An isolated parallel plate capacitor contains charge Q = +10 C on its plates, co

Question

An isolated parallel plate capacitor contains charge Q = +10 C on its plates, corresponding to a potential difference of 10 V. The initial separation between the plates is d. Later, the distance between the plates is increased by a factor of two while the charge on each plate remains unchanged. What is the voltage between the two plates after the separation becomes 2d?

Could you please explain how you obtained the answer as well... Thanks!

Could you please explain how you obtained the answer as well... Thanks!

Explanation / Answer

Q = CV

as distance between increases capacitance changes.

capacitance C = e0 A / d

initial capacitane is C then new capacitance C' = e0A / (2d) = C / 2

charge is same = (C x 10) = ((C/2) x V' )

V' = 20 Volt

ANs(D)

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