Question

Two charges are separated by 2 m and repel each other with a force of 20 N. If they are moved to a separation of 4 m, what will be the repulsive force?

Answer 1

Step-by-step explanation:

It is given that,

Separation between the charges, r₁ = 2 m

Force between the charges, F₁ = 20 N

Separation between the charges finally, r₂ = 4 m

Let F₂ is the repulsive force. The formula for the repulsive force is given by :

`F=k(q_1q_2)/(r^2)`

`(F_1)/(F_2)=(r_2^2)/(r_1^2)`

`(F_1)/(F_2)=(r_2^2)/(r_1^2)`

`F_2=(F_1r_1^2)/(r_2^2)`

`F_2=(20* 2^2)/(4^2)`

`F_2=5\ N`

So, the repulsive force is 5 newton. Hence, this is the required solution.