Q. Two point charges exert on each other a force F 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
A) $\frac{r}{k}$
B) $rK$
C) $r\sqrt{K}$
D) $\frac{r}{\sqrt{K}}$
Explanation
(d) $F = \frac{q_{1}q_{2}}{4\pi\varepsilon_{0}r^{2}}\ $
In a medium of dielectric constant, $\ K = \frac{\varepsilon}{\varepsilon_{0}}$s
$F' = \frac{q_{1}q_{2}}{4\pi\varepsilon_{0}r^{2}}$
As $F' = F$
$\therefore\ \ \ \ \frac{q_{1}q_{2}}{4\pi\varepsilon R^{2}} = \frac{q_{1}q_{2}}{4\pi\varepsilon_{0}r^{2}}\ $
$R^{2} = \frac{\ \varepsilon_{0}}{\varepsilon_{\ }}r^{2} = \frac{r^{2}}{K}$
$\therefore\ \ \ R = \frac{r}{\sqrt{K}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.