The problems are on the coulomb's law worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Electric charge comes in positive and negative kinds, is conserved and is quantised in units of e = 1.60 × 10⁻¹⁹ C. These worked answers correct the 20 problems of the AP Physics 2 worksheet on coulomb's law, from counting electrons to unequal charges and motion: each solution shows the formula, the substitution and the final answer with its unit.
Compare each answer with your own working. For full credit on AP-style problems, name the law, substitute with units, give the direction of any force and write the final answer with a sensible number of significant figures.
Level: AP Physics 2 (algebra-based) · Curriculum: AP P2 Unit 10
Problem 1 Counting electrons
1. Extra electrons make it negative.2. Q = Ne = 2.0 × 10¹³ × 1.60 × 10⁻¹⁹ = 3.2 × 10⁻⁶ C.
3. N = 1.0 × 10⁻⁶ / 1.60 × 10⁻¹⁹ = 6.3 × 10¹² electrons.
Problem 2 Charging by friction
1. Q = 5.0 × 10¹⁰ × 1.60 × 10⁻¹⁹ = −8.0 × 10⁻⁹ C.2. The wool lost the same electrons: +8.0 × 10⁻⁹ C.
3. The total is zero, an example of conservation of charge.
Problem 3 Attract or repel
1. Repulsive: like charges repel.2. Attractive: unlike charges attract.
3. Attractive: the rod polarises the ball, bringing the opposite charge nearer to the rod than the like charge.
Problem 4 A simple Coulomb force
1. F = (8.99 × 10⁹)(4.0 × 10⁻⁶)(6.0 × 10⁻⁶) / (0.50)² = 0.86 N on each sphere (Newton’s third law).2. Repulsive.
3. Directly away from the +6.0 μC sphere, along the line joining them.
Problem 5 Unit conversions
1. 3.5 × 10⁻⁶ C: 3.5 × 10⁻⁶ C.2. 2.5 × 10⁻⁷ C.
3. 0.040 m.
4. 0.75 mC = 0.75 × 10⁻³ C = 750 μC.
Problem 6 Inverse-square scaling
1. Twice the distance: 12 / 4 = 3.0 N.2. Half the distance: 12 × 4 = 48 N.
3. Three times the distance: 12 / 9 = 1.3 N.
Problem 7 Changing charge and distance
1. Factor = 2 / 3² = 2/9, so F = 0.90 × 2/9 = 0.20 N.2. Factor = (2 × 2) / (½)² = 16, so F = 14 N (14.4 N).
3. With one charge zero, F = 0 N.
Problem 8 Charging by induction
1. Free electrons move towards the rod, so the near side is negative and the far side is positive.2. Electrons flow from the earth, through Bilal, onto the sphere, attracted by the positive rod.
3. The sphere now has extra electrons and, with the earth path gone, keeps them: the final charge is negative, opposite to the rod.
Problem 9 Electroscope
1. Electrons are repelled down to the leaves, which become negative and repel each other, while the disc becomes positive.2. The leaves fall together because the electrons return and the electroscope is neutral overall.
3. Some electrons transfer to the electroscope, so it gains a net negative charge and the leaves stay apart.
Problem 10 Comb and paper
1. The comb polarises the paper: charge shifts slightly within each atom so the side nearest the comb has the opposite sign.2. The opposite charge is closer to the comb than the like charge, and the force falls with distance, so attraction wins.
3. The pieces gain charge from the comb on contact and then carry the same sign, so they are repelled.
Problem 11 Finding an unknown charge
1. F = kq²/r², so q = r√(F/k).2. q = 0.10 × √(2.7 / 8.99 × 10⁹) = 0.10 × 1.733 × 10⁻⁵ = 1.7 × 10⁻⁶ C (1.7 μC).
Problem 12 Net force on a line
1. Attraction towards −x: F = (8.99 × 10⁹)(6.0 × 10⁻⁶)(1.0 × 10⁻⁶) / (0.10)² = 5.4 N.2. Attraction towards +x, distance 0.30 m: F = (8.99 × 10⁹)(2.0 × 10⁻⁶)(1.0 × 10⁻⁶) / (0.30)² = 0.20 N.
3. Net = 5.39 − 0.20 = 5.2 N towards −x.
Problem 13 Sharing charge
1. F = (8.99 × 10⁹)(8.0 × 10⁻⁶)(2.0 × 10⁻⁶) / (0.20)² = 3.6 N, attractive.2. Total +6.0 μC shared equally: +3.0 μC each.
3. F = (8.99 × 10⁹)(3.0 × 10⁻⁶)² / (0.20)² = 2.0 N, repulsive.
Problem 14 Electric force in hydrogen
1. Fe = (8.99 × 10⁹)(1.60 × 10⁻¹⁹)² / (5.3 × 10⁻¹¹)² = 8.2 × 10⁻⁸ N.2. Fg = Gmemp/r² = 3.6 × 10⁻⁴⁷ N.
3. Fe/Fg = 2.3 × 10³⁹, so gravity is negligible inside atoms.
Problem 15 Force against 1/r² graph
1. F = (kq²)(1/r²), which has the form y = mx, so F is proportional to 1/r².2. Gradient = 0.90 / 100 = 9.0 × 10⁻³ N·m².
3. kq² = gradient, so q = √(9.0 × 10⁻³ / 8.99 × 10⁹) = 1.0 × 10⁻⁶ C.
Problem 16 Zero-force point
1. Between them, the two forces point in opposite directions, so they can cancel. Outside, both push the same way.2. 4.0 / x² = 1.0 / (0.30 − x)², so 2 / x = 1 / (0.30 − x), giving 0.60 − 2x = x and x = 0.20 m.
3. No: q cancels in the equation, so the balance point is the same for any third charge.
Problem 17 Hanging charged balls
1. sinθ = 0.050 / 1.00, so θ = 2.9°.2. Vertically T cosθ = mg and horizontally T sinθ = F, so F = mg tanθ = 0.0020 × 9.81 × 0.0501 = 9.8 × 10⁻⁴ N.
3. q = d√(F/k) = 0.10 × √(9.84 × 10⁻⁴ / 8.99 × 10⁹) = 3.3 × 10⁻⁸ C (33 nC).
Problem 18 Forces in two dimensions
1. r = √(0.20² + 0.15²) = 0.25 m.2. F = (8.99 × 10⁹)(4.0 × 10⁻⁶)(2.0 × 10⁻⁶) / (0.25)² = 1.15 N each.
3. The vertical components (±0.6 × 1.15 N) cancel and the horizontal parts (0.8 × 1.15 N each) add: net = 1.84 N = 1.8 N in the +x direction.
Problem 19 Hovering sphere
1. The upward electric repulsion equals its downward weight, so the net force is zero.2. W = mg = 4.0 × 10⁻⁴ × 9.81 = 3.9 × 10⁻³ N.
3. r = √(kq²/F) = √(8.99 × 10⁹ × (2.5 × 10⁻⁸)² / 3.92 × 10⁻³) = 0.038 m (3.8 cm).
Problem 20 Unequal charges and motion
1. The forces are equal in size and opposite, by Newton’s third law, even though the charges differ.2. a = F/m, so the +Q charge has an acceleration four times larger than the +4Q charge.
3. The net external force on the system is zero and it started at rest, so the centre of mass stays at rest.
