Q. A force of $40\, N$ is acting between two charges in air. If the space between them is filled with glass with $\varepsilon_{r}=8$, what will be the force?
J & K CETJ & K CET 2002
Solution:
From Coulombs law the force between two charges is
$F=\frac{1}{4 \pi \varepsilon_{0} K} \cdot \frac{q_{1} q_{2}}{r^{2}}$
where $r$ is distance between the charges and $K$ the dielectric constant.
In air: $F_{1}=\frac{1}{4 \pi \varepsilon_{0} K} \cdot \frac{q_{1} q_{2}}{r^{2}}=40\, N$
In glass: $F_{2}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1} q_{2}}{8 r^{2}}$
$\therefore \frac{F_{1}}{F_{2}}=\frac{40}{F_{2}}=8$
$\Rightarrow F_{2} =5\, N$
