Physics - Capacitor Question with Solution | TestHub

The formula to calculate the capacitance of a parallel plate capacitor having thickness $d$ and with a dielectric of thickness $x$ between the plates is given by

$C=\frac{{\epsilon }_{0}A}{\left(d-x\right)+{\displaystyle \frac{x}{K}}} ...\left(1\right)$

Substitute the values of the parameters for first case into equation (1) to obtain the capacitance.

$\begin{array}{rcl}{C}_1& =& \frac{{\epsilon }_{0}A}{{\displaystyle \frac{2d}{3}}+\frac{{\displaystyle \frac{d}{3}}}{4}}\\ & =& \frac{4{\epsilon }_{0}A}{3d} ...\left(2\right)\end{array}$

Similarly, substitute the parameters into equation (1) for the second case to obtain the capacitance

$\begin{array}{rcl}{C}_2& =& \frac{{\epsilon }_{0}A}{{\displaystyle \frac{d}{3}}+\frac{{\displaystyle \frac{2d}{3}}}{4}}\\ & =& \frac{4{\epsilon }_{0}A}{2d}\\ & =& \frac{2{\epsilon }_{0}A}{d} ...\left(3\right)\end{array}$

Divide equation (3) by equation (2) to obtain the capacitance in the second case.

$\begin{array}{rcl}\frac{C_2}{C_1}& =& \frac{\frac{2{\epsilon }_{0}A}{d}}{\frac{4{\epsilon }_{0}A}{3d}}\\ & =& \frac{3}{2}\\ C_2& =& \frac{3}{2}{C}_1 ...\left(4\right)\end{array}$

Substitute the value of the capacitance in the first case into equation (4) to obtain the value of the capacitance in the second case.

$\begin{array}{rcl}{C}_2& =& \frac{3}{2}\times 2 \mathrm{\mu F}\\ & =& 3 \mathrm{\mu F}\end{array}$