A capacitor consists of two large parallel plates of area A separated by a very small distance d.This capacitor is connected to a battery and charged until its plates carry charges +Q and -Q,and then disconnected from the battery.If the separation between the plates is now doubled,the potential difference between the plates will

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Quizplus · Accepted Answer

The potential difference (V) between the plates of a capacitor is given by V = Q/C, where Q is the charge on the plates and C is the capacitance. The capacitance of a parallel plate capacitor is given by C = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of the plates, and d is the separation between the plates. When the separation (d) is doubled, the new capacitance (C') becomes ε₀A/(2d), which is half of the original capacitance. Since the charge (Q) remains the same (because the capacitor is disconnected from the battery), the new potential difference (V') is twice the original potential difference (V), because V' = Q/C' and C' is half of C.

A Capacitor Consists of Two Large Parallel Plates of Area