Question

The force between two identical charges placed at a distance of $r$ in vacuum is $F$. Now, a slab of dielectric of 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. $F$
• B. $\frac{3}{5}F$
• C. $\frac{4}{9}F$
• D. $\frac{F}{2}$

Answer 1

The correct option is C $\frac{4}{9}F$

in vacuum, $F=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}.....\left(i\right)$
Suppose, force between the charges is same when charges are $r^{\prime }$ distance apart in dielectric.
$F^{\prime }=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{{k{r}^{\prime }}^2}.......\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$, $k{r^{\prime }}^2=r^2$ or $r=\sqrt{k{r}^{\prime }}$
In the given situation, force between the charges would be
$F^{\prime }=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{{(\frac{r}{2}+\sqrt{4}\frac{r}{2})}^2}=\frac{4}{9}\frac{q^2}{4\pi {\epsilon }_0{r}^2}=\frac{4F}{9}$