If two charges q_1 and q_2 are separated by a | vedclass

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Question

(A) The electrostatic force between two charges in a medium with dielectric constant $K$ is given by: $F = \frac{1}{4 \pi \varepsilon_0 K} \frac{q_1 q

Vedclass · Accepted Answer

$d \sqrt{K}$

Vedclass · Answer

(A) The electrostatic force between two charges in a medium with dielectric constant $K$ is given by: $F = \frac{1}{4 \pi \varepsilon_0 K} \frac{q_1 q_2}{d^2}$.
The electrostatic force between the same charges in air at an equivalent distance $d'$ is: $F_{	ext{air}} = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{d'^2}$.
For the forces to be equal $(F = F_{	ext{air}})$,we equate the expressions:
$\frac{1}{4 \pi \varepsilon_0 K} \frac{q_1 q_2}{d^2} = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{d'^2}$.
Simplifying the equation:
$\frac{1}{K d^2} = \frac{1}{d'^2}$.
Taking the square root of both sides:
$d'^2 = K d^2 \implies d' = d \sqrt{K}$.

If two charges $q_1$ and $q_2$ are separated by a distance '$d$' and placed in a medium of dielectric constant $K$,what will be the equivalent distance between the charges in air for the same electrostatic force?

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