A400 µf capacitor is constructed out of two parallel plates of very large area which are separated by 1 mm. a battery is used to charge the capacitor to a potential of 100 v . (a) if the capacitor is disconnected from the battery, how much work (in joules) must be done to pull the plates apart to a separation of 2 mm?

W=2 J

Explanation:

Given that

C= 400 µF

V= 100 V

d= 1 mm

Energy before disconnected from the battery

U= 1/2 CV²=Q²/(2C)

Energy after  disconnected from the battery

U'=1/2 C'V²=Q²/(2C')

The work done W

W= U' - U

W= Q²/(2C') - Q²/(2C)

1/2 CV²=Q²/(2C)

We know that

Given that

d'=2mm ,d= 1mm

C/C'= d'/d= 2

By putting the values

W=2 J

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