Question

A 400 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?

Answer 1

Answer:

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)

$W=\dfrac{Q^2}{2C}\left ( \dfrac{C}{C'} -1\right )$

1/2 CV²=Q²/(2C)

$W=\dfrac{CV^2}{2}\left ( \dfrac{C}{C'} -\right )$

We know that

$C=\dfrac{\varepsilon _oA}{d}$

$C'=\dfrac{\varepsilon _oA}{d'}$

Given that

d'=2mm ,d= 1mm

C/C'= d'/d= 2

$W=\dfrac{CV^2}{2}\left ( \dfrac{C}{C'} -\right )$

By putting the values

$W=\dfrac{400\times 10^{-6}\times 100^2}{2}( 2 -1 )$

W=2 J