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

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 :

$F=k\dfrac{q_1q_2}{r^2}$

$\dfrac{F_1}{F_2}=\dfrac{r_2^2}{r_1^2}$

$\dfrac{F_1}{F_2}=\dfrac{r_2^2}{r_1^2}$

$F_2=\dfrac{F_1r_1^2}{r_2^2}$

$F_2=\dfrac{20\times 2^2}{4^2}$

$F_2=5\ N$

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