Question

What happens to the force between two charges when each charge is doubled and the distance between them is 1/4 its original

magnitude?

Answer 1

F' = 64 F

Step-by-step explanation:

The electric force between charges is given by :

`F=(kq_1q_2)/(r^2)`

Where

q₁ and q₂ are charges

r is the distance between charges

When each charge is doubled and the distance between them is 1/4 its original magnitude such that,

q₁' = 2q₁, q₂' = 2q₂ and r' = (r/4)

New force,

`F'=(kq_1'q_2')/(r'^2)`

Apply new values,

`F'=(k* 2q_1* 2q_2)/(((r)/(4))^2)\\\\=(k* 4q_1q_2)/((r^2)/(16))\\\\=64* (kq_1q_2)/(r^2)\\\\=64F`

So, the new force becomes 64 times the initial force.