Question

At what separation will two charges, each of magnitude 6.0 μC, exert a force of 0.70 N on

each other?

Answer 1

The given magnitude of the two charges are 6μc and they exert a force of 0.7N on each other. We need to find the distance of separation between the particles.

By Coloumb's Law :-

`\sf\longrightarrow F =(1)/(4\pi\epsilon_0) (q_1q_2)/(r^2)`

where

• q1 and q2 are the two charges
• r is the distance of separation
• 1/4πe0 = 9 * 10⁹

Substitute the respective values,

`\sf\longrightarrow` 0.7 = {(9*10⁹) (6*6) (10-⁶ *10-⁶) }/ r²

`\sf\longrightarrow` r² = (36*9 * 10-³)/0.7

`\sf\longrightarrow` r =√{ 36*9*10-³)/√7 m

`\sf\longrightarrow` r = 6* 3* 10-¹ /√70 m

`\sf\longrightarrow` r = 0.21 m

And we are done!