Question

When two charges are kept in air medium, at certain distance $d$ apart, the force between them is $F$. When they are kept in a dielectric medium at the same distance of separation, the force between them becomes $F / 2$. Then the dielectric constant of the medium is

• A. 5
• B. 2
• C. 4
• D. 3
• E. 8

Answer 1

Answer: (B)

According to Coulomb's law, force between two charges at a certain distance $d$ apart in air medium is given as,
$F=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{d^2} \ldots \text {.(i) }$
where, $\varepsilon_0$ is the permittivity of free space.
When these charges are kept in a dielectric medium, then force between them is given as
$F_D=\frac{1}{4 \pi \varepsilon_0 K} \frac{q_1 q_2}{d^2} .$.....(ii)
where, $K$ is the dielectric constant.
Given, $F_D=\frac{F}{2}$
$ =\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{2 d^2}$
$=\frac{1}{4 \pi \varepsilon_0 K} \frac{q_1 q_2}{d^2} \text { [from Eqs. (i) and (ii)] } $
$\Rightarrow K=2$