Question

A parallel plate condenser with oil (dielectric constant $2$) between the plates has capacitance $C$. If the oil is removed, the capacitance of capacitor becomes :

• A. $\sqrt{2C}$
• B. $2C$
• C. $\frac{C}{\sqrt{2}}$
• D. $\frac{C}{2}$

Answer 1

The correct option is D $\frac{C}{2}$

The capacitance of a parallel plate capacitor with dielectric (oil) between its plates is
$C=\frac{K{\epsilon }_{0}A}{d}........\left(i\right)$
Where ${\epsilon }_0=$ electric permitivity of free space
$K=$dielectric constant
$A=$ area of each plate of capacitor
$d=$distance between two plates
When dielectric (oil) is removed, so capacitance
$C_0=\frac{{\epsilon }_{0}A}{d}............\left(ii\right)$
Comparing Eqs. $\left(i\right)$ and $\left(ii\right)$ we get
$C=K{C}_0$
$\Rightarrow$ $C_0=\frac{C}{K}=\frac{C}{2}$ ($\because K=2$)