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Assume you have a simple battery-bulb-capacitor circuit. If you replace the capacitor with one that has twice the area as the old capacitor, the bulb will stay on longer because there has to be more charge to accumulate on the plates of the capacitor to zero out the net electric field. Mathematically speaking in (Q/A)/2εo (equation of the fringe field of a capacitor) if A increases, Q has to also to get a net fringe equivalent to the electric field inside the wire. Since current stays the same and more charge has to accumulate, it takes longer.