Physics - Capacitor Question with Solution | TestHub

The two regions of capacitor can be considered in series to calculate total capacitance of the arrangement.

$\frac{1}{C_{\text{eq }}}=\frac{d-y}{{\epsilon }_{0}A}+\frac{y}{K{\epsilon }_{0}A}=\frac{d-y+y/2}{{\epsilon }_{0}A}$

$\Rightarrow C_{eq}=\frac{{\epsilon }_{0}A}{d-y/2}$

Here $y=\left(\frac{d}{3}-vt\right)$

$\therefore {\mathrm{C}}_{\text{eq }}=\frac{{\epsilon }_0\mathrm{A}}{\mathrm{d}-\frac{\mathrm{d}}{6}+\frac{\mathrm{vt}}{2}}=\frac{6{\epsilon }_0\mathrm{A}}{5\mathrm{d}+3\mathrm{vt}}$

Hence time constant $(\tau )=\mathrm{RC}$

$\Rightarrow \tau =\frac{6{\epsilon }_0\mathrm{AR}}{5\mathrm{d}+3\mathrm{vt}}$

Hence option (A) is correct for area $=\mathrm{A}=1$ unit.