1.Easy
Two point charges repel each other with a force of 54 N. The first charge is halved and the distance between the charges is made four times as large. Find the new force F′.
F = 54 Nq₁′ = 0.5·q₁r′ = 4·r
F′ = N
Coulomb’s law & electric fields worksheet with answers
2.Easy
Two point charges of +5 µC and −10 µC are 10 cm apart. Find the magnitude of the electric force F between them. Use k = 8.99 × 10⁹ N·m²/C².
q1 = +5 µCq2 = −10 µCr = 10 cm
F = N
3.Medium
Charges of +10 µC and +5 µC are 50 cm apart. At what distance x from the +10 µC charge, on the line between them, is the electric field zero?
q1 = +10 µCq2 = +5 µCd = 50 cm
x = cm
4.Medium
A charge of +1 µC and a charge of −3 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 = +1 µCq2 = −3 µCr1 = 5 cmr2 = 15 cm
E = N/C
5.Hard
Charges of +10 µC and +10 µC are 12 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 = +10 µCq2 = +10 µCd = 12 cmh = 3 cm
E = N/C
6.Hard
An electron enters the uniform field between two parallel plates moving parallel to them at 1 × 10⁷ m/s. The plates are 5 cm long and the field between them is 4400 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 × 10⁷ m/sL = 5 cmE = 4400 N/Ce = 1.602 × 10⁻¹⁹ Cme = 9.109 × 10⁻³¹ kg
y = mm
Answer key
- 1. F′ = 1.688 N
- 2. F = 44.95 N
- 3. x = 29.29 cm
- 4. E = 4.795 × 10⁶ N/C
- 5. E = 1.787 × 10⁷ N/C
- 6. y = 9.673 mm
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