1.Easy
Two point charges repel each other with a force of 30 N. The first charge is halved and the distance between the charges is made four times as large. Find the new force F′.
F = 30 Nq₁′ = 0.5·q₁r′ = 4·r
F′ = N
Coulomb’s law & electric fields worksheet with answers
2.Easy
Two point charges repel each other with a force of 32 N. The first charge is doubled and the distance between the charges is tripled. Find the new force F′.
F = 32 Nq₁′ = 2·q₁r′ = 3·r
F′ = N
3.Medium
Two identical small metal spheres carry charges of +3 µC and −8 µC. They are touched together and then put back 10 cm apart. Find the magnitude of the force F between them now. Use k = 8.99 × 10⁹ N·m²/C².
q1 = +3 µCq2 = −8 µCr = 10 cm
F = N
4.Medium
A charge of +3 µC and a charge of −5 µC are 50 cm apart. Point P lies on the line between them, 20 cm from the first charge. Find the magnitude of the electric field E at P.
q1 = +3 µCq2 = −5 µCr1 = 20 cmr2 = 30 cm
E = N/C
5.Hard
Charges of +8 µC and −8 µC are 16 cm apart. Point P lies 8 cm from the midpoint between them, on the perpendicular bisector. Find the magnitude of the electric field E at P. Use k = 8.99 × 10⁹ N·m²/C².
q1 = +8 µCq2 = −8 µCd = 16 cmh = 8 cm
E = N/C
6.Hard
An electron enters the uniform field between two parallel plates moving parallel to them at 5 × 10⁷ m/s. The plates are 10 cm long and the field between them is 2100 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 = 5 × 10⁷ m/sL = 10 cmE = 2100 N/Ce = 1.602 × 10⁻¹⁹ Cme = 9.109 × 10⁻³¹ kg
y = mm
Answer key
- 1. F′ = 0.9375 N
- 2. F′ = 7.111 N
- 3. F = 5.619 N
- 4. E = 1.174 × 10⁶ N/C
- 5. E = 7.946 × 10⁶ N/C
- 6. y = 0.7387 mm
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