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WAEC - Physics (2001 - No. 33)

As the plates of a charged variable capacitor are moved closer together, the potential difference between them
increased
decrease
remains the same
is doubled

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As the plates of a charged variable capacitor are moved closer together, the potential difference between them decreases.

This is because the capacitance increases as the distance between the plates decreases. For a constant charge, the potential difference (V = \(\frac{\text{Q}}{\text{C}}\)) will decrease when capacitance (C) increases.

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