Two equal point charges exert a force F on ea | vedclass

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(A) According to Coulomb's law, the force between two point charges $q_1$ and $q_2$ separated by a distance $r$ in a medium with dielectric constant $

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$\frac{d}{\sqrt{K}}$

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(A) According to Coulomb's law, the force between two point charges $q_1$ and $q_2$ separated by a distance $r$ in a medium with dielectric constant $K$ is given by $F = \frac{1}{4\pi\epsilon_0 K} \frac{q_1 q_2}{r^2}$.In air, the dielectric constant $K = 1$. Thus, the force $F$ between two equal charges $q$ at distance $d$ is $F = \frac{1}{4\pi\epsilon_0} \frac{q^2}{d^2}$.In a medium with dielectric constant $K$ at distance $D$, the force is $F' = \frac{1}{4\pi\epsilon_0 K} \frac{q^2}{D^2}$.Given that the force remains the same $(F = F')$, we equate the two expressions:$\frac{1}{4\pi\epsilon_0} \frac{q^2}{d^2} = \frac{1}{4\pi\epsilon_0 K} \frac{q^2}{D^2}$.Canceling the common terms $\frac{q^2}{4\pi\epsilon_0}$, we get $\frac{1}{d^2} = \frac{1}{K D^2}$.Rearranging for $D^2$, we get $D^2 = \frac{d^2}{K}$.Taking the square root of both sides, we find $D = \frac{d}{\sqrt{K}}$.

Two equal point charges exert a force $F$ on each other when they are placed distance $d$ apart in air. When they are placed distance $D$ apart in a medium of dielectric constant $K$, they exert the same force. The distance $D$ is equal to

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