Question

Two point charges placed at a distance r in air expert 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 $K=16$ is:

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

Answer 1

The correct option is C $\frac{r}{8}$

Two points charge placed at a distance $=r$

force $=F$

value of distance $=R$

$F=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}$

$4F=\frac{1}{4\pi {ϵ}_0\times K}\frac{q_1{q}_2}{R^2}$

or, $F=\frac{1}{4\pi {ϵ}_0\times 4\times 16}\times \frac{q_1{q}_2}{R^2}$

or, $\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}$ $=\frac{1}{64}\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{R^2}$

$=\frac{1}{64}\left(\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{R^2}\right)$

or, $\frac{1}{r^2}=\frac{1}{64}\times \left(\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{R^2}\right)$

or, $\frac{1}{r^2}=\frac{1}{64}\times \frac{1}{R^2}$

$\frac{1}{r}=\frac{1}{8}\frac{1}{R}$

or, $R=\frac{r}{8}$