The problems are on the electric potential worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Electric potential is potential energy per unit charge, V = U/q, measured in volts. These worked answers correct the 20 problems of the AP Physics C worksheet on electric potential, from units of potential to spot the error: each solution shows the formula, the substitution and the final answer with its unit.
Full methods follow for every problem: formula, substitution and the result with its unit in bold. Check each line against your own working, and look for the reasoning about signs and directions, which is where most marks are lost.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:E&M Unit 9
Problem 1 Units of potential
1. V = J/C, so 1 V = 1 J/C.2. W = qΔV = (3.0 × 10⁻⁶)(200) = 6.0 × 10⁻⁴ J.
Problem 2 Point charge potential
1. V = kq/r = (8.99 × 10⁹)(5.0 × 10⁻⁹)/0.30 = 150 V (149.8 V).2. Doubling r halves V: 75 V.
Problem 3 Uniform field gap
1. ΔV = Ed = (1500)(0.020) = 30 V.2. The field points toward lower potential, so plate A is higher: plate A.
Problem 4 Electron accelerated
1. KE = eΔV = (1.60 × 10⁻¹⁹)(250) = 4.0 × 10⁻¹⁷ J.2. v = √(2KE/m) = √(8.0 × 10⁻¹⁷/9.11 × 10⁻³¹) = 9.4 × 10⁶ m/s.
Problem 5 Inside a conductor
1. E = 0.2. With E = 0 everywhere inside, ΔV = −∫E·dl = 0 between any two points, and the surface is part of the same body, so the conductor is one equipotential.
Problem 6 Reading a V–x graph
1. |E| = |ΔV/Δx| = 12/0.040 = 300 V/m.2. V falls as x increases, so E points in the +x direction.
3. W = q(Vi − Vf) = (2.0 × 10⁻⁶)(12 − 0) = 2.4 × 10⁻⁵ J.
Problem 7 Two charges, one point
1. V = k(q₁ + q₂)/r = (8.99 × 10⁹)(3.0 − 1.0)(10⁻⁹)/0.20 = 17.98/0.20 = 90 V.2. Potential is a scalar, so contributions add algebraically, with sign.
Problem 8 Field given by x²
1. V(x) = −∫ from 0 to x of 600x′² dx′ = −200x³.2. V(0.30) = −200(0.027) = −5.4 V.
Problem 9 Field from potential
1. Ex = −dV/dx = −(40 − 300x) = 300x − 40 (V/m).2. Ex(0.10) = 30 − 40 = −10 V/m.
3. 300x = 40 gives x = 0.133 m.
Problem 10 Ring on its axis
1. s = √(0.12² + 0.16²) = 0.20 m, V = kQ/s = (8.99 × 10⁹)(4.0 × 10⁻⁹)/0.20 = 180 V.2. Ex = kQx/s³ = (35.96)(0.16)/0.0080 = 7.2 × 10² V/m.
Problem 11 Assembling two charges
1. W = kq₁q₂/r = (8.99 × 10⁹)(2.0 × 10⁻⁶)(3.0 × 10⁻⁶)/0.50 = 0.11 J.2. W is positive because the charges repel, so Wanjiru does positive work and the pair stores that energy.
Problem 12 Charged sphere
1. Q = VR/k = (3000)(0.15)/(8.99 × 10⁹) = 5.0 × 10⁻⁸ C.2. E = V/R = 3000/0.15 = 2.0 × 10⁴ V/m.
3. V = 3000(0.15/0.45) = 1000 V.
4. Inside, V = 3000 V and E = 0.
Problem 13 Joined spheres
1. The wire is a conductor, so charge flows until there is no field along it, which means equal potentials.2. kQ₁/R₁ = kQ₂/R₂ gives Q ∝ R, so Q₁ : Q₂ = 1 : 3, hence Q₁ = 20 nC and Q₂ = 60 nC.
3. V = kQ₁/R₁ = (8.99 × 10⁹)(2.0 × 10⁻⁸)/0.10 = 1.8 × 10³ V.
4. E = V/R, so E₁/E₂ = R₂/R₁ = 3: the small sphere has the stronger surface field.
Problem 14 Spacing of equipotentials
1. E = 20/0.020 = 1.0 × 10³ V/m.2. E = 20/0.010 = 2.0 × 10³ V/m.
3. The field is stronger where the lines are closer, between 40 V and 60 V.
4. The field points toward lower potential, to the left, perpendicular to the lines.
Problem 15 Potential near a long wire
1. V(0.10) − V(0.40) = −∫ from 0.40 to 0.10 of (2kλ/r) dr = 2kλ ln(0.40/0.10).2. 2kλ = 2(8.99 × 10⁹)(3.0 × 10⁻⁹) = 53.94 V, times ln 4 = 1.386 gives 75 V. The point 0.10 m from the wire is at the higher potential.
Problem 16 Charged rod on its axis
1. Piece dq = (Q/L)dx at distance x from P: V = k(Q/L)∫ from d to d+L of dx/x = k(Q/L) ln[(L + d)/d].2. V = (8.99 × 10⁹)(1.2 × 10⁻⁸) ln(0.60/0.10) = 107.9 × 1.792 = 1.9 × 10² V (193 V).
Problem 17 Charged disc
1. σ/2ε₀ = 5.0 × 10⁻⁹/(1.77 × 10⁻¹¹) = 282.5 V/m, and √(0.15² + 0.20²) − 0.15 = 0.10 m, so V = 28 V.2. Ex = −dV/dx = (σ/2ε₀)(1 − x/√(x² + R²)) = 282.5(1 − 0.60) = 1.1 × 10² V/m.
3. For x ≫ R, √(x² + R²) ≈ x + R²/2x, so V ≈ (σ/2ε₀)(R²/2x) = σπR²/(4πε₀x) = kQ/x.
Problem 18 Uniform charged ball
1. V(r) = V(R) + ∫ from r to R of kQr′/R³ dr′ = kQ/R + kQ(R² − r²)/(2R³) = kQ(3R² − r²)/(2R³).2. Centre: V = 3kQ/2R = 1.5(89.9)/0.050 = 2.7 × 10³ V.
3. At R/2: V = 1.375 kQ/R = 1.375(1798) = 2.5 × 10³ V.
Problem 19 Particle released near a charge
1. U = kQq/r: U(0.10) = 0.7192 J and U(0.50) = 0.1438 J, so KE = 0.5754 J and v = √(2KE/m) = 15 m/s.2. All 0.7192 J becomes kinetic: v = √(2 × 0.7192/0.0050) = 17 m/s.
3. The potential energy cannot fall below zero, so no more energy is available to convert.
Problem 20 Spot the error
1. Zero field inside only means V is constant there, not zero. Inside the shell V equals its surface value.2. V = kQ/R = (8.99 × 10⁹)(−30 × 10⁻⁹)/0.25 = −1.1 × 10³ V.
3. V = kQ/r = −269.7/0.50 = −5.4 × 10² V.
