A capacitor consists of two 7.0-cm-diameter circular plates separated by 1.0 mm. The plates are charged to 160 V, then the battery is removed.

- How much energy is stored in the capacitor?
- How much work must be done to pull the plates apart to where the distance between them is 2.0 mm?

## Question 1.

[latex]

d=1 mm=1*10^{-3}m \\

D=7cm=7*10^{-2}m \\

V=160V

[/latex]

The Capacitance of Parallel Plate capacitor is given by:

[latex]

C=\frac{(8.85*10^{-12}\frac{F}{m})}{1*10^{-3}m}*\pi(\frac{7*10^{-2}}{2}*m)^2=3.4*10^{-11}*F=34pF

[/latex]

The energy stored in parallel plate Capacitor is given by:

[latex]

U=1/2*CV^{2}=4.35*10^{-7}FV^{2}=4.35*10^{-7}J

[/latex]

Hence energy stored in Capacitor is

[latex]

U=4.35*10^{-7}J

[/latex]

## Question 2.

When the plates are pulled apart by distance:

[latex]

(d_{n}=2mm=2*10^{-3}m=2d

[/latex]

The new capacitance is given by:

[latex]

C_{n}=\frac{C}{2}

[/latex]

Since the battery is removed, the process takes place with a constant charge on the capacitor.

Therefore new energy given by:

[latex]

U_{n}=\frac{Q^{2}}{2C_{n}}=2U

[/latex]

This change is energy of the capacitor is accounted for the work done to pull plates apart,

therefore work done is:

[latex]

W=U_{n}-U=2U-U=U=4.35*10^{-7}J

[/latex]