Question

Two identical metal plates are given positive charge $Q_1$ and $Q_2$ $(<Q_1)$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance $C$, the potential difference between them is

• A. $(Q_1+Q_2)/2C$
• B. $(Q_1+Q_2)/C$
• C. $(Q_1-Q_2)/C$
• D. $(Q_1-Q_2)/\left(2C\right)$

Answer 1

The correct option is D $(Q_1-Q_2)/\left(2C\right)$

The potential difference between the two identical metal plates is given as

$C=\frac{{\epsilon }_{0}A}{d}$

Let the surface charge density is given as

${\sigma }_1={\sigma }_{21}=\frac{Q}{A}$

The net electric field is

$E_{net}=\frac{{\sigma }_1-{\sigma }_2}{2{\epsilon }_0}$

We know the potential difference is given as

$V=E.d$

By substituting the above values we get

$V=\frac{Q_1-Q_2}{2C}$