1.Easy
Two point charges repel each other with a force of 28 N. The first charge is halved and the distance between the charges is tripled. Find the new force F′.
F = 28 Nq₁′ = 0.5·q₁r′ = 3·r
F′ = N
Coulomb’s law & electric fields worksheet with answers
2.Easy
An electron (charge −1.602 × 10⁻¹⁹ C, mass 9.109 × 10⁻³¹ kg) is in a uniform electric field of 1700 N/C. Ignoring gravity, find the magnitude of its acceleration a.
E = 1700 N/C|q| = 1.602 × 10⁻¹⁹ Cm = 9.109 × 10⁻³¹ kg
a = m/s²
3.Medium
A charge of +4 µC and a charge of −2 µC are 40 cm apart. Point P lies on the line between them, 30 cm from the first charge. Find the magnitude of the electric field E at P.
q1 = +4 µCq2 = −2 µCr1 = 30 cmr2 = 10 cm
E = N/C
4.Medium
A charge of +2 µC and a charge of +10 µC are 10 cm apart. Point P lies on the line between them, 4 cm from the first charge. Find the magnitude of the electric field E at P.
q1 = +2 µCq2 = +10 µCr1 = 4 cmr2 = 6 cm
E = N/C
5.Hard
Charges of +4 µC and +4 µC are 10 cm apart. Point P lies 3 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 = +4 µCq2 = +4 µCd = 10 cmh = 3 cm
E = N/C
6.Hard
Three positive point charges sit at the corners of a right triangle: q₃ = 6 µC is at the right-angle corner, q₁ = 8 µC is 5 cm from it along one side and q₂ = 3 µC is 5 cm from it along the other side. Find the magnitude of the net electric force F on q₃.
q1 = 8 µCq2 = 3 µCq3 = 6 µCr13 = 5 cmr23 = 5 cm
F = N
Answer key
- 1. F′ = 1.556 N
- 2. a = 2.99 × 10¹⁴ m/s²
- 3. E = 2.198 × 10⁶ N/C
- 4. E = 1.373 × 10⁷ N/C
- 5. E = 1.088 × 10⁷ N/C
- 6. F = 184.3 N
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