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

Electrostatics PYQ — JEE Main 2024 · 31 Jan · Shift 2 · Q76 with Solution

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Correct answer: A

By Coulomb’s law, the force between two point charges in vacuum is

F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}

Here, the separation is $r$ cm, and this force is given as $F$ .

When the charges are placed in a medium of dielectric constant $K$ , Coulomb’s law becomes

F' = \frac{1}{4\pi\varepsilon_0 K}\frac{q_1 q_2}{(r')^2}

where $r'$ is the new separation.

Given:

So,

F' = \frac{1}{4\pi\varepsilon_0 \cdot 5}\frac{q_1 q_2}{\left(\frac{r}{5}\right)^2}

Since

\left(\frac{r}{5}\right)^2 = \frac{r^2}{25}

we get

F' = \frac{1}{4\pi\varepsilon_0 \cdot 5} \cdot \frac{q_1 q_2}{r^2/25}

F' = \frac{1}{4\pi\varepsilon_0 \cdot 5} \cdot \frac{25 q_1 q_2}{r^2}

F' = 5 \left( \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} \right)

But

\frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} = F

Therefore,

F' = 5F

Hence, the correct answer is A.