Physics - Capacitor Question with Solution | TestHub

$\mathrm{C}=\frac{\mathrm{A}{\epsilon }_0}{\mathrm{d}}$

$$\begin{array}{rlr}& \frac{1}{C_{eq}}=\frac{1}{C}+\frac{2}{3C}=\frac{5}{3C}& C_{eq}=\frac{3C}{5}=\frac{3A{\epsilon }_0}{5d}\end{array}$$

Alternative Method : $\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}=\frac{\mathrm{x}+\mathrm{y}}{{\mathrm{V}}_{\mathrm{AB}}}$

$\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}=\frac{\mathrm{x}+\mathrm{y}}{{\mathrm{V}}_{\mathrm{AB}}}$

Potential difference between 1 and 4 is same

$\frac{\mathrm{y}}{\mathrm{A}{\epsilon }_0}=\frac{2\mathrm{x}}{\mathrm{A}{\epsilon }_0}$

$\mathrm{y}=2\mathrm{x}$

$\mathrm{V}=\left(\frac{2\mathrm{y}+\mathrm{x}}{\mathrm{A}{\epsilon }_0}\right)\mathrm{d}$ $\mathrm{C}=\frac{(\mathrm{x}+2\mathrm{x})\mathrm{xA}{\epsilon }_0}{(5\mathrm{x})\mathrm{d}}=\frac{3\mathrm{A}{\epsilon }_0}{5\mathrm{d}}$.