Question

Two identical thin metal plates has charge q₁ and q₂ respectively such that q₁ > q₂. The plates were brought close to each other to form a parallel plate capacitor of capacitance C. The potential difference between them is

• A. (q1+q2/C)
• B. (q1-q2/C)
• C. (q1-q2/2C)
• D. 2(q1-q2/C)

Answer 1

The Correct Option is C

To find the potential difference between two identical thin metal plates with charges \(q_1\) and \(q_2\), we consider the formation of a parallel plate capacitor. Given the situation, the potential difference \(V\) between the plates is determined by the formula:

\(V = \frac{q_{\text{net}}}{C}\) 

where:

• \(q_{\text{net}}\) is the net charge on the plates.
• \(C\) is the capacitance of the capacitor formed.

Initially, the charges on the two plates are \(q_1\) and \(q_2\), respectively, with \(q_1 > q_2\). When two plates are brought closer, they induce charges on each other and redistribute. The net charge that causes the potential difference is the difference in charges, that is:

\(q_{\text{net}} = q_1 - q_2\)

However, due to the induction process, the effective potential difference caused by the charge difference is halved for a parallel plate capacitor. Therefore, the actual potential difference \(V\) would be:

\(V = \frac{q_1 - q_2}{2C}\)

This expression matches with one of the provided options. Therefore, the correct answer is as follows:

Correct Answer: \(\frac{q_1-q_2}{2C}\)

Conclusion: The potential difference between the two plates, considering the induction effects in a parallel plate configuration, is given by the expression \(\frac{q_1-q_2}{2C}\).

Answer 2

E=q1−q2/2ε0A\(E=\frac{q_1−q_2}{2ε_0A}\)
\(v=\frac{(q1−q2)d}{2ε0A}\)
= \(\frac{q_1-q_2}{2C}\)
So, the correct option is (C): (q1-q2)/2C