Question

A capacitance is formed by two identical metal plates. The plates are given charges $Q_1$ and $Q_2(Q_2<Q_1)$ . If capacitance of the capacitor is $C,$ what is the p.d. between the plates?

• A. $\frac{Q_1+Q_2}{2C}$
• B. $\frac{Q_1+Q_2}{C}$
• C. $\frac{Q_1-Q_2}{C}$
• D. $\frac{Q_1-Q_2}{2C}$

Answer 1

The correct option is D $\frac{Q_1-Q_2}{2C}$

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

${\sigma }_1=\frac{Q_1}{A}$

${\sigma }_2=\frac{Q_2}{A}$

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

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