1.Easy
An electron (charge −1.602 × 10⁻¹⁹ C, mass 9.109 × 10⁻³¹ kg) is in a uniform electric field of 4200 N/C. Ignoring gravity, find the magnitude of its acceleration a.
E = 4200 N/C|q| = 1.602 × 10⁻¹⁹ Cm = 9.109 × 10⁻³¹ kg
a = m/s²
Coulomb’s law & electric fields worksheet with answers
2.Easy
Find the magnitude of the electric field E at a distance of 40 cm from a point charge of −10 µC. Use k = 8.99 × 10⁹ N·m²/C².
q = −10 µCr = 40 cm
E = N/C
3.Medium
A charge of +6 µC and a charge of −2 µC are 50 cm apart. Point P lies on the line between them, 12.5 cm from the first charge. Find the magnitude of the electric field E at P.
q1 = +6 µCq2 = −2 µCr1 = 12.5 cmr2 = 37.5 cm
E = N/C
4.Medium
A charge of +8 µC and a charge of −4 µC are 20 cm apart. Point P lies on the line between them, 5 cm from the first charge. Find the magnitude of the electric field E at P.
q1 = +8 µCq2 = −4 µCr1 = 5 cmr2 = 15 cm
E = N/C
5.Hard
Charges of +4 µC and +4 µC are 16 cm apart. Point P lies 6 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 = 16 cmh = 6 cm
E = N/C
6.Hard
Three positive point charges sit at the corners of a right triangle: q₃ = 2 µC is at the right-angle corner, q₁ = 10 µC is 8 cm from it along one side and q₂ = 1 µC is 10 cm from it along the other side. Find the magnitude of the net electric force F on q₃.
q1 = 10 µCq2 = 1 µCq3 = 2 µCr13 = 8 cmr23 = 10 cm
F = N
Answer key
- 1. a = 7.387 × 10¹⁴ m/s²
- 2. E = 561875 N/C
- 3. E = 3.58 × 10⁶ N/C
- 4. E = 3.037 × 10⁷ N/C
- 5. E = 4.315 × 10⁶ N/C
- 6. F = 28.15 N
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