Question

For charges q1 and q2 if force between them for some separation in air is F then force between them in a medium of permittivity e will be

Answer 1

Kq1q2/r2=F

in medium

the permitivitiy will be divided by F
F/e

Answer 2

Answer:

if the charges are in medium of permittivity, $\varepsilon$, then the force between the charges is $F=\frac{1}{4\pi \varepsilon_{0} k} \frac{q_{1} q_{2} }{r^{2}}$

Explanation:

• According to Coulomb's law, if two charges $q_{1}$ and $q_{2}$ are kept at a distance $r$, then the force acting between the charges is given by

                                      $F=k\frac{q_{1} q_{2} }{r^{2}}$

       where $k$ is a proportionality constant.

• If the charges are in air or vacuum, $k=\frac{1}{4\pi \varepsilon_{0} } =9\times 10^{9} N.m^{2} /C^{2}$
• For a medium,  the dielectric constant, $k={\varepsilon_{r}= \frac{\varepsilon }{\varepsilon_{0} }$

       where ${\varepsilon_{r}, \varepsilon$ and $\varepsilon_{0}$ are relative permittivity, permittivity of medium and permittivity of vacuum respectively.

• Therefore if the charges are placed in medium of permittivity, $\varepsilon$, then  

                                 $F=\frac{1}{4\pi \varepsilon_{0} k} \frac{q_{1} q_{2} }{r^{2}}$ or  $F=\frac{1}{4\pi \varepsilon_{0} \varepsilon_{r}} \frac{q_{1} q_{2} }{r^{2}}$