Question

if the charge on two point charged bodies and medium surrounding them are kept on change and the distance between them is reduced by 50% then the force between them
(a) is doubled
(b) is quadrupled
(c) becomes half
(d) decreases to 1/4th of their original force

Answer 1

(d) decreases to 1/4th of their original force
Let's idealize these charges as point charges, so that we can use the formula for electrostatic force between two charges q1q1 and q2q2 (expressed in Coulombs), called Coulomb's law:

F1=keq1q2r2F1=keq1q2r2

Where keke is the Coulomb's constant, approximately equal to 8.9875∗109Nm2C−2.8.9875∗109Nm2C−2.

When we double the distance rr, the formula above becomes:

F2=keq1q2(2r)2=keq1q24r2F2=keq1q2(2r)2=keq1q24r2

If we divide this result by the original formula, we can see how many times the force has decreased.

F2F1=14r21r2=r24r2=14F2F1=14r21r2=r24r2=14

So, the force decreases in a way that is linearly proportional to the square of the distance between the charges.

An equivalent way of saying it is that the electrostatic force is inversely proportional to the square of the distance between the charges.