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(C) The force between two point charges $q_1$ and $q_2$ separated by a distance $r$ in air is given by Coulomb's law: $F = \frac{1}{4\pi\epsilon_0} \f

Vedclass · Accepted Answer

$r/\sqrt{K}$

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(C) The force between two point charges $q_1$ and $q_2$ separated by a distance $r$ in air is given by Coulomb's law: $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$.In a medium with dielectric constant $K$,the force $F'$ is given by $F' = \frac{1}{4\pi\epsilon_0 K} \frac{q_1 q_2}{r'^2}$,where $r'$ is the new distance.Given that the force remains the same,$F = F'$,so:$\frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} = \frac{1}{4\pi\epsilon_0 K} \frac{q_1 q_2}{r'^2}$.Canceling the common terms,we get: $\frac{1}{r^2} = \frac{1}{K r'^2}$.Rearranging for $r'$,we get: $r'^2 = \frac{r^2}{K}$.Taking the square root on both sides,we find: $r' = \frac{r}{\sqrt{K}}$.

Two point charges placed at a distance $r$ in air experience a certain force. Then the distance at which they will experience the same force in a medium of dielectric constant $K$ is

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