Question

Two charges attract each other with a force of 3.0 n. what will be the force if the distance between them is reduced to one-ninth of its original value?

Answer 1

The electrostatic force between two charges is given by
$F= k_e \frac{q_1 q_2}{d^2}$
where ke is the Coulomb's constant, q1 and q2 are the two charges and d is the distance between the two charges.
We can see from the formula that the force is proportional to $\frac{1}{d^2}$. This means that if we reduce the distance to one-ninth, i.e. the new distance is
$d_{new} = \frac{1}{9}d$
then the force will scale as
 $\frac{1}{(d_{new})^2}= \frac{1}{( \frac{1}{9} )^2d^2}= \frac{81}{d^2}$
So, the new force will be 81 times stronger than the initial value, therefore the new force is
$F_{new}=81 F=81 \cdot 3.0 N=243 N$