Question

For charges $q_1$ and $q_2$, if force between them for some separation in air is $F$, then force between them in a medium of permittivity $\epsilon$ will be

• A. $\frac{{\epsilon }_0}{\epsilon }F$
• B. $\frac{\epsilon }{{\epsilon }_0}F$
• C. $\epsilon {\epsilon }_{0}F$
• D. $\frac{F}{{\epsilon }_0\epsilon }$

Answer 1

Answer: (A)

Force in air, i.e.,
$F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$
and force in medium,
i.e., $F_m=\frac{1}{4\pi {\epsilon }_0{\epsilon }_r}\frac{q_1{q}_2}{r^2}\left(\because \frac{\epsilon }{{\epsilon }_0}={\epsilon }_r\right)$
$=\frac{1}{4\pi \epsilon }\frac{q_1{q}_2}{r^2}$
$=\frac{{\epsilon }_0}{\epsilon }\cdot \frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$
$\Rightarrow F_m=\frac{{\epsilon }_0}{\epsilon }F$