Question

A capacitor, with air in between the plates, is charged to a potential $V_0$, now the battery is removed and a di-electric medium with di-electric constant $k$ is filled in the space between the plates, due to this the potential difference is dropped to $\frac{V_0}{10}$. Find the value of $k?$

• A. $1.6$
• B. $5$
• C. $8$
• D. $10$

Answer 1

The correct option is D $10$

Let the new capacitance after the insertion of dielectric be $C^{\prime }$ and final voltage be $V^{\prime }$.

If $K$ is the dielectric constant of the glass slab,

Then new capacitance, $C^{\prime }=KC$

As the capacitor is isolated, the charge on the plates will remain the same. Due to an increase in capacitance, voltage across the capacitor will decrease.

So, $Q=CV=C^{\prime }{V}^{\prime }$

$\Rightarrow \frac{C^{\prime }}{C}=\frac{V}{V^{\prime }}$

$\Rightarrow \frac{KC}{C}=\frac{V}{V/10}(\because V^{\prime }=\frac{V}{10})$

$\therefore K=10$

Hence, option $\left(d\right)$ is correct.

Key concept- $1.)$ The charge remains the same on the plates of an isolated capacitor. $2.)$ With insertion of a dielectric, capacitance increases.