Question

Two charges when kept at a distance of $1m$ apart in vacuum have some force of repulsion. If the force of repulsion between these two charges be same, when placed in an oil of dielectric constant $4$, the distance of separation is :

• A. $0.25m$
• B. $0.4m$
• C. $0.5m$
• D. $0.6m$

Answer 1

Answer: (C)

$0.5m$

In vacuum force of repulsion $=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{(1{)}^2}$

given the force of repulsion is same in oil so let the distance in oil be X.
$\Rightarrow \frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{(1{)}^2}=\frac{1}{4\pi K{\epsilon }_0}\frac{q_1{q}_2}{X^2}$

$\Rightarrow \frac{K}{1}=\frac{1}{X^2}$

$\Rightarrow 4=\frac{1}{X^2}$

$\Rightarrow 2=\frac{1}{X}$

$\Rightarrow X=0.5m$