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
$F=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0}[r-t +t \sqrt{K}]^{2}}$
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 $K=9$
$\therefore F_{2}=q_{1} q_{2} / 4 \pi \varepsilon_{0}\left[r-\frac{r}{3}+\frac{r}{3} \sqrt{3}\right]^{2}=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0}\left(\frac{25}{9}\right) r^{2}}$
$\therefore \frac{F_{1}}{F_{2}}=\frac{q_{1} q_{2}}{9 \pi \varepsilon_{0} r^{2}} \times \frac{4 \pi \varepsilon_{0}\left(\frac{25}{9}\right) r^{2}}{q_{1} q_{2}}=\frac{100}{81}$