Question

Two points charges $Q_1$ and $Q_2$ are placed at separation $d$ in vacuum and force acting between them is $F$. Now a dielectric slab of thickness $\frac{d}{2}$ and dielectric constant $K=4$ is placed between them. The new force between the charges will be

• A. $\frac{4F}{9}$
• B. $\frac{2F}{9}$
• C. $\frac{F}{9}$
• D. $\frac{5F}{9}$

Answer 1

The correct option is A $\frac{4F}{9}$

We know that, force between charges separated by distance $r$ in dielectric medium is equivalent to that of charges placed at distance $r\sqrt{K}$ of vacuum
(here $K$ is dielectric constant of medium)



Now, from coulombs law $F\propto \frac{1}{d^2}$
Ratio of force when charges are placed in dielectric to when charges are placed in vaccum is
$\frac{F_d}{F}=\frac{d^2}{{(\frac{d}{2}+\frac{d\sqrt{4}}{2})}^2}=\frac{4}{9}$
$F_d=\frac{4F}{9}$