Two charges q_1 and q_2 are separated with distance 'd' and placed in a medium of dielectric constant 'k'. The equivalent distance between charges in air for the same electrostatic force is

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Correct answer: C. d k

Force between two charges in a medium of dielectric constant

k

is given by

F_{m} = \dfrac{1}{4\pi\epsilon_{0} k} \dfrac{q_{1} q_{2}}{d^{2}}

Force between the same charges in air at an equivalent distance

d'

is given by

F_{a} = \dfrac{1}{4\pi\epsilon_{0}} \dfrac{q_{1} q_{2}}{d'^{2}}

Given that the electrostatic force remains the same,

F_{m} = F_{a}

\dfrac{1}{4\pi\epsilon_{0} k} \dfrac{q_{1} q_{2}}{d^{2}} = \dfrac{1}{4\pi\epsilon_{0}} \dfrac{q_{1} q_{2}}{d'^{2}}

\dfrac{1}{k d^{2}} = \dfrac{1}{d'^{2}}

d'^{2} = k d^{2}

d' = d\sqrt{k}

Answer:

d\sqrt{k}

Two charges q_1 and q_2 are separated with distance 'd' and placed...

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