1.Easy
Find the magnitude of the electric field E at a distance of 25 cm from a point charge of +50 nC. Use k = 8.99 × 10⁹ N·m²/C².
q = +50 nCr = 25 cm
E = N/C
Coulomb’s law & electric fields worksheet with answers
2.Easy
Two point charges repel each other with a force of 4 N. The first charge is tripled and the distance between the charges is halved. Find the new force F′.
F = 4 Nq₁′ = 3·q₁r′ = 0.5·r
F′ = N
3.Medium
A charge of +2 µC and a charge of +8 µC are 20 cm apart. Point P lies on the line between them, 15 cm from the first charge. Find the magnitude of the electric field E at P.
q1 = +2 µCq2 = +8 µCr1 = 15 cmr2 = 5 cm
E = N/C
4.Medium
Two identical small metal spheres carry charges of −8 µC and +6 µC. They are touched together and then put back 5 cm apart. Find the magnitude of the force F between them now. Use k = 8.99 × 10⁹ N·m²/C².
q1 = −8 µCq2 = +6 µCr = 5 cm
F = N
5.Hard
Three positive point charges sit at the corners of a right triangle: q₃ = 10 µC is at the right-angle corner, q₁ = 8 µC is 5 cm from it along one side and q₂ = 4 µC is 10 cm from it along the other side. Find the magnitude of the net electric force F on q₃.
q1 = 8 µCq2 = 4 µCq3 = 10 µCr13 = 5 cmr23 = 10 cm
F = N
6.Hard
An electron enters the uniform field between two parallel plates moving parallel to them at 1.5 × 10⁷ m/s. The plates are 8 cm long and the field between them is 4300 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 = 1.5 × 10⁷ m/sL = 8 cmE = 4300 N/Ce = 1.602 × 10⁻¹⁹ Cme = 9.109 × 10⁻³¹ kg
y = mm
Answer key
- 1. E = 7192 N/C
- 2. F′ = 48 N
- 3. E = 2.797 × 10⁷ N/C
- 4. F = 3.596 N
- 5. F = 289.9 N
- 6. y = 10.76 mm
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