Question

When two charged particles are placed at the certain distance apart in vacuum, then the force experienced by one of the charges is equal to $F.$ If vacuum is replaced by a medium of dielectric constant $6,$ then the net force experienced by the charged particle will be equal to

$[$0.77 Mark$]$

• A. $6F$
• B. $F$
• C. $\frac{F}{6}$
• D. $\frac{F}{12}$

Answer 1

The correct option is C $\frac{F}{6}$

From Coulomb’s Law, electrostatic force between two point charges in vacuum
$F_v=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$
In a medium, other than vacuum,
$\epsilon =K{\epsilon }_0$ (Where $K$ is the dielctric constant)
$\Rightarrow \epsilon =6{\epsilon }_0$
$\Rightarrow F^{\prime }=\frac{1}{4\pi \epsilon }\frac{q_1{q}_2}{r^2}$
$\Rightarrow F^{\prime }=\frac{1}{4\pi \left(6{\epsilon }_0\right)}\frac{q_1{q}_2}{r^2}=\frac{F}{6}$