Question

The force between two identical charges placed at a distance of $r$ in a vacuum is $F$ . Now a slab of dielectric constant $4$ is inserted between these two charges. If the thickness of the slab is $\frac{r}{2}$ , then the force between the charges will become

• A. $\frac{3}{5}F$
• B. $\frac{4}{9}F$
• C. $\frac{F}{4}$
• D. $\frac{F}{2}$

Answer 1

Answer: (B)

$F=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}$
For a material of dielectric constant $K$ between the charges, the effective separation in air, $r_{eff}$ is given by
$\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r_{eff}^2}=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{K{r}^2}$
$\Rightarrow r_{eff}=\sqrt{K}r$
$\therefore F^{{}^{\prime }}=\frac{1}{4\pi {\left(ϵ\right)}_0}\frac{q_1{q}_2}{{\left(\frac{r}{2}+\frac{\sqrt{K}r}{2}\right)}^2}$
$\Rightarrow \frac{F^{{}^{\prime }}}{F}=\frac{1}{{\left(\frac{1}{2}+\frac{\sqrt{4}}{2}\right)}^2}=\frac{4}{9}$