Question

Two point charges placed at a distance $r$ in air exert a force $F$ on each other. The distance at which they experience force $4F$ when placed in a medium of dielectric constant $K=16$ is -

• A. $r$
• B. $\frac{r}{8}$
• C. $\frac{r}{4}$
• D. $\frac{r}{2}$

Answer 1

Answer: (B)

Force in air at a distance $r,$
$F= \, \frac{q_{1} q_{2}}{4 \pi ϵ_{0} r^{2}}$ ......(i)
Let $r’$ be the distance in the medium, therefore
$4 F=\frac{q_{1} q_{2}}{4 \pi \epsilon_{0}(16)\left(r^{\prime}\right)^{2}} \quad \ldots \ldots$ (ii)
From (i) $\div$ (ii)
$\frac{1}{4}=\frac{16\left(r^{\prime}\right)^{2}}{r^{2}}$
$\therefore r^{\prime}=\frac{r}{8}$