Question

Two point charges are placed at separation $d$ in vacuum and the force between them is $F$. Now a dielectric slab of thickness $t=\frac{d}{3}$ and dielectric constant $K$ is placed between the charges and the force becomes $\frac{9F}{25}$. Find the value of $K$.

Answer 1

From the coulomb's law of electrostatic force, the electrostatic force acting between two charges are inversly proportional to the square of the seperation 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 $\frac{d}{3}$ in slab medium is equivalent to a distance $\frac{d}{3}\sqrt{K}$ of vacuum. Now force acting between the charges, will be

$\frac{9F}{25}\propto \frac{1}{{(\frac{2d}{3}+\frac{d}{3}\sqrt{K})}^2}......\left(2\right)$

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

$\frac{9}{25}=\frac{1}{{(\frac{2}{3}+\frac{\sqrt{K}}{3})}^2}$

$\Rightarrow \frac{2}{3}+\frac{\sqrt{K}}{3}=\frac{5}{3}$

$\Rightarrow \frac{\sqrt{K}}{3}=\frac{5}{3}-\frac{2}{3}$

$\therefore K=9$

Accepted answers: 9,9.0,9.00