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Question

(a) $F' = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}{q_2}}}{{{r^2}}} = \frac{F}{K}$
If $F$ is the force in air, then $F'$ is less than $F$

Vedclass · Accepted Answer

become ${k^{ - 1}}$ times

Vedclass · Answer

(a) $F' = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}{q_2}}}{{{r^2}}} = \frac{F}{K}$
If $F$ is the force in air, then $F'$ is less than $F$ since $K > 1$.

When air is replaced by a dielectric medium of constant $k$, the maximum force of attraction between two charges separated by a distance

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