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 $\varepsilon$ will be

• A. $\frac{\varepsilon_{0}}{\varepsilon} F$
• B. $\frac{\varepsilon}{\varepsilon_{0}} F$
• C. $\varepsilon \varepsilon_{0} F$
• D. $\frac{F}{\varepsilon_{0} \varepsilon}$

Answer 1

Answer: (A)

Force in air, i.e.,
$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}$
and force in medium,
i.e., $F_{m}=\frac{1}{4 \pi \varepsilon_{0} \varepsilon_{r}} \frac{q_{1} q_{2}}{r^{2}} \left(\because \frac{\varepsilon}{\varepsilon_{0}}=\varepsilon_{r}\right)$
$=\frac{1}{4 \pi \varepsilon} \frac{q_{1} q_{2}}{r^{2}}$
$=\frac{\varepsilon_{0}}{\varepsilon} \cdot \frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}$
$\Rightarrow F_{m}=\frac{\varepsilon_{0}}{\varepsilon} F$