The problems are on the electric field worksheet. Try them on your own first, then compare with the answers below.
Quick answer: An electric field E = F/q is the force per unit positive charge, a vector drawn with field lines. These worked answers correct the 20 problems of the AP Physics 2 worksheet on electric field, from field from a force to reading an equipotential map: each solution shows the formula, the substitution and the final answer with its unit.
Compare your working with each answer. Credit in AP-style marking goes to the correct relation, a substitution with units and a conclusion that states direction or sign where relevant. Small rounding differences are acceptable.
Level: AP Physics 2 (algebra-based) · Curriculum: AP P2 Unit 10
Problem 1 Field from a force
1. E = F/q = 6.0 × 10⁻⁶ / 2.0 × 10⁻⁹ = 3.0 × 10³ N/C.2. East, the direction of the force on a positive charge.
3. Same size, 6.0 × 10⁻⁶ N, but west, opposite to the field.
Problem 2 Field of a point charge
1. E = kQ/r² = (8.99 × 10⁹)(3.0 × 10⁻⁹) / (0.15)² = 1.2 × 10³ N/C.2. Towards the charge, since it is negative.
Problem 3 Reading field lines
1. Charge A: the closer the lines, the stronger the field.2. The direction of the force on a positive test charge.
3. At a crossing point the field would have two directions at once, but the field at a point has only one direction.
Problem 4 Field between plates
1. E = V/d = 300 / 0.015 = 2.0 × 10⁴ V/m.2. F = qE = 4.0 × 10⁻⁹ × 2.0 × 10⁴ = 8.0 × 10⁻⁵ N.
Problem 5 Energy from a voltage
1. ΔU = qΔV = 2.0 × 10⁻⁶ × 12 = 2.4 × 10⁻⁵ J.2. One volt is one joule of energy per coulomb of charge: 1 V = 1 J/C.
Problem 6 Electron-volts
1. 250 eV.2. 250 × 1.60 × 10⁻¹⁹ = 4.0 × 10⁻¹⁷ J.
3. 3.2 × 10⁻¹⁷ / 1.60 × 10⁻¹⁹ = 200 eV.
Problem 7 Electron between plates
1. E = 150 / 0.030 = 5.0 × 10³ V/m.2. F = eE = 8.0 × 10⁻¹⁶ N, so a = F/m = 8.8 × 10¹⁴ m/s².
3. ½mv² = eΔV, so v = √(2 × 1.60 × 10⁻¹⁹ × 150 / 9.11 × 10⁻³¹) = 7.3 × 10⁶ m/s.
Problem 8 Potential of a point charge
1. V = kQ/r: at 0.20 m, 3.6 × 10² V; at 0.40 m, 1.8 × 10² V.2. ΔV = 360 − 180 = 1.8 × 10² V.
3. W = qΔV = 1.0 × 10⁻⁹ × 180 = 1.8 × 10⁻⁷ J, supplied from outside since the charges repel.
Problem 9 Equipotential rings
1. V at the rings: 360 V (0.10 m) and 90 V (0.40 m). W = qΔV = 2.0 × 10⁻⁹ × 270 = 5.4 × 10⁻⁷ J.2. Zero: the ring is an equipotential, so ΔV = 0.
3. Radially outward, perpendicular to the ring.
Problem 10 Proton down a potential
1. ΔU = qΔV = 1.60 × 10⁻¹⁹ × (−300) = −4.8 × 10⁻¹⁷ J (300 eV lost).2. The kinetic energy gained is 4.8 × 10⁻¹⁷ J, so v = √(2 × 4.8 × 10⁻¹⁷ / 1.67 × 10⁻²⁷) = 2.4 × 10⁵ m/s.
Problem 11 Field from a potential graph
1. Gradient = (20 − 80) / 6.0 = −10 V/cm = −1000 V/m.2. E = −dV/dx = 1.0 × 10³ V/m in the +x direction (towards lower potential).
3. F = eE = 1.60 × 10⁻¹⁹ × 1000 = 1.6 × 10⁻¹⁶ N in +x.
Problem 12 Midpoint between opposite charges
1. Each: E = (8.99 × 10⁹)(3.0 × 10⁻⁹) / (0.10)² = 2.7 × 10³ N/C. Away from the positive charge and towards the negative: both point towards +x.2. The fields add: 5.4 × 10³ N/C towards +x.
3. V = kQ/0.10 − kQ/0.10 = 0 V.
Problem 13 Droplet held at rest
1. E = 600 / 0.0050 = 1.2 × 10⁵ V/m.2. qE = mg, so q = 5.9 × 10⁻¹⁵ × 9.81 / 1.2 × 10⁵ = 4.8 × 10⁻¹⁹ C.
3. N = 4.8 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 3 electrons. The force is upward, towards the positive plate, so the droplet is negative.
Problem 14 Energy of two charges
1. U = kq₁q₂/r = (8.99 × 10⁹)(2.0 × 10⁻⁶)(3.0 × 10⁻⁶) / 0.30 = 0.18 J.2. All the potential energy becomes kinetic: ½mv² = 0.1798, so v = √(2 × 0.1798 / 0.010) = 6.0 m/s.
Problem 15 Charge inside a conductor
1. Zero.2. Free charges move until their own field cancels the external field inside, which leaves the excess charge on the outer surface.
3. The charge flows around the outside of the body and the field inside the cavity is zero, so no current flows through the people.
Problem 16 Electron beam deflection
1. E = 100 / 0.010 = 1.0 × 10⁴ V/m; a = eE/m = 1.76 × 10¹⁵ m/s².2. t = 0.040 / 2.0 × 10⁷ = 2.0 × 10⁻⁹ s.
3. y = ½at² = 0.5 × 1.756 × 10¹⁵ × (2.0 × 10⁻⁹)² = 3.5 mm, less than the 5.0 mm half-gap, so it escapes.
4. vy = at = 3.5 × 10⁶ m/s; tanθ = 3.5 × 10⁶ / 2.0 × 10⁷, so θ = 10°.
Problem 17 Opposite charges on a line
1. V = kQ/0.20 − kQ/0.20 = 0 V. Each field is 449.5 N/C pointing the same way, so E = 9.0 × 10² N/C towards the negative charge.2. V = kQ(1/0.10 − 1/0.30) = 17.98 × 6.667 = 1.2 × 10² V.
3. W = qΔV = 1.0 × 10⁻⁹ × 120 = 1.2 × 10⁻⁷ J.
Problem 18 Ball in a horizontal field
1. Weight down, tension along the thread, electric force horizontal in the direction of the field (positive charge).2. T cos20° = mg and T sin20° = qE, so E = mg tan20° / q = 0.0020 × 9.81 × 0.364 / 5.0 × 10⁻⁶ = 1.4 × 10³ N/C.
3. ΔV = Ed = 1428 × 0.050 = 71 V.
Problem 19 Closest approach
1. K = 5.0 × 10⁶ × 1.60 × 10⁻¹⁹ = 8.0 × 10⁻¹³ J.2. At closest approach all K has become potential energy: K = k(2e)(79e)/r.
3. r = (8.99 × 10⁹)(158)(1.60 × 10⁻¹⁹)² / 8.0 × 10⁻¹³ = 4.5 × 10⁻¹⁴ m.
Problem 20 Reading an equipotential map
1. 20 / 0.020 = 1000 V/m, 20 / 0.020 = 1000 V/m, and 20 / 0.010 = 2000 V/m.2. Perpendicular to the lines, from high to low potential (from the 60 V side towards the 0 V side).
3. W = qΔV = 2.0 × 10⁻⁶ × 60 = 1.2 × 10⁻⁴ J.
4. Between the 40 V and 60 V lines, where the lines are closest: the field and the force are strongest there.
