1.Easy
Two point charges repel each other with a force of 8 N. The first charge is made four times as large and the distance between the charges is halved. Find the new force F′.
F = 8 Nq₁′ = 4·q₁r′ = 0.5·r
F′ = N
Coulomb’s law & electric fields worksheet with answers
2.Easy
A proton (charge 1.602 × 10⁻¹⁹ C, mass 1.673 × 10⁻²⁷ kg) is in a uniform electric field of 1400 N/C. Ignoring gravity, find the magnitude of its acceleration a.
E = 1400 N/C|q| = 1.602 × 10⁻¹⁹ Cm = 1.673 × 10⁻²⁷ kg
a = m/s²
3.Medium
A charge of +10 µC and a charge of −6 µC are 40 cm apart. Point P lies on the line between them, 20 cm from the first charge. Find the magnitude of the electric field E at P.
q1 = +10 µCq2 = −6 µCr1 = 20 cmr2 = 20 cm
E = N/C
4.Medium
Two identical small metal spheres carry charges of +6 µC and −3 µC. They are touched together and then put back 20 cm apart. Find the magnitude of the force F between them now. Use k = 8.99 × 10⁹ N·m²/C².
q1 = +6 µCq2 = −3 µCr = 20 cm
F = N
5.Hard
Charges of +4 µC and −4 µC are 16 cm apart. Point P lies 5 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 = 5 cm
E = N/C
6.Hard
An electron enters the uniform field between two parallel plates moving parallel to them at 4 × 10⁷ m/s. The plates are 10 cm long and the field between them is 4100 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 = 4 × 10⁷ m/sL = 10 cmE = 4100 N/Ce = 1.602 × 10⁻¹⁹ Cme = 9.109 × 10⁻³¹ kg
y = mm
Answer key
- 1. F′ = 128 N
- 2. a = 1.341 × 10¹¹ m/s²
- 3. E = 3.596 × 10⁶ N/C
- 4. F = 0.5057 N
- 5. E = 6.853 × 10⁶ N/C
- 6. y = 2.253 mm
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