Question

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

• 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

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

Step 1: Calculation of the net electric field

The net electric field due to the positive charges can be calculated as follows:

$$\begin{gathered} E=E_1-E_2 \\ E=\frac{Q_1}{2{\epsilon }_{0}A}-\frac{Q_2}{2{\epsilon }_{0}A} \\ E=\frac{1}{2{\epsilon }_{0}A}\left(Q_1-Q_2\right) \end{gathered}$$

Where, $E$ is the net electric field, $Q_1$ and $Q_2$ are the charges, ${\epsilon }_0$ is the proportionality constant, and $A$ is the area of each plate.

Step 2: Calculation of the potential difference

The potential difference between the plates can be calculated as follows:

$V=Ed$

Where, $V$ is the potential difference, and $d$ is the distance between the plates.

$$\begin{gathered} V=\frac{d}{2{\epsilon }_{0}A}\left(Q_1-Q_2\right) \\ V=\frac{Q_1-Q_2}{2C} \end{gathered}$$

Hence, option (D) is correct.