AP Physics C: E&M 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.
Question 1
A parallel-plate capacitor with plate separation d is charged to a magnitude Q and then isolated from the voltage source. If the plate separation is increased to 2d without discharging the plates, what is the new potential difference between the plates?
It decreases to one-half the initial potential difference.
It remains equal to the initial potential difference.
It increases to two times the initial potential difference.
It increases to four times the initial potential difference.
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Answer: (C)
For an isolated capacitor, charge Q is conserved. Since capacitance C is inversely proportional to separation d (C = εA/d), doubling d halves C. Using V = Q/C, halving the capacitance while keeping Q constant doubles the potential difference.
Question 2
A positively charged solid conductor in electrostatic equilibrium has an irregular shape with a sharp end of radius of curvature R and a blunt end of radius of curvature 3R. Which of the following correctly compares the electric potential V and surface charge density σ at the sharp end (subscript S) to the blunt end (subscript B)?
The potentials are equal (V_S = V_B), but the charge density is greater at the sharp end (σ_S > σ_B)
The potentials are equal (V_S = V_B), but the charge density is greater at the blunt end (σ_S < σ_B)
The potential is greater at the sharp end (V_S > V_B), and the charge density is equal (σ_S = σ_B)
The potential is greater at the blunt end (V_S < V_B), and the charge density is equal (σ_S = σ_B)
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Answer: (A)
Connecting the equipotential nature of conductors with the geometric dependence of charge density.
Question 3
A point charge +Q is placed at the center of a neutral conducting spherical shell with inner radius R and outer radius 2R. Which of the following correctly compares the magnitude of the surface charge density on the inner surface (σ_in) to the magnitude on the outer surface (σ_out)?
The magnitude of σ_in is four times the magnitude of σ_out due to the smaller surface area
The magnitude of σ_in is two times the magnitude of σ_out due to the smaller surface radius
The magnitude of σ_in is equal to the magnitude of σ_out due to charge conservation
The magnitude of σ_in is one-fourth the magnitude of σ_out due to the smaller surface area
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Answer: (A)
Analyzing how fixed induced charge magnitudes distribute over different spherical areas.
Question 4
Two conducting spheres with radii R and 3R are separated by a large distance and connected by a thin conducting wire. If the system carries a net static charge Q, which statement correctly compares the electric field magnitude at the surface of the small sphere (E_S) to the large sphere (E_L)?
E_S is three times greater than E_L because charge density is higher on the smaller sphere
E_S is nine times greater than E_L because field strength scales with the inverse square of radius
E_S is equal to E_L because the wire ensures both spheres remain at the same electric potential
E_S is one-third of E_L because the larger sphere accumulates more total charge than the smaller one
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Answer: (A)
Combining the equipotential condition with the electric field formula for spheres.
Question 5
An uncharged conducting slab is placed between two parallel infinite non-conducting sheets with charge densities +σ and -σ. Which of the following correctly compares the magnitude of the induced charge density on the slab's surface (σ_ind) to the source charge density σ?
σ_ind is equal to σ to create a net zero electric field inside the conductor
σ_ind is greater than σ to create a net zero electric field inside the conductor
σ_ind is less than σ because the slab is not grounded to a charge reservoir
σ_ind is zero because the slab is initially uncharged and electrically isolated
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Answer: (A)
Applying the condition of electrostatic equilibrium inside a conductor to determine induced charge.
Question 6
A parallel-plate capacitor with plate separation is charged to a total charge and then disconnected from the battery. If the plate separation is increased to while the charge remains constant, how does the stored potential energy change?
The stored energy doubles to
The stored energy halves to
The stored energy quadruples to
The stored energy remains at
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Answer: (A)
Doubling the separation halves the capacitance (). Since and is constant, halving doubles the stored energy.
Question 7
A parallel-plate capacitor with plate separation remains connected to a battery with constant voltage . If the separation is increased to , what is the new magnitude of the electric field between the plates?
The electric field decreases to
The electric field increases to
The electric field decreases to
The electric field remains at
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Answer: (A)
For a parallel-plate capacitor, . Since voltage is held constant by the battery, doubling the separation reduces the electric field magnitude by half.
Question 8
A parallel-plate capacitor connected to a battery stores energy with air between its plates. If a dielectric with constant is inserted while the battery remains connected, what is the new stored energy?
The energy increases to
The energy decreases to
The energy increases to
The energy remains at
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Answer: (A)
Inserting a dielectric increases capacitance to . Since and is constant, the stored energy triples.
Question 9
A parallel-plate capacitor is charged to potential difference and then disconnected from the source. If a slab with dielectric constant is inserted filling the space between plates, what is the new potential difference?
The potential difference decreases to
The potential difference increases to
The potential difference decreases to
The potential difference remains at
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Answer: (A)
Inserting the dielectric increases capacitance to . Since charge is constant (isolated), decreases by a factor of 4.
Question 10
A parallel-plate capacitor has circular plates of radius and separation . If the radius is doubled to and the separation is doubled to , by what factor does the capacitance change?
The capacitance increases by a factor of 2
The capacitance increases by a factor of 4
The capacitance decreases by a factor of 2
The capacitance remains the same value
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Answer: (A)
Area depends on , so doubling quadruples . Doubling halves . The net effect is times the original capacitance.
What's on the AP Physics C: E&M 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 628 multiple-choice questions and 59 FRQs written for Unit 10, Conductors and Capacitors.
Practice AP Physics C: E&M Unit 10 FRQs
Answer the unit's free-response questions one at a time and get a score on every part.
- FRQ 1 – Mathematical Routines (18 questions)
- FRQ 2 – Translation Between Representations (6 questions)
- FRQ 3 – Experimental Design (5 questions)
- FRQ 4 – Qualitative/Quantitative Translation (30 questions)
Review Unit 10 before you start
- 10.1 Electrostatics with Conductors
- 10.2 Redistribution of Charge Between Conductors
- 10.3 Capacitors
- 10.4 Dielectrics