Consider a capacitor that consists of two parallel plates,one of which can move relative to the other.Air fills the space between the plates,the capacitance is 45.2 pF when the separation between plates is d = 0.360 cm.A battery with potential difference V=12.0 V is connected to the plates to fully charge them and then it is disconnected from the plates.The plates are moved to a separation of 0.180 cm.What is the potential difference V' between the plates now?

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Quizplus · Accepted Answer

The capacitance of a parallel-plate capacitor is given by:C = ε₀A/dwhere ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.If the capacitance is doubled, then the distance between the plates must be halved. Therefore, the new distance between the plates is:d' = d/2 = 0.180 cmThe potential difference between the plates is given by:V = Q/Cwhere Q is the charge on the plates.Since the charge on the plates is constant, the potential difference between the plates will be halved when the capacitance is doubled. Therefore, the new potential difference between the plates is:V' = V/2 = 6 V

Consider a Capacitor That Consists of Two Parallel Plates,one of Which