Question

A parallel plate capacitor with plate area $A$ and plate separation $d=2\text{m}$ has a capacitance of $4\mu F$. The new capacitance of the system, if half of the space between them is filled with a dielectric material of dielectric constant $K=3$ (as shown in figure) will be:

Answer 1

Original capacitance, $C=\frac{{\in }_{0}A}{d}$

Capacitance for other half, $C_1=\frac{{\in }_{0}A}{\frac{d}{2}}=\frac{2{\in }_{0}A}{d}$

Capacitance of half of the capacitor, $C_2=\frac{K{\in }_{0}A}{\frac{d}{2}}=3\times \frac{2{\in }_{0}A}{d}$

Now both capacitors are in series connection. Their equivalent capacitance will be,

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$

$\frac{1}{C_{eq}}=\frac{d}{2{\in }_{0}A}+\frac{d}{6{\in }_{0}A}$

$C_{eq}=\frac{12}{8}\frac{{\in }_{0}A}{d}$

Equivalent circuit is


Now $\frac{{\in }_{0}A}{d}=4\mu F$
$\Rightarrow \frac{12}{8}\frac{{\in }_{0}A}{d}=6\mu F$