Question

The force with which the plates of a parallel plate capacitor, having charge Q and area of each plate A, attract each other is

• A. zero
• B. $\frac{Q^2}{{\epsilon }_{0}A}$
• C. $\frac{Q^2}{\sqrt{2}{\epsilon }_{0}A}$
  
• D. $\frac{Q^2}{2{\epsilon }_{0}A}$

Answer 1

The correct option is D $\frac{Q^2}{2{\epsilon }_{0}A}$

To find the force of attraction between the charged capacitor plates we will use a method called the method of virtual displacement. We simply equate the work $△W$ required to make a small change $△d$ in the plate separation d to the resulting change $△U$ in the stored energy, i.e.,
$△W$=$△U\left(i\right)$
If F is the magnitude of the force between the plates, then the work $△W$ done to increase the plate separation by $△d$ is given by
$△W$=F$△d\left(ii\right)$
Now we know that the energy U of a parallel-plate capacitor of plate area A and capacitance C is
$U=\frac{Q^2}{2C}=\frac{Q^{2}d}{2{\epsilon }_{0}A}$
Where Q is the charge on the capacitor plates. The increase $△U$ in U due to an increase $△d$ in d is, therefore, given by
$△U=\frac{Q^2△d}{2{\epsilon }_{0}A}\left(iii\right)$
Equating (ii) and (iii) we get
$F=\frac{Q^2}{2{\epsilon }_{0}A},whichischoice\left(d\right)$