Question

Two point charges $Q_1$ and $Q_2$ 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 $\frac{4F}{9}$

From the coulomb's law , the electrostatic force acting between two charges are inversly proportional to the square of the distance between them,

$F\propto \frac{1}{d^2}......\left(1\right)$

When a dielectric slab of dielectric constant $K$ is placed between the charges the distance $\frac{d}{2}$ in a medium is equivalent to a distance $\frac{d}{2}\sqrt{K}$ of vacuum. Now force acting between the charges, will be

$F{}^{\prime }\propto \frac{1}{{(\frac{d}{2}+\frac{d}{2}\sqrt{K})}^2}......\left(2\right)$

Now from equation $\left(1\right)$ and $\left(2\right)$, we get,

$\frac{F{}^{\prime }}{F}=\frac{1}{{(\frac{1}{2}+\frac{\sqrt{K}}{2})}^2}$

Now substituting $K=4$, we get,

$\frac{F^{{}^{\prime }}}{F}=\frac{1}{\frac{1}{4}(1+\sqrt{4}{)}^2}$

$\Rightarrow \frac{F^{{}^{\prime }}}{F}=\frac{F}{\left(\frac{9}{4}\right)}$

$\Rightarrow F^{{}^{\prime }}=\frac{4F}{9}$

Hence, option (a) is the correct answer.