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 made four times as large. Find the new force F′.
F = 28 Nq₁′ = 0.5·q₁r′ = 4·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 2400 N/C. Ignoring gravity, find the magnitude of its acceleration a.
E = 2400 N/C|q| = 1.602 × 10⁻¹⁹ Cm = 9.109 × 10⁻³¹ kg
a = m/s²
3.Medium
A charge of +1 µC and a charge of +3 µC are 20 cm apart. Point P lies on the line between them, 12 cm from the first charge. Find the magnitude of the electric field E at P.
q1 = +1 µCq2 = +3 µCr1 = 12 cmr2 = 8 cm
E = N/C
4.Medium
Three point charges lie on a straight line: −5 µC on the left, −2 µC on the right 10 cm away, and +6 µC between them, 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 = −5 µCqR = −2 µCqM = +6 µCrL = 5 cmrR = 5 cm
F = N
5.Hard
Two identical small balls, each of mass 2 g, hang from the same point on insulating threads 60 cm long. They are given equal charges and each thread settles at 10° from the vertical. Use g = 9.81 m/s². Find the charge q on each ball in nC. Use k = 8.99 × 10⁹ N·m²/C².
m = 2 gL = 60 cmθ = 10°g = 9.81 m/s²
q = nC
6.Hard
An electron enters the uniform field between two parallel plates moving parallel to them at 2.5 × 10⁷ m/s. The plates are 5 cm long and the field between them is 3100 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 = 2.5 × 10⁷ m/sL = 5 cmE = 3100 N/Ce = 1.602 × 10⁻¹⁹ Cme = 9.109 × 10⁻³¹ kg
y = mm
Answer key
- 1. F′ = 0.875 N
- 2. a = 4.221 × 10¹⁴ m/s²
- 3. E = 3.59 × 10⁶ N/C
- 4. F = 64.73 N
- 5. q = 129.3 nC
- 6. y = 1.09 mm
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