Question

A parallel-plate capacitor has capacitance 7.50 ?F.

(a) How much energy is stored in the capacitor if it is connected to a 15.00-V battery?
?J

(b) If the battery is disconnected and the distance between the charged plates doubled, what is the energy stored?
?J

(c) The battery is subsequently reattached to the capacitor, but the plate separation remains as in part (b). How much energy is stored?
?J

Answer 1

Note that

E = 1/2 C V^2

Thus,

E = 8.4375*10^-4 J or 844 uJ [ANSWER, PART A]

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The capacitance will be halved if the distance is doubled. As the charge remains constant and

E = Q^2 / 2C

energy will be doubled as C is halved. Thus,

E = 1688 uJ   [ANSWER, PART B]

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Being attached again to the battery,

E = 1/2 C V^2

As the capacitance is now halved, the energy is half that of part A,

E = 422 uJ   [ANSWER, PART C]