Physics - Capacitor Question with Solution | TestHub

The initial capacitance of the parallel plate air capacitor is given by:

$C_0=\frac{{\epsilon }_{0}A}{d}=9\text{ pF}$

When the space is filled with two dielectrics of thicknesses $d_1$ and $d_2$, it can be considered as two capacitors connected in series.

The capacitance of the first part is:

$C_1=\frac{K_1{\epsilon }_{0}A}{d_1}=\frac{3{\epsilon }_{0}A}{d/3}=\frac{9{\epsilon }_{0}A}{d}=9C_0$

The capacitance of the second part is:

$C_2=\frac{K_2{\epsilon }_{0}A}{d_2}=\frac{6{\epsilon }_{0}A}{2d/3}=\frac{9{\epsilon }_{0}A}{d}=9C_0$

The equivalent capacitance $C$ of the series combination is given by:

$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$

$\frac{1}{C}=\frac{1}{9C_0}+\frac{1}{9C_0}=\frac{2}{9C_0}$

$C=\frac{9C_0}{2}$

Substituting the value of $C_0=9\text{ pF}$:

$C=\frac{9\times 9}{2}=\frac{81}{2}=40.5\text{ pF}$

Answer: 40.5 pF