Question

two point charges separated by a distance d repel each other with a force 8N . if the separation is doubled force of repulsion will be.....??!!!

a) 2N
b) 8N
c) 16N
d) 4N

Answer 1

the answer is a)2N because the force varies inversely as the square of the distance between the charges

Answer 2

Answer:

If the separation is doubled the force of repulsion will be 2 N. Option A is the right answer.

Explanation:

The two point charges repel each other from a distance of d with the repulsive force as 8 N. on doubling the separation between the two point charges as 2d, the repulsive force is calculated as follows -

As we know that, $F=k q \times q / r^{2}$

$8=\mathrm{k} \mathrm{q} \times \mathrm{q} / \mathrm{d}^{2}$

Now when the separation is doubled, r = 2d  

$\begin{array}{l}{F=k q \times q / r^{2}} \\ {F^{\prime}=k q \times q /(2 d)^{2}} \\ {F^{\prime}=k q \times q /(2 d)^{2}} \\ {F^{\prime}=k q \times q / 4 d^{2}} \\ {F^{\prime}=1 / 4\left(k q \times q / d^{2}\right)}\end{array}$

F’ = 1/ 4 x 8

F’ = 2 N