1.Easy
Two point charges repel each other with a force of 34 N. The first charge is halved and the distance between the charges is tripled. Find the new force F′.
F = 34 Nq₁′ = 0.5·q₁r′ = 3·r
F′ = N
Coulomb’s law & electric fields worksheet with answers
2.Easy
Two point charges of −8 µC and +4 µC are 6 cm apart. Find the magnitude of the electric force F between them. Use k = 8.99 × 10⁹ N·m²/C².
q1 = −8 µCq2 = +4 µCr = 6 cm
F = N
3.Medium
Three point charges lie on a straight line: −2 µC on the left, −1 µC on the right 20 cm away, and −3 µC between them, 5 cm from the left charge. Find the magnitude of the net electric force F on the middle charge. Use k = 8.99 × 10⁹ N·m²/C².
qL = −2 µCqR = −1 µCqM = −3 µCrL = 5 cmrR = 15 cm
F = N
4.Medium
Charges of +4 µC and +2 µC are 30 cm apart. At what distance x from the +4 µC charge, on the line between them, is the electric field zero?
q1 = +4 µCq2 = +2 µCd = 30 cm
x = cm
5.Hard
Two identical small balls, each of mass 3 g, hang from the same point on insulating threads 80 cm long. They are given equal charges and each thread settles at 20° from the vertical. Use g = 9.81 m/s². Find the charge q on each ball in nC. Use k = 8.99 × 10⁹ N·m²/C².
m = 3 gL = 80 cmθ = 20°g = 9.81 m/s²
q = nC
6.Hard
An electron enters the uniform field between two parallel plates moving parallel to them at 3 × 10⁷ m/s. The plates are 5 cm long and the field between them is 2200 N/C. Electron charge 1.602 × 10⁻¹⁹ C, mass 9.109 × 10⁻³¹ kg; ignore gravity. How far y is it deflected sideways when it leaves the plates? Give the answer in mm.
v₀ = 3 × 10⁷ m/sL = 5 cmE = 2200 N/Ce = 1.602 × 10⁻¹⁹ Cme = 9.109 × 10⁻³¹ kg
y = mm
Answer key
- 1. F′ = 1.889 N
- 2. F = 79.91 N
- 3. F = 20.38 N
- 4. x = 17.57 cm
- 5. q = 597.3 nC
- 6. y = 0.5374 mm
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