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 $2d$. What is the work required to separate the plates?

• A. $2{\epsilon }_{0}A{V}^2/d$
• B. ${\epsilon }_{0}A{V}^2/d$
• C. $3{\epsilon }_{0}A{V}^2/2d$
• D. ${\epsilon }_{0}A{V}^2/2d$

Answer 1

Answer: (D)

$W=U_2-U_1=\frac{q^2}{2}[\frac{1}{C_2}-\frac{1}{C_1}]$
$C_1=\frac{{\epsilon }_{0}A}{d},C_2=\frac{C_1}{2}=\frac{{\epsilon }_{0}A}{2d}$
$q=C_{1}V=\frac{{\epsilon }_{0}AV}{d}$
Solve to get $W=\frac{1}{2}\frac{{\epsilon }_{0}A{V}^2}{d}$
Alternatively:
$W=Fd=\frac{Q^2}{2A{\epsilon }_0}d$
$=\frac{C_1^2{V}^2}{2{\epsilon }_{0}A}d=\frac{1}{2}\frac{{\epsilon }_{0}A{V}^2}{d}$