# 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?

Part 1:


\varepsilon_{0}- Dielectric constant\\
A - Area of plate\\
d -Distance between plates\\
V - Voltage\\
D -Diameter of circular plate\\
Given:\\
d=1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m}\\
D=7 \mathrm{~cm}=7 \times 10^{-2} \mathrm{~m}\\
V=160 \mathrm{~V}\\

\text{For circular plate, Area is calculated by.}\\
A=\pi r^{2}=\pi\left(\frac{D}{2}\right)^{2}\\
=\pi\left(\frac{7 \times 10^{-2}}{2}\right)^{2} \mathrm{~m}^{2}\\
\text{The Capacitance of parallel Plate capacitor is given by,}\\
C=\frac{\varepsilon_{0} A}{d}\\
=\frac{\left(8.85 \times 10^{-12} \frac{F}{m}\right)}{1 \times 10^{-3} \mathrm{~m}} \pi\left(\frac{7 \times 10^{-2}}{2} m\right)^{2}\\
=3.4 \times 10^{-11} F\\
=34 \mathrm{pF}\\
\text{The energy stored in parallel plate Capacitor is given by,}\\
U=\frac{1}{2} C V^{2}\\
=\frac{1}{2}\left(3.4 \times 10^{-11} F\right)(160 \mathrm{~V})^{2}\\
=4.35 \times 10^{-7} \mathrm{FV}^{2}\\
=4.35 \times 10^{-7} J\\
\text{Hence energy stored in Capacitor is } U=4.35 \times 10^{-7} \mathrm{D}\\

Part 2:

\text{When the plates are pulled apart by distance } d_{n}.\\
d_{n}=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}=2 \mathrm{~d}\\
\text{The new capacitance is given by.}\\
\begin{aligned} C_{n} &=\frac{\varepsilon_{0} A}{d_{n}} \\ &=\frac{\varepsilon_{0} A}{2 d}=\frac{C}{2} \end{aligned}\\
\text{Since the battery is removed, the process takes place with constant charge on the capacitor.}\\
\text{Therefore new energy given by,}\\
\begin{aligned} U_{n} &=\frac{Q^{2}}{2 C_{n}} \\ &=\frac{Q^{2}}{2\left(\frac{C}{2}\right)} \\ &=2\left(\frac{Q^{2}}{2 C}\right) \\ &=2 U \end{aligned}\\
\text{This change is energy of the capacitor is accounted for the work done to pull plates apart,}\\
\text{Therefore work done is,}\\

W =U_{n}-U \\
=2 U-U \\
=U \\
=4.35 \times 10^{-7} \mathrm{~J}\\

\text{Hence require work to pull plates apart for given case is, } W=4.35 \times 10^{-7} \mathrm{~J}

### Raymond Puzio

Raymond Puzio has a PhD in Physics from Yale University. I have been creating PlanetPhysics with Aaron Krowne and Ben Loftin since 2005.