Question

Two identical metal plates are given positive charges ${Q_1}$ 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. $\dfrac{{({Q_1} + {Q_2})}}{{2C}}$
B. $\dfrac{{({Q_1} + {Q_2})}}{C}$
C. $\dfrac{{({Q_1} - {Q_2})}}{C}$
D. $\dfrac{{({Q_1} - {Q_2})}}{{2C}}$

Answer 1

When we bring two charged metals together, they touch each other then charge transfer takes place between the metals unless the charge on both become equal. but when these charged metals are kept close to each other (not attached) then charges get induced on each. The formula of the potential across the capacitor plates in terms of capacitance is to be used here.

Complete step by step solution:
It is given that the charges on the metals are ${Q_1}$ and ${Q_2}$. Also ${Q_1} > {Q_2}$. Now, when these charged plates are brought together to form a parallel plate capacitor, the charge will induce the charge on both the metal plates.
Therefore, the charge induced on the first plate is $- {Q_2}$. And the total charge on the first plate is now
${Q_1} - {Q_2}$.
Similarly, the charge induced on the second plate is $- {Q_1}$. And the total charge on the second plate is no ${Q_2} - {Q_1} = - \left( {{Q_1} - {Q_2}} \right)$.
So, now both plates have equal and opposite charges which is the condition for a parallel plate capacitor to form.
Now, we have the following relation to finding the potential across the capacitor plates.
$V = \dfrac{{{Q_1} - {Q_2}}}{C}$
Here, $C$ is the capacitance of the parallel plate capacitor.
Hence, the correct option is (C) $\dfrac{{{Q_1} - {Q_2}}}{C}$.

Note:
The capacitor is an electronic component that is used to store energy in the form of an electric field. The capacitor is used in those devices that need a large amount of energy to get started. A very common example is the ceiling fan in our house.