q)(-3Q)| = 3Q^2. Upon contact, the net charge −2Q-2Q is shared equally, leaving each with −Q-Q. The final force is proportional to ∣(−Q)(−Q)∣=Q2|(-Q)(-Q)| = Q^2, which is one-third the initial magnitude.

  6. Question 6

    A proton and an alpha particle (charge +2e+2e, mass 4mp4m_p) are held at rest a distance rr apart. When released, the proton experiences an initial acceleration of magnitude apa_p. What is the magnitude of the initial acceleration of the alpha particle?

    ap/4a_p/4

    ap/2a_p/2

    2ap2a_p

    4ap4a_p

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    Answer: (A)

    Using Newton's Third Law, the electrostatic force FF on both particles is equal in magnitude. Since a=F/ma = F/m and the alpha particle has 4 times the mass of the proton, its acceleration is F/(4mp)=ap/4F/(4m_p) = a_p/4.

  7. Question 7

    Two point charges, +q+q and +9q+9q, are fixed at positions x=0x=0 and x=Lx=L, respectively. At what position xx along the line connecting them is the net electric force on a third positive test charge equal to zero?

    L/4L/4

    L/3L/3

    L/2L/2

    3L/43L/4

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    Answer: (A)

    Setting the forces equal, kqQx2=k9qQ(L−x)2k\frac{qQ}{x^2} = k\frac{9qQ}{(L-x)^2}, leads to 1x=3L−x\frac{1}{x} = \frac{3}{L-x}. Solving for xx yields L−x=3xL-x = 3x, so 4x=L4x = L and x=L/4x = L/4.

  8. Question 8

    Three identical point charges +q+q are placed at the vertices of an equilateral triangle with side length aa. The magnitude of the net electric force on any one charge is FF. If the side length of the triangle is increased to 2a2a, what is the new magnitude of the net force on one charge?

    F/4F/4

    F/2F/2

    2F2F

    4F4F

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    Answer: (A)

    Recognizing that the geometry (angles) remains constant, the net force depends only on the distance scaling. Since Coulomb's force scales as 1/r21/r^2, doubling the distance reduces the force by a factor of 22=42^2=4.

  9. Question 9

    Two point charges exert a repulsive force of magnitude FF on each other. If the charge of one particle is tripled and the distance between them is reduced to one-third of its original value, what is the new force magnitude?

    27F27F

    9F9F

    3F3F

    F/3F/3

    Show answer

    Answer: (A)

    Applying Coulomb's Law, the new force is proportional to (3)(1)(1/3)2\frac{(3)(1)}{(1/3)^2}. The denominator becomes 1/91/9, so the total factor is 3×9=273 \times 9 = 27.

  10. Question 10

    An electron travels in a circular orbit of radius rr around a proton. If the radius of the orbit is increased to 4r4r, by what factor does the orbital speed of the electron change?

    It decreases by a factor of 2

    It decreases by a factor of 4

    It increases by a factor of 2

    It increases by a factor of 4

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    Answer: (A)

    Equating electric force to centripetal force (ke2/r2=mv2/rk e^2/r^2 = mv^2/r) shows that v2∝1/rv^2 \propto 1/r, so v∝1/rv \propto 1/\sqrt{r}. Increasing rr by 4 decreases vv by 4=2\sqrt{4}=2.

What's on the AP Physics C: E&M Unit 8 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 1062 multiple-choice questions and 8 FRQs written for Unit 8, Electric Charges & Fields: Gauss's Law.

Practice AP Physics C: E&M Unit 8 FRQs

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

Review Unit 8 before you start

Other AP Physics C: E&M unit tests