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 is the region around a charge where other charges feel a force, with strength is E = F / q in N/C. These worked answers correct the 20 problems of the Physics 11-12 worksheet on electric field, from field from a measured force to spark from a metal sphere: each solution shows the formula, the substitution and the final answer with its unit.
Each answer gives the formula, the substitution and a result with its unit. Remember that charges must be in coulombs and distances in metres, and that small differences in the last digit are acceptable when you round at a different step.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB D.2
Problem 1 Field from a measured force
1. E = F / q = 6.0 × 10⁻⁵ / 2.0 × 10⁻⁹ = 3.0 × 10⁴ N/C.2. F = qE = 5.0 × 10⁻⁹ × 3.0 × 10⁴ = 1.5 × 10⁻⁴ N.
Problem 2 Distance and force
1. F = kQ₁Q₂ / r² = (8.99 × 10⁹ × 1.0 × 10⁻⁶ × 2.0 × 10⁻⁶) / (0.30)² = 0.200 N, so F ≈ 0.20 N, repulsive (like charges).2. Doubling r divides the force by 2² = 4, so F ≈ 0.050 N.
Problem 3 Capacitor-style plates
d = 3.0 mm = 3.0 × 10⁻³ m. E = V / d = 12 / 3.0 × 10⁻³ = 4.0 × 10³ V/m.Problem 4 Electronvolts
1. The gain is 200 eV.2. E = 200 × 1.60 × 10⁻¹⁹ = 3.2 × 10⁻¹⁷ J.
Problem 5 Direction of field and force
1. Field lines point away from a positive charge and towards a negative charge.2. The force on a negative charge is opposite to the field, so it points west.
3. Closer field lines mean a stronger field, so a larger force per unit charge.
Problem 6 Field of a point charge
E = kQ / r² = (8.99 × 10⁹ × 4.0 × 10⁻⁹) / (0.30)² = 399.6 N/C, so E ≈ 4.0 × 10² N/C, directed away from the sphere.Problem 7 Proton in a field
1. F = eE = 1.60 × 10⁻¹⁹ × 2.5 × 10⁴ = 4.0 × 10⁻¹⁵ N.2. a = F / m = 4.0 × 10⁻¹⁵ / 1.67 × 10⁻²⁷ = 2.4 × 10¹² m/s².
3. Weight = mg = 1.67 × 10⁻²⁷ × 9.81 = 1.6 × 10⁻²⁶ N. The ratio F / W = 2.4 × 10¹¹, so the weight is negligible.
Problem 8 Work against a charge
1. V = kQ / r: at 0.15 m, V = 360 V; at 0.060 m, V = 899 V.2. W = qΔV = 2.0 × 10⁻⁹ × (899 − 360) = 1.1 × 10⁻⁶ J, so W ≈ 1.1 × 10⁻⁶ J, done by the person pushing the charges together.
Problem 9 Suspended droplet
1. E = V / d = 600 / 0.010 = 6.0 × 10⁴ V/m.2. The electric force balances the weight: qE = mg, so q = mg / E = (2.0 × 10⁻¹⁵ × 9.81) / 6.0 × 10⁴ = 3.3 × 10⁻¹⁹ C.
3. n = q / e = 2.04 ≈ 2 electrons. The force on a negative droplet must point up, so the field points down and the top plate is positive.
Problem 10 Midpoint of two charges
Each charge is 0.10 m from the midpoint. Each field = kQ / r² = (8.99 × 10⁹ × 4.0 × 10⁻⁹) / (0.10)² = 3596 N/C. Away from + and towards − point in the same direction, so the fields add: E = 7192 N/C ≈ 7.2 × 10³ N/C, directed towards the negative charge.Problem 11 Reading a potential graph
1. V = kQ / r = (kQ) × (1 / r), which has the form y = mx, so V is proportional to 1/r.2. Gradient = 270 / 10 = 27 V m (1/r = 10 m⁻¹ at 0.10 m).
3. Gradient = kQ, so Q = 27 / 8.99 × 10⁹ = 3.0 × 10⁻⁹ C (3.0 nC).
Problem 12 Accelerating through a gun
1. KE = eV = 1.60 × 10⁻¹⁹ × 1500 = 2.4 × 10⁻¹⁶ J.2. v = √(2KE / m) = √(2 × 2.4 × 10⁻¹⁶ / 9.11 × 10⁻³¹) = 2.3 × 10⁷ m/s.
3. v = √(2 × 2.4 × 10⁻¹⁶ / 1.67 × 10⁻²⁷) = 5.4 × 10⁵ m/s, smaller by a factor of about 43 because the proton is far heavier.
Problem 13 Electric against gravity
1. Fe = ke² / r² = (8.99 × 10⁹ × (1.60 × 10⁻¹⁹)²) / (1.0 × 10⁻⁹)² = 2.3 × 10⁻¹⁰ N.2. Fg = Gm² / r² = (6.67 × 10⁻¹¹ × (1.67 × 10⁻²⁷)²) / (1.0 × 10⁻⁹)² = 1.9 × 10⁻⁴⁶ N.
3. Fe / Fg = 1.2 × 10³⁶. Both forces follow 1/r², so the ratio does not change with distance.
Problem 14 Spot the error
1. The distance was not converted to metres and not squared, and the force actually falls to one quarter, not one half, when r doubles.2. F = (8.99 × 10⁹ × 25 × 10⁻¹⁸) / (0.050)² = 2.2475 × 10⁻⁷ / 2.5 × 10⁻³ = 9.0 × 10⁻⁵ N.
3. Force at 10 cm = 9.0 × 10⁻⁵ / 4 = 2.2 × 10⁻⁵ N.
Problem 15 Reading a field pattern
1. Lines leave A and end on B, so A is positive and B is negative.2. More lines leave A than arrive at B, with extra lines escaping to the left, so A has the larger magnitude of charge.
3. The field at P points from A to B (to the right). An electron is negative, so the force is opposite: towards A (to the left).
Problem 16 Ball on a thread
1. Weight mg (down), electric force qE (horizontal, towards the negative plate) and tension T (along the thread).2. In equilibrium tan θ = qE / mg, so E = mg tan θ / q = (1.0 × 10⁻³ × 9.81 × tan 20°) / 2.0 × 10⁻⁶ = 1785 N/C ≈ 1.8 × 10³ N/C.
3. T cos θ = mg, so T = mg / cos 20° = 9.81 × 10⁻³ / 0.9397 = 10.4 × 10⁻³ N ≈ 1.04 × 10⁻² N.
Problem 17 Beam between plates
1. E = V / d = 400 / 0.020 = 2.0 × 10⁴ V/m.2. The horizontal speed is constant: t = L / v = 0.040 / 3.0 × 10⁷ = 1.3 × 10⁻⁹ s.
3. a = eE / m = 3.5 × 10¹⁵ m/s². y = ½at² = 0.5 × 3.5 × 10¹⁵ × (1.3 × 10⁻⁹)² = 3.1 mm. vy = at = 4.7 × 10⁶ m/s, so tan θ = vy / v = 0.156 and θ ≈ 8.9°, towards the positive plate.
Problem 18 Alpha particle and gold nucleus
1. KE = 5.0 × 1.60 × 10⁻¹³ = 8.0 × 10⁻¹³ J.2. U = kQ₁Q₂ / r = k(2e)(79e) / r.
3. At closest approach all the kinetic energy has become potential energy: r = k(2e)(79e) / KE = (8.99 × 10⁹ × 158 × (1.60 × 10⁻¹⁹)²) / 8.0 × 10⁻¹³ = 4.5 × 10⁻¹⁴ m, about 45 fm.
Problem 19 Zero potential but not zero field
Let x be the distance from the +6.0 nC charge. V = 0 requires k(6.0 × 10⁻⁹)/x = k(2.0 × 10⁻⁹)/(0.40 − x), so 6(0.40 − x) = 2x, 2.4 = 8x and x = 0.30 m.Field from the positive charge: kQ/r² = 599 N/C, towards the negative charge. Field from the negative charge (0.10 m away): 1798 N/C, also towards the negative charge. Total E = 2397 N/C ≈ 2.4 × 10³ N/C towards the −2.0 nC charge.
Problem 20 Spark from a metal sphere
1. V = Ed = 3.0 × 10⁶ × 2.0 × 10⁻³ = 6.0 × 10³ V.2. Outside a sphere the field is E = kQ / R², so Q = ER² / k = (3.0 × 10⁶ × (0.050)²) / 8.99 × 10⁹ = 8.3 × 10⁻⁷ C (0.83 µC).
3. V = kQ / R = ER = 3.0 × 10⁶ × 0.050 = 1.5 × 10⁵ V.
