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
$4F=\frac{q_1{q}_2}{4\pi {ϵ}_0(16){\left(r^{\prime }\right)}^2}\dots \dots$ (ii)
From (i) $÷$ (ii)
$\frac{1}{4}=\frac{16{\left(r^{\prime }\right)}^2}{r^2}$
$\therefore r^{\prime }=\frac{r}{8}$