Electrostatics Question 636
Question: A parallel plate capacitor has plates of area $ A $ and separation $ d $ and is charged to a potential difference$ V $ . The charging battery is then disconnected and the plates are pulled apart until their separation is 2$ d $ . What is the work required to separate the plates?
Options:
A) $ 2{\varepsilon _{0}}AV^{2}/d $
B) $ {\varepsilon _{0}}AV^{2}/d $
C) $ 3{\varepsilon _{0}}AV^{2}/d $
D) $ {\varepsilon _{0}}AV^{2}/2d $
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Answer:
Correct Answer: D
Solution:
[d] $ W=U _{2}-U _{1}=\frac{q^{2}}{2}[ \frac{1}{C _{2}}-\frac{1}{C _{1}} ] $
$ C _{1}=\frac{{\varepsilon _{0}}A}{d},C _{2}=\frac{C _{1}}{2}=\frac{{\varepsilon _{0}}A}{2d} $
$ q=C _{1}V=\frac{{\varepsilon _{0}}AV}{d} $
Solve to get $ W=\frac{1}{2}\frac{{\varepsilon _{0}}AV^{2}}{d} $
Alternatively: $ W=Fd=\frac{Q^{2}}{2A{\varepsilon _{0}}}d=\frac{C^{2} _{1}V^{2}}{2{\varepsilon _{0}}A}d=\frac{1}{2}\frac{{\varepsilon _{0}}AV^{2}}{d} $
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