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  4. Two point charges exert a force F on each other when they are placed r distance apart in air. When they are placed R distance apart in a medium of dielectric constant K, they exert the same force. The distance R equals

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Q. Two point charges exert a force $F$ on each other when they are placed $r$ distance apart in air. When they are placed $R$ distance apart in a medium of dielectric constant $K$, they exert the same force. The distance $R$ equals

Electric Charges and Fields

Solution:

$ F_{1}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r^{2}}$ or
$F_{2}=\frac{1}{4 \pi \varepsilon_{0} K} \frac{q^{2}}{R^{2}}$
As $F_{1}=F_{2}$,
hence $\frac{q^{2}}{r^{2}}=\frac{1}{K} \frac{q^{2}}{R^{2}} $
$\therefore R=r / \sqrt{K}$