A parallel-plate capacitor has capacitance 3.50 µF. (a) How much energy is store

Question

A parallel-plate capacitor has capacitance 3.50 µF. (a) How much energy is stored in the capacitorif it is connected to a 15.00 Vbattery?
1 µJ

(b) If the battery is disconnected and the distance between thecharged plates doubled, what is the energy stored?
2 µJ

(c) The battery is subsequently reattached to the capacitor, butthe plate separation remains as in part (b). How much energy isstored?
3 µJ (a) How much energy is stored in the capacitorif it is connected to a 15.00 Vbattery?
1 µJ

(b) If the battery is disconnected and the distance between thecharged plates doubled, what is the energy stored?
2 µJ

(c) The battery is subsequently reattached to the capacitor, butthe plate separation remains as in part (b). How much energy isstored?
3 µJ

Explanation / Answer

(a)   energy stored = (1/2) CV2 = (1/2) * 3.50 *15.002 =    393.75 J . (b)   if the distance between the plates is doubled,the stored energy increases by a factor of 2... it isnow .          393.75 *2   =     787.5 J . (c)   if the battery is reattached, the energyactually decreases!! The capacitor recharges the battery... and theenergy stored becomes one-half of the original,or     (1/2) * 393.75 =    196.88 J

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