Question

Two point charges are kept in air with a separation between them. The force between them is $F_1$. if half of the space between the charges is tilled with a dielectric of dielectric constant $4$ and the force between them is $F_2$, if $\frac{1}{3}$ rd of the space between the charges is filled with dielectric of dielectric constant 9. Then $\frac{F_1}{F_2}$ is

• A. $\frac{27}{64}$
• B. $\frac{16}{81}$
• C. $\frac{81}{16}$
• D. $\frac{100}{81}$

Answer 1

Answer: (D)

When dielectric of thickness $t$ is introduced in two charges at distance $r$, the effective force between the charges is given by
$F=\frac{q_1{q}_2}{4\pi {\epsilon }_0[r-t+t\sqrt{K}{]}^2}$
where, $K=$ dielectric constant of medium
In first case, $t=r/2$ and $K=4$
$\therefore F_1=\frac{q_1{q}_2}{4\pi {\epsilon }_0{\left[r-r/2+\frac{r}{2}\sqrt{4}\right]}^2}=\frac{q_1{q}_2}{4\pi {\epsilon }_0\frac{9}{4}{r}^2}=\frac{q_1{q}_2}{9\pi {\epsilon }_0{r}^2}$
In second case, $t=r/3$ and $K=9$
$\therefore F_2=q_1{q}_2/4\pi {\epsilon }_0{\left[r-\frac{r}{3}+\frac{r}{3}\sqrt{3}\right]}^2=\frac{q_1{q}_2}{4\pi {\epsilon }_0\left(\frac{25}{9}\right)r^2}$
$\therefore \frac{F_1}{F_2}=\frac{q_1{q}_2}{9\pi {\epsilon }_0{r}^2}\times \frac{4\pi {\epsilon }_0\left(\frac{25}{9}\right)r^2}{q_1{q}_2}=\frac{100}{81}$