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# How much work must be done to pull the plates apart to where the distance between them is 2.0 mm?

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.

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

## Question 1.

$d=1 mm=1*10^{-3}m \\ D=7cm=7*10^{-2}m \\ V=160V$

The Capacitance of Parallel Plate capacitor is given by:

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

The energy stored in parallel plate Capacitor is given by:

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

Hence energy stored in Capacitor is

$U=4.35*10^{-7}J$

## Question 2.

When the plates are pulled apart by distance:
$(d_{n}=2mm=2*10^{-3}m=2d$

The new capacitance is given by:

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

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

Therefore new energy given by:

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

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

therefore work done is:

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