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steposvetlana [31]

3 years ago

10

An ideal parallel-plate capacitor consists of a set of two parallel plates of area Separated by a very small distance 푑. This ca

pacitor is connected to a battery that charges the capacitor such that the energy stored in the capacitor is 푈'. Now the battery is disconnected and the separation between the plates is doubled, how much energy is stored in the capacitor?

1 answer:

dolphi86 [110]3 years ago

8 0

Answer:

doubled the initial value

Explanation:

Let the area of plates be A and the separation between them is d.

Let V be the potential difference of the battery.

The energy stored in the capacitor is given by

U = Q^2/2C ...(1)

Now the battery is disconnected, it means the charge is constant.

the separation between the plates is doubled.

The capacitance of the parallel plate capacitor is inversely proportional to the distance between the plates.

C' = C/2

the new energy stored

U' = Q^2 / 2C'

U' = Q^2/C = 2 U

The energy stored in the capacitor is doubled the initial amount.

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