Electrostatics

JEE Main level, single correct, medium

22 of 311

Question 22 of 311

If two charges q1{q}_{1}and q2{q}_{2} are separated with distance dd and placed in a medium of dielectric constant kk. What will be the equivalent distance between charges in air for the same electrostatic force?

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Answer: (A) dkd\sqrt{k}

The force between the two charges when they are placed in dielectric is
F1=14πϵokq1q2d2{F}_{1}=\dfrac{1}{4\pi {\epsilon }_{o}k}\dfrac{{q}_{1}{q}_{2}}{{d}^{2}}
The force between the two charges when they are placed at separation d′d' is
F2=14πϵoq1q2d′2{F}_{2}=\dfrac{1}{4\pi {\epsilon }_{o}}\dfrac{{q}_{1}{q}_{2}}{d{'}^{2}}
Since, the force between the two charges is same in both case
F1=F2{F}_{1}={F}_{2}
⇒14πϵokq1q2d2=14πϵoq1q2d′2\Rightarrow \dfrac{1}{4\pi {\epsilon }_{o}k}\dfrac{{q}_{1}{q}_{2}}{{d}^{2}}=\dfrac{1}{4\pi {\epsilon }_{o}}\dfrac{{q}_{1}{q}_{2}}{d{'}^{2}}
⇒d′=dk\Rightarrow d'=d\sqrt{k}

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