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)

$F = \frac{1}{4 \pi \epsilon _{0}} \frac{q_{1} q_{2}}{r^{2}}$
For a dielectric of dielectric constant K between the charges, the effective separation in air r_eff is given by
$\frac{1}{4 \pi \epsilon _{0}} \frac{q_{1} q_{2}}{r_{eff}^{2}} = \frac{1}{4 \pi \epsilon _{0}} \frac{q_{1} q_{2}}{K r^{2}}$
$⇒ \, \, r_{eff} = \sqrt{K} r$
$\therefore \quad F^{\prime}=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{\left(\frac{r}{2}+\frac{\sqrt{K} r}{2}\right)^2}$
$⇒ \, \, \frac{F^{'}}{F} = \frac{1}{\left(\frac{1}{2} + \frac{\sqrt{4}}{2}\right)^{2}} = \frac{4}{9}$