Download AP Electricity Quiz Multiple Choice Solutions

AP Electricity Quiz Multiple Choice Solutions:
1. C—Electric field is a vector. Look at the field at the center due to each charge. The field due to the lefthand charge points away from the positive charge; i.e., to the right; the field due to the right-hand charge
points to the left. Because the charges are equal and are the same distance from the center point, the fields
due to each charge have equal magnitudes. So the electric field vectors cancel! E = 0.
2. D—Electric potential is a scalar. Look at the potential at the center due to each charge: Each charge is
distance a/2 from the center point, so the potential due to each iskQ/(a/2), which works out to 2kQ/a. The
potentials due to both charges are positive, so add these potentials to get 4kQ/a.
3. B—If the potential difference between plates is, say, 100 V, then we could say that one plate is at +100 V
and the other is at zero V. So, the potential must change at points in between the plates. The electric field
is uniform and equal to V/d (d is the distance between plates). Thus, the potential increases linearly
between the plates, and A must have twice the potential as B.
4. A—The electric field by definition is uniform between parallel plates. This means the field must be the
same everywhere inside the plates.
5. A - Total electric flux from a charge q is q/ε0. Since the charge q is situated at the centre of the cube (fig.),
all the six faces of the cube have equal share of the flux so that the flux through one face is q/6ε0.
6. B - Joules=Coulombs x Volts - eV = charge x voltage  energy
7. E - Definition and properties of conductor
8. E - When the battery is disconnected, Q remains constant. Since C decreases when d increases and Q =
CV, V will increase.
9. B - Loop rule = voltage rule. Voltage is electric potential, which is energy per charge: conservation of
energy
10. B – both equal 2 Ω
11. B - R = ρL/A. If L ÷ 2, R ÷ 2 and is r ÷ 2 then A ÷ 4 and R × 4 making the net effect R ÷ 2 × 4
12. D - In parallel V1 = V2. Q1 = C1V1 and Q2 = C2V2 so Q1/Q2 = C1/C2 = 1.5
13. A - 1 kW-h = 1000 W × 60 min = 60,000 W-min = I2Rt = I2(20 Ω)(30 min)
14. D - E = kQ/r2
15. C - V = kQ/r
16. A - Since both charges are positive, the electric field vectors point in opposite directions at points
between the two. At point A, the magnitudes of the electric field vectors are equal and therefore
cancel out, making E = 0 at point A
17. A - V = ΣkQ/r and since both charges are positive, the largest potential is at the closest point to the
two charges (it is more mathematically complex than that, but this reasoning works for the
choices given)
18. C - C = ε0A/d; if A × 2, C × 2 and if d × 2, C ÷ 2 so the net effect is C is unchanged
19. E - The net charge on the two spheres is +Q so when they touch and separate, the charge on each sphere
(divided equally) is ½ Q. F ∝ Q1Q2 so before contact F ∝ (2Q)(Q) = 2Q2 and after contact F ∝ (½
Q)(½ Q) = ¼ Q2 or 1/8 of the original force
20. B - R = ρL/A. Greatest resistance is the longest, narrowest resistor.
21. A – junction rule = current rule. Current is moving charge: charge in = charge out.
22. D - In series, the equivalent capacitance is calculated using reciprocals, like resistors in parallel. This
results in an equivalent capacitance smaller than the smallest capacitor
23. A - Current is greatest where resistance is least. The resistances are, in order, 1 Ω, 2 Ω, 4 Ω, 2 Ω and 6
Ω.
24. E – See above
25. E- Least power is for the greatest resistance (P = E2/R)
26. B - R = ρL/A. If L ÷ 2, R ÷ 2 and is r ÷ 2 then A ÷ 4 and R × 4 making the net effect R ÷ 2 × 4
1
1
27. E - In series 𝐶 = ∑ 𝐶
𝑇
*Bonus:
1. C - Amperes = I (current); Volts = V (potential difference); Seconds = t (time): IVt = energy
2. C - Assuming C remains constant and UC = ½ CV2, for UC to double V must increase by √2