Electrostatics

JEE Main level, single correct, medium

15 of 311

Question 15 of 311

Force between two point charges q1{q}_{1} and q2{q}_{2} placed in vacuum at rr cmcm apart is FF. Force between them when placed in a medium having dielectric K=5K=5 at r5\dfrac{r}{5} cmcm apart will be:

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Answer: (B) 5F5F

Using Coulomb's law, we can write:
In air F=14πϵ0q1q2r2F=\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{{q}_{1}{q}_{2}}{{r}^{2}}
In medium F′=14π(Kϵ0)q1q2(r′)2=14π(5ϵ0)25q1q2(r)2=5FF'=\dfrac{1}{4\pi \left(K{\epsilon }_{0}\right)}\dfrac{{q}_{1}{q}_{2}}{{\left(r'\right)}^{2}}=\dfrac{1}{4\pi \left(5{\epsilon }_{0}\right)}\dfrac{25{q}_{1}{q}_{2}}{{\left(r\right)}^{2}}=5F

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