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Question

(B) The force between two point charges $q_1$ and $q_2$ separated by a distance $d$ in air is given by Coulomb's Law: $F_{	ext{air}} = \frac{1}{4\pi\

Vedclass · Accepted Answer

decreases $K$ times

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(B) The force between two point charges $q_1$ and $q_2$ separated by a distance $d$ in air is given by Coulomb's Law: $F_{	ext{air}} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{d^2}$.When the medium is replaced by a dielectric of constant $K$ (also known as relative permittivity $\epsilon_r$),the permittivity of the medium becomes $\epsilon = K\epsilon_0$.The force in the medium is given by: $F_{	ext{medium}} = \frac{1}{4\pi\epsilon} \frac{q_1 q_2}{d^2} = \frac{1}{4\pi(K\epsilon_0)} \frac{q_1 q_2}{d^2}$.Therefore,$F_{	ext{medium}} = \frac{1}{K} \left( \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{d^2} ight) = \frac{F_{	ext{air}}}{K}$.Thus,the force decreases $K$ times.

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

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