1.Easy
Two point charges repel each other with a force of 22 N. The first charge is doubled and the distance between the charges is halved. Find the new force F′.
F = 22 Nq₁′ = 2·q₁r′ = 0.5·r
F′ = N
Coulomb’s law & electric fields worksheet with answers
2.Easy
Find the magnitude of the electric field E at a distance of 3 cm from a point charge of −1 µC. Use k = 8.99 × 10⁹ N·m²/C².
q = −1 µCr = 3 cm
E = N/C
3.Medium
Charges of +8 µC and +5 µC are 100 cm apart. At what distance x from the +8 µC charge, on the line between them, is the electric field zero?
q1 = +8 µCq2 = +5 µCd = 100 cm
x = cm
4.Medium
A charge of +10 µC and a charge of +1 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 = +10 µCq2 = +1 µCr1 = 4 cmr2 = 6 cm
E = N/C
5.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 2200 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 = 2200 N/Ce = 1.602 × 10⁻¹⁹ Cme = 9.109 × 10⁻³¹ kg
y = mm
6.Hard
An electron enters the uniform field between two parallel plates moving parallel to them at 3 × 10⁷ m/s. The plates are 8 cm long and the field between them is 1100 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 = 8 cmE = 1100 N/Ce = 1.602 × 10⁻¹⁹ Cme = 9.109 × 10⁻³¹ kg
y = mm
Answer key
- 1. F′ = 176 N
- 2. E = 9.989 × 10⁶ N/C
- 3. x = 55.85 cm
- 4. E = 5.369 × 10⁷ N/C
- 5. y = 0.5374 mm
- 6. y = 0.6878 mm
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