Physics - Capacitor Question with Solution | TestHub

${\mathrm{C}}_1=\frac{\mathrm{A}{\epsilon }_0}{\mathrm{d}}$ ...(i) ${\mathrm{C}}_2=\frac{8\mathrm{A}{\epsilon }_0}{{\mathrm{d}}^{\prime }}\Rightarrow 16{\mathrm{C}}_1=\frac{8\mathrm{A}{\epsilon }_0}{{\mathrm{d}}^{\prime }}$ ...(iii) where ${\mathrm{d}}^{\prime }$ the unknown separation. Dividing (i) by (ii) $\frac{C_1}{16C_1}=\frac{A{\epsilon }_0}{d}\times \frac{d^{\prime }}{A{\epsilon }_0}$ $\frac{8\mathrm{d}}{16}={\mathrm{d}}^{\prime }$ $\therefore {\mathrm{d}}^{\prime }=\frac{\mathrm{d}}{2}$