If two charges q 1 and q 2 are separated with distance d and placed in a medium of dielectric constant k . What will be the equivalent distance between charges in air for the same electrostatic force?

Question

Aicharya · Accepted Answer

The force between the two charges when they are placed in dielectric is

$F_{1} = \frac{1}{4 π ε_{o} k} \frac{q_{1} q_{2}}{d^{2}}$

The force between the two charges when they are placed at separation $d'$ is

$F_{2} = \frac{1}{4 π ε_{o}} \frac{q_{1} q_{2}}{d'^{2}}$

Since, the force between the two charges is same in both case

$F_{1} = F_{2}$

$\Rightarrow \frac{1}{4 π ε_{o} k} \frac{q_{1} q_{2}}{d^{2}} = \frac{1}{4 π ε_{o}} \frac{q_{1} q_{2}}{d'^{2}}$

$\Rightarrow d' = d \sqrt{k}$

If two charges q 1 and q 2 are separated with distance d and placed in a medium of dielectric constant k .…

Solution