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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 capacitor of capacitance C. The potential difference between them is:

1. (q1 + q2) / C
2. (q1 - q2) / C
3. (q1 - q2) / 2C
4. 2(q1 - q2) / C

Correct answer: (q1 - q2) / 2C

Solution

The potential difference between the plates of a capacitor is determined by the difference in charge between them divided by the capacitance. Since the plates have charges q1 and q2, the effective potential difference is (q1 - q2) divided by the total capacitance, which in this case is 2C due to the configuration of the plates, leading to the correct expression being (q1 - q2) / 2C.

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