What is the mutual force of electrostatic repulsion, if the charge on each is 6.5 × 10^-7 C?

Question

i. Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm.

What is the mutual force of electrostatic repulsion, if the charge on each is 6.5 × 10^-7 C?

The radii of A and B are negligible compared to the distance of separation,

ii. What is the force of repulsion if each sphere is charged double the above amount and the distance between them is halved?

Answer 1

Given: q₁ = 6.5 × 10^-7 C q₂ = 6.5 × 10^-7 C

r = 50 cm = 0.50 m

To find: Force of repulsion (F)

Formula: F = \(\frac{1}{4\pi \epsilon_0}\,\frac{q_1 q_2}{r^2}\)

Calculation:

From formula,

F = 1.52 × 10^-2 N

ii. When each charge is doubled and the distance between them is reduced to half, then

∴ F = 0.24 N