1.Easy
Two point charges repel each other with a force of 14 N. The first charge is made four times as large and the distance between the charges is tripled. Find the new force F′.
F = 14 Nq₁′ = 4·q₁r′ = 3·r
F′ = N
Coulomb’s law & electric fields worksheet with answers
2.Easy
Two point charges repel each other with a force of 58 N. The first charge is halved and the distance between the charges is doubled. Find the new force F′.
F = 58 Nq₁′ = 0.5·q₁r′ = 2·r
F′ = N
3.Medium
A charge of +2 µC and a charge of +6 µC are 60 cm apart. Point P lies on the line between them, 36 cm from the first charge. Find the magnitude of the electric field E at P.
q1 = +2 µCq2 = +6 µCr1 = 36 cmr2 = 24 cm
E = N/C
4.Medium
A charge of +6 µC and a charge of +10 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 = +6 µCq2 = +10 µCr1 = 12 cmr2 = 8 cm
E = N/C
5.Hard
Three positive point charges sit at the corners of a right triangle: q₃ = 8 µC is at the right-angle corner, q₁ = 4 µC is 6 cm from it along one side and q₂ = 8 µC is 10 cm from it along the other side. Find the magnitude of the net electric force F on q₃.
q1 = 4 µCq2 = 8 µCq3 = 8 µCr13 = 6 cmr23 = 10 cm
F = N
6.Hard
An electron enters the uniform field between two parallel plates moving parallel to them at 3 × 10⁷ m/s. The plates are 5 cm long and the field between them is 4500 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 = 3 × 10⁷ m/sL = 5 cmE = 4500 N/Ce = 1.602 × 10⁻¹⁹ Cme = 9.109 × 10⁻³¹ kg
y = mm
Answer key
- 1. F′ = 6.222 N
- 2. F′ = 7.25 N
- 3. E = 797724 N/C
- 4. E = 1.03 × 10⁷ N/C
- 5. F = 98.47 N
- 6. y = 1.099 mm
Full step-by-step solutions: practise this topic on examproblems.com.