Question

A parallel-plate capacitor has plates of unequal area. The larger plate is connected to the positive terminal of the battery and the smaller plate to its negative terminal. Let $Q_{+}$ and $Q_{-}$ be the charges appearing on the positive and negative plates respectively. Calculate the potential difference between the two plates.

Answer 1

Let, the area of the larger plate be $A_{+}$ and that of the smaller plate be $A_{-}$

Capacitance ($C$)$=\frac{Q}{|\Delta V|}$

Potential difference $\Delta V=-\underset{0}{\overset{d}{\int }}E.dl$ ....($1$)

Let $d$ be the distance between the plates and $E_{+}$ be electric field due to large plate and $E_{-}$ be electric field due to smaller plate.

$\begin{array}{c}{E}_{+}=\frac{{\sigma }_{+}}{2{\epsilon }_o}=\frac{Q}{2A_{+}{\epsilon }_o}\\ E_{-}=\frac{{\sigma }_{-}}{2{\epsilon }_o}=\frac{Q}{2A_{-}{\epsilon }_o}\end{array}$

Where, $\sigma$ is charge density

Now, from ($1$)

$\begin{array}{c}\Delta V=-{\int }_0^{d}E.dl\\ =-{\int }_0^d\left(E_{+}+E_{-}\right).dl\\ =\frac{Q}{2{\epsilon }_o}\left(\frac{1}{A_{+}}+\frac{1}{A_{-}}\right){\int }_0^{d}dl\\ =\frac{Q}{2{\epsilon }_o}\left(\frac{A_{-}+A_{+}}{A_{+}{A}_{-}}\right)d.\end{array}$