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 (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
`(1)/(d^2)`. This means that if we reduce the distance to one-ninth, i.e. the new distance is

`d_(new) = (1)/(9)d`
then the force will scale as

`(1)/((d_(new))^2)= (1)/(( (1)/(9) )^2d^2)= (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`