Question

The capacitance of a parallel plate capacitor in air is $2\mu F$. If a dielectric medium is placed between the plates then the potential difference reduces to $\frac{1}{6}$ of the original value. The dielectric constant of the medium is :

• A. $6$
• B. $3$
• C. $2.2$
• D. $4.4$

Answer 1

The correct option is A $6$

As no battery is connected to the capacitor, the charge remains constant.
Before inserting the dielectric, the charge is $Q=CV$
After inserting the dielectric, the capacitance become $C^{\prime }=KC$ and hence charge $Q^{\prime }=\frac{C^{\prime }V}{6}=\frac{KCV}{6}$
As the charge is constant , $Q=Q^{\prime }$

$\Rightarrow CV=\frac{KCV}{6}$
$\therefore K=6$