Question

Two point charges placed at a distance $r$ in air exert a force $F$ on each other. The value of distance $R$ at which they experience force $4F$ when placed in a medium of dielectric constant $K=9$ is :

• A. $r$
• B. $\frac{r}{6}$
• C. $\frac{r}{8}$
• D. $2r$

Answer 1

Answer: (B)

Initially force, $F=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}.......\left(1\right)$

Finally, $4F=\frac{1}{4\pi ϵ}\frac{q_1{q}_2}{R^2}$ where $K=\frac{ϵ}{{ϵ}_0}\Rightarrow ϵ=9{ϵ}_0$

now,$4F=\frac{1}{4\pi \left(9{ϵ}_0\right)}\frac{q_1{q}_2}{R^2}......\left(2\right)$

$\left(1\right)/\left(2\right),\frac{1}{4}=\frac{9R^2}{r^2}\Rightarrow R=\frac{r}{6}$