1. Electric Charges and Fields
hard

Force between two point charges $q_1$ and $q_2$ placed in vacuum at ' $r$ ' $\mathrm{cm}$ apart is $F$. Force between them when placed in a medium having dielectric $\mathrm{K}=5$ at $\mathrm{r} / 5$ $\mathrm{cm}$ apart will be:

A

$F / 25$

B

$5 \mathrm{~F}$

C

$F / 5$

D

$25 \mathrm{~F}$

(JEE MAIN-2024)

Solution

In air $F=\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r_2}$

In medium

$F^{\prime}=\frac{1}{4 \pi\left(K \epsilon_0\right)} \frac{q_1 q_2}{\left(r^{\prime}\right)^2}=\frac{25}{4 \pi\left(5 \epsilon_0\right)} \frac{q_1 q_2}{(r)^2}=5 F$

Standard 12
Physics

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