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A parallel plate capacitor has 1 μF capacitance. One of its two plates is given +2 μC charge and the other plate, +4 μC charge. The potential difference developed across the capacitor is-

1. 1 V
2. 2 V
3. 3 V
4. 5 V

Correct answer: 1 V

Solution

The potential difference across a capacitor is determined by the formula V = Q/C, where Q is the charge on one plate and C is the capacitance. In this case, using the charge of +2 μC and the capacitance of 1 μF, we find V = 2 μC / 1 μF = 2 V, but since the charges are not equal, we consider the effective charge difference, leading to a potential difference of 1 V.

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