If two charges and are separated by a distance and placed in a medium of dielectric constant , what will be the equivalent distance between the charges in air for the same electrostatic force? [2023]

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Solution

Q.
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 the charges in air for the same electrostatic force [2023]

1 $K \sqrt{d}$

2 $1.5 d \sqrt{K}$

3 $2 d \sqrt{K}$

4 $d \sqrt{K}$

Ans.
(4)

$F = \frac{1}{(4 π ε_{0})} \frac{q_{1} q_{2}}{k d^{2}} (in medium)$

$F_{air} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{d^{2}}$

$F = F_{air}$

$\frac{q_{1} q_{2}}{4 π ε_{0} k d^{2}} = \frac{q_{1} q_{2}}{4 π ε_{0} d^{2}}$

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

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