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)
[tex]W=\dfrac{Q^2}{2C}\left ( \dfrac{C}{C'} -1\right )[/tex]
1/2 CV²=Q²/(2C)
[tex]W=\dfrac{CV^2}{2}\left ( \dfrac{C}{C'} -\right )[/tex]
We know that
[tex]C=\dfrac{\varepsilon _oA}{d}[/tex]
[tex]C'=\dfrac{\varepsilon _oA}{d'}[/tex]
Given that
d'=2mm ,d= 1mm
C/C'= d'/d= 2
[tex]W=\dfrac{CV^2}{2}\left ( \dfrac{C}{C'} -\right )[/tex]
By putting the values
[tex]W=\dfrac{400\times 10^{-6}\times 100^2}{2}( 2 -1 )[/tex]
W=2 J
