A parallel-plate capacitor has plates with an area of 395 cm 2 and an air-filled

Question

A parallel-plate capacitor has plates with an area of 395 cm2 and an air-filled gap between the plates that is 2.29 mm thick. The capacitor is charged by a battery to 575 V and then is disconnected from the battery.   (a) How much energy is stored in the capacitor?
1      J

(b) The separation between the plates is now increased to 4.58 mm. How much energy is stored in the capacitor now?
2      J

(c) How much work is required to increase the separation of the plates from 2.29 mm to 4.58 mm?
3      J

Explain your reasoning.
4 (a) How much energy is stored in the capacitor?
1      J

(b) The separation between the plates is now increased to 4.58 mm. How much energy is stored in the capacitor now?
2      J

(c) How much work is required to increase the separation of the plates from 2.29 mm to 4.58 mm?
3      J

Explain your reasoning.
4

Explanation / Answer

C=eo*A/d..

C=152.6 pF......


energy stored u=(1/2)*C*V^2=2.52*10^-5 J.....


B)U1*d1=U2*d2....
U2=1.26*10^-5 J.....

C) W=U1-U2=(2.52-1.26)=1.26*10^-5 J

4) explanation : work done is positive because if separation incresed means capacitance is decreased.....
we are doing work to move from high capacitance to low capacitance

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