--- title: "AP Physics C: E&M Unit 10 Practice Test" description: "A timed AP Physics C: E&M Unit 10 (Conductors and Capacitors) practice test in the AP format, drawn from 628 multiple-choice questions and 59 FRQs written for the unit. The standard version takes about 88 minutes." canonical: "https://fiveable.me/ap-physics-c-e-m/unit-test/10" type: "unit-test" subject: "AP Physics C: E&M" unit: "Unit 10" --- # AP Physics C: E&M Unit 10 Practice Test ## Overview A timed AP Physics C: E&M Unit 10 (Conductors and Capacitors) practice test in the AP format, drawn from 628 multiple-choice questions and 59 FRQs written for the unit. The standard version takes about 88 minutes. ## Test Format - 20 Multiple Choice Questions (40 minutes) - 1 FRQ – FRQ 1 – Mathematical Routines (24 minutes) - 1 FRQ – FRQ 2 – Translation Between Representations (24 minutes) ## Unit FRQs by Type - [FRQ 1 – Mathematical Routines](/ap-physics-c-e-m/frqs/mathematical-routines?unit=10): 18 questions - [FRQ 2 – Translation Between Representations](/ap-physics-c-e-m/frqs/translation-between-representations?unit=10): 6 questions - [FRQ 3 – Experimental Design](/ap-physics-c-e-m/frqs/experimental-design?unit=10): 5 questions - [FRQ 4 – Qualitative/Quantitative Translation](/ap-physics-c-e-m/frqs/qualitative-quantitative-translation?unit=10): 30 questions ## Practice Questions ### 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? - (A) It decreases to one-half the initial potential difference. - (B) It remains equal to the initial potential difference. - (C) It increases to two times the initial potential difference. - (D) It increases to four times the initial potential difference. **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)? - (A) The potentials are equal (V_S = V_B), but the charge density is greater at the sharp end (σ_S > σ_B) - (B) The potentials are equal (V_S = V_B), but the charge density is greater at the blunt end (σ_S < σ_B) - (C) The potential is greater at the sharp end (V_S > V_B), and the charge density is equal (σ_S = σ_B) - (D) The potential is greater at the blunt end (V_S < V_B), and the charge density is equal (σ_S = σ_B) **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)? - (A) The magnitude of σ_in is four times the magnitude of σ_out due to the smaller surface area - (B) The magnitude of σ_in is two times the magnitude of σ_out due to the smaller surface radius - (C) The magnitude of σ_in is equal to the magnitude of σ_out due to charge conservation - (D) The magnitude of σ_in is one-fourth the magnitude of σ_out due to the smaller surface area **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)? - (A) E_S is three times greater than E_L because charge density is higher on the smaller sphere - (B) E_S is nine times greater than E_L because field strength scales with the inverse square of radius - (C) E_S is equal to E_L because the wire ensures both spheres remain at the same electric potential - (D) E_S is one-third of E_L because the larger sphere accumulates more total charge than the smaller one **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? - (A) σ_ind is equal to σ to create a net zero electric field inside the conductor - (B) σ_ind is greater than σ to create a net zero electric field inside the conductor - (C) σ_ind is less than σ because the slab is not grounded to a charge reservoir - (D) σ_ind is zero because the slab is initially uncharged and electrically isolated **Answer: (A)** Applying the condition of electrostatic equilibrium inside a conductor to determine induced charge. ### Question 6 A parallel-plate capacitor with plate separation $$d$$ is charged to a total charge $$Q$$ and then disconnected from the battery. If the plate separation is increased to $$2d$$ while the charge remains constant, how does the stored potential energy $$U$$ change? - (A) The stored energy doubles to $$2U$$ - (B) The stored energy halves to $$U/2$$ - (C) The stored energy quadruples to $$4U$$ - (D) The stored energy remains at $$U$$ **Answer: (A)** Doubling the separation $$d$$ halves the capacitance $$C$$ ($$C \propto 1/d$$). Since $$U = Q^2/2C$$ and $$Q$$ is constant, halving $$C$$ doubles the stored energy. ### Question 7 A parallel-plate capacitor with plate separation $$d$$ remains connected to a battery with constant voltage $$V$$. If the separation is increased to $$2d$$, what is the new magnitude of the electric field $$E$$ between the plates? - (A) The electric field decreases to $$E/2$$ - (B) The electric field increases to $$2E$$ - (C) The electric field decreases to $$E/4$$ - (D) The electric field remains at $$E$$ **Answer: (A)** For a parallel-plate capacitor, $$E = V/d$$. Since voltage $$V$$ is held constant by the battery, doubling the separation $$d$$ reduces the electric field magnitude by half. ### Question 8 A parallel-plate capacitor connected to a battery stores energy $$U_0$$ with air between its plates. If a dielectric with constant $$\kappa = 3$$ is inserted while the battery remains connected, what is the new stored energy? - (A) The energy increases to $$3U_0$$ - (B) The energy decreases to $$U_0/3$$ - (C) The energy increases to $$9U_0$$ - (D) The energy remains at $$U_0$$ **Answer: (A)** Inserting a dielectric increases capacitance to $$3C$$. Since $$U = 1/2 CV^2$$ and $$V$$ is constant, the stored energy triples. ### Question 9 A parallel-plate capacitor is charged to potential difference $$V_0$$ and then disconnected from the source. If a slab with dielectric constant $$\kappa = 4$$ is inserted filling the space between plates, what is the new potential difference? - (A) The potential difference decreases to $$V_0/4$$ - (B) The potential difference increases to $$4V_0$$ - (C) The potential difference decreases to $$V_0/2$$ - (D) The potential difference remains at $$V_0$$ **Answer: (A)** Inserting the dielectric increases capacitance to $$4C$$. Since charge $$Q$$ is constant (isolated), $$V = Q/C$$ decreases by a factor of 4. ### Question 10 A parallel-plate capacitor has circular plates of radius $$r$$ and separation $$d$$. If the radius is doubled to $$2r$$ and the separation is doubled to $$2d$$, by what factor does the capacitance $$C$$ change? - (A) The capacitance increases by a factor of 2 - (B) The capacitance increases by a factor of 4 - (C) The capacitance decreases by a factor of 2 - (D) The capacitance remains the same value **Answer: (A)** Area $$A$$ depends on $$r^2$$, so doubling $$r$$ quadruples $$A$$. Doubling $$d$$ halves $$C$$. The net effect is $$4/2 = 2$$ times the original capacitance. ## Review the Unit - [10.1 Electrostatics with Conductors](/ap-physics-c-e-m/unit-10/1-electrostatics-with-conductors/study-guide/4Vb5LzwBQm2HSChq) - [10.2 Redistribution of Charge Between Conductors](/ap-physics-c-e-m/unit-10/2-redistribution-of-charge-between-conductors/study-guide/3zelmsMupFfJh7VP) - [10.3 Capacitors](/ap-physics-c-e-m/unit-10/3-capacitors/study-guide/jGOEgdPDnNfNGUeR) - [10.4 Dielectrics](/ap-physics-c-e-m/unit-10/4-dielectrics/study-guide/94aiEgDjuJxhK3Px) ## Other Unit Tests - [Unit 8: Electric Charges & Fields: Gauss's Law](/ap-physics-c-e-m/unit-test/8) - [Unit 9: Electric Potential](/ap-physics-c-e-m/unit-test/9) - [Unit 11: Electric Circuits](/ap-physics-c-e-m/unit-test/11) - [Unit 12: Magnetic Fields & Electromagnetism](/ap-physics-c-e-m/unit-test/12) - [Unit 13: Electromagnetic Induction](/ap-physics-c-e-m/unit-test/13) ## About This Document Canonical Fiveable pages are available as Markdown at the same path plus `.md`. - [llms.txt](https://fiveable.me/llms.txt): index of Fiveable's sections and URL patterns - [llms-full.txt](https://fiveable.me/llms-full.txt): complete subject and unit listing - [MCP server](https://fiveable.me/mcp): call Fiveable as tools instead of fetching pages (`https://fiveable.me/api/mcp`) - [MCP server for AP teachers](https://fiveable.me/mcp/teachers): a teacher's classes, assignments and AP-rubric grading (`https://fiveable.me/api/mcp/teacher`)