1.Easy
Two point charges repel each other with a force of 26 N. The first charge is halved and the distance between the charges is doubled. Find the new force F′.
F = 26 Nq₁′ = 0.5·q₁r′ = 2·r
F′ = N
Coulomb’s law & electric fields worksheet with answers
2.Easy
Two point charges repel each other with a force of 42 N. The first charge is halved and the distance between the charges is doubled. Find the new force F′.
F = 42 Nq₁′ = 0.5·q₁r′ = 2·r
F′ = N
3.Medium
Three point charges lie on a straight line: +3 µC on the left, +5 µC on the right 30 cm away, and +5 µC between them, 22.5 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 = +3 µCqR = +5 µCqM = +5 µCrL = 22.5 cmrR = 7.5 cm
F = N
4.Medium
Two identical small metal spheres carry charges of −4 µC and +5 µC. They are touched together and then put back 25 cm apart. Find the magnitude of the force F between them now. Use k = 8.99 × 10⁹ N·m²/C².
q1 = −4 µCq2 = +5 µCr = 25 cm
F = N
5.Hard
Three positive point charges sit at the corners of a right triangle: q₃ = 4 µC is at the right-angle corner, q₁ = 8 µC is 5 cm from it along one side and q₂ = 2 µC is 10 cm from it along the other side. Find the magnitude of the net electric force F on q₃.
q1 = 8 µCq2 = 2 µCq3 = 4 µCr13 = 5 cmr23 = 10 cm
F = N
6.Hard
An electron enters the uniform field between two parallel plates moving parallel to them at 3.5 × 10⁷ m/s. The plates are 10 cm long and the field between them is 1800 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 = 3.5 × 10⁷ m/sL = 10 cmE = 1800 N/Ce = 1.602 × 10⁻¹⁹ Cme = 9.109 × 10⁻³¹ kg
y = mm
Answer key
- 1. F′ = 3.25 N
- 2. F′ = 5.25 N
- 3. F = 37.29 N
- 4. F = 0.03596 N
- 5. F = 115.3 N
- 6. y = 1.292 mm
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