Two points charges are kept in air with a separation between them. The force between them is $F_1$, if half…
Two points 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 filled with a dielectric constant 4 and the force between them is $F_2$. If $\frac{1}{3} \mathrm{rd}$ of the space between the charges is filled with dielectric of dielectric constant 9. Then $\frac{F_1}{F_2}$ is
$\frac{27}{64}$
$\frac{16}{81}$
$\frac{81}{64}$
$\frac{100}{81}$
Solution
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 \varepsilon_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 \varepsilon_0\left[r-r / 2+\frac{r}{2} \sqrt{4}\right]^2}=\frac{q_1 q_2}{4 \pi \varepsilon_0 \frac{9}{4} r^2}=\frac{q_1 q_2}{9 \pi \varepsilon_0 r^2}
$
In second case, $t=r / 3$ and $K=9$
$
\begin{aligned}
& \therefore F_2=q_1 q_2 / 4 \pi \varepsilon_0\left[r-\frac{r}{3}+\frac{r}{3} \sqrt{3}\right]^2=\frac{q_1 q_2}{4 \pi \varepsilon_0\left(\frac{25}{9}\right) r^2} \\
& \therefore \frac{F_1}{F_2}=\frac{q_1 q_2}{9 \pi \varepsilon_0 r^2} \times \frac{4 \pi \varepsilon_0\left(\frac{25}{9}\right) r^2}{q_1 q_2}=\frac{100}{81}
\end{aligned}
$