A parallel plate capacitor has 1μF capacitance. One of its two plates is given +2μC charge and the other plate, +4μC charge. Find the potential difference developed across the capacitor.

$$\begin{gathered} \because \text{ Electric field due to cap plate is }\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0} \\ \mathrm{Since} \mathrm{both} \mathrm{the} \mathrm{plate} \mathrm{has} \mathrm{given} \mathrm{positive} \mathrm{charges}. \\ \mathrm{E}=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0} \\ \text{ Net Electric .field. }\frac{{\mathrm{\sigma }}_2}{2{\mathrm{\epsilon }}_0}-\frac{{\mathrm{\sigma }}_1}{2{\mathrm{\epsilon }}_0}\mathrm{\sigma }=\frac{\mathrm{Q}}{\mathrm{A}} \\ {\mathrm{E}}_{\mathrm{net}}=\frac{{\mathrm{Q}}_2}{2{\mathrm{AE}}_0}-\frac{{\mathrm{Q}}_1}{2{\mathrm{AE}}_0} \\ {\mathrm{E}}_{\text{net }}=\frac{1}{1{\mathrm{AE}}_0}\left({\mathrm{Q}}_2-{\mathrm{Q}}_1\right)\text{ here }{\mathrm{Q}}_2=4\mathrm{\mu c}\text{ Here } \\ {\mathrm{Q}}_2=2\mathrm{\mu c} \\ {\mathrm{E}}_{\text{net }}=\frac{1}{2{\mathrm{AE}}_0}(4-2)=\frac{1}{{\mathrm{AE}}_0} \\ \text{ Also }\mathrm{V}=\mathrm{ED} \\ ={\mathrm{E}}_{\text{net }}\times \mathrm{d}=\frac{1}{{\mathrm{AE}}_0}\times \mathrm{d}=\frac{1}{\mathrm{c}} \\ \because \mathrm{C}=\frac{{\mathrm{E}}_0\mathrm{A}}{\mathrm{d}} \\ \because \mathrm{C}=1\mathrm{\mu F} \\ \mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}}=\frac{1\mathrm{\mu C}}{1\mathrm{\mu F}}=1\mathrm{Volt} \end{gathered}$$