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Two identical conducting plates carry positive charges Q1 and Q2, with Q2 < Q1. When they are placed close together so that they act as a parallel-plate capacitor of capacitance C, what is the potential difference between the plates?

1. (Q1 + Q2) / 2C
2. (Q1 + Q2) / C
3. (Q1 - Q2) / C
4. (Q1 - Q2) / 2C

Correct answer: (Q1 - Q2) / 2C

Solution

For two charged conducting plates forming a capacitor, the potential difference is V = (Q1 - Q2)/(2C).

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