Physics - Capacitor Question with Solution | TestHub

Upon inserting dielectric we get two capacitors connected in parallel. Plate area of each capacitor is half of the old capacitor while the plate separation remains same New capacitance ${\mathrm{C}}^{\prime }={\mathrm{C}}_1+{\mathrm{C}}_2$ $\Rightarrow {\mathrm{C}}^{\prime }=\frac{{\epsilon }_0{\mathrm{k}}_1(\mathrm{A}/2)}{\mathrm{d}}+\frac{{\epsilon }_0{\mathrm{k}}_2(\mathrm{A}/2)}{\mathrm{d}}$ $\Rightarrow {\mathrm{C}}^{\prime }=\frac{{\mathrm{k}}_1+{\mathrm{k}}_2}{2}\left(\frac{{\epsilon }_0\mathrm{A}}{\mathrm{d}}\right)=\frac{{\mathrm{k}}_1+{\mathrm{k}}_2}{2}\mathrm{C}$ (Option A) Here C is the old capacitance (without dielectrics)

Since both ${\mathrm{C}}_1$ and ${\mathrm{C}}_2$ are in parallel they have same potential difference (V)

${\mathrm{E}}_1=\frac{\mathrm{V}}{\mathrm{d}}={\mathrm{E}}_2$

Hence, option (D) is correct. Charge on both the capacitors will be different

Hence ${\sigma }_1\ne {\sigma }_2$