Question

When you double the distance between a pair of charged particles what happens to the force between them?

Answer 1

Answer:

Force between the two charges becomes one fourth of the initial force.

Explanation:

The electrostatic force acting between any two charges is given as,

$F = k\frac{q_{1}q_{2}}{r^{2}}$

Here,

F = force

k = Coulomb's constant

$q_{1}$ = magnitude of charge of the first particle

$q_{2}$ = magnitude of charge of the second particle

$r$ = separation between the two charges

From the above relation,

$F \propto \frac{1}{r^{2}}$

Thus,

$\frac{F_{1}}{F_{2}} = \left ( \frac{r_{2}}{r_{1}} \right )^{2}$

$\frac{F_{1}}{F_{2}} = \left ( \frac{2r}{r} \right )^{2}$

$\Rightarrow \ F_{2} = \frac{1}{4}F_{4}.$