Question

When an insulator of thickness equal to the separation between the plates is introduced between the plates, the potential difference reduces to one-tenth of the original value. The dielectric constant of the material is

• A. $1.6$
• B. $5$
• C. $10$
• D. $40$

Answer 1

The correct option is C $10$

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

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

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

$\frac{KC}{C}=\frac{V}{\left(\frac{V}{10}\right)}=10$

$\therefore K=10$

Hence, option $\left(c\right)$ 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.