Two point charges placed at a certain distance r in air exert a force F on each other. Then, the distance r' at which these charges will exert the same force in a medium of dielectric constant K is given by

A
r
B
r/K
C
$r / \sqrt{K}$
D
$r \sqrt{K}$

Solution:

Force exerted between two point charges in air $F = \frac{q_{1} q_{2}}{4 {πε}_{0} r^{2}}$
So, force exerted between two point charges in a medium of dielectric constant
K is $F^{'} = \frac{q_{1} q_{2}}{4 {πε}_{0} K {(r')}^{2}}$
Where, r' is the distance.
Given, F = F'
$\Rightarrow \frac{q_{1} q_{2}}{4 {πε}_{0} r^{2}} = \frac{q_{1} q_{2}}{4 {πε}_{0} K {(r')}^{2}} \Rightarrow \frac{1}{r^{2}} = \frac{1}{K {(r')}^{2}}$
$\begin{array}{l} \Rightarrow r^{2} = K {(r')}^{2} \Rightarrow r = \sqrt{K} r' \\ ∴ r' = \frac{r}{\sqrt{K}} \end{array}$

Two point charges placed at a certain distance r in air exert a force F on each other. Then, the distance r' at which these charges will exert the same force in a medium of dielectric constant K is given by

Solution:

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