An uncharged parallel plate capacitor having its lower end fixed and upper end is attached with spring having spring constant K. Upper plate is in equilibrium before switch is closed. After switch is closed, the condition on the potential of battery so that the system can acquire new equilibrium position is  $V\le \sqrt{{\left(\frac{p}{q}\right)}^r\frac{L^{3}K}{{\epsilon }_{0}A}}.$ Then  $p+q+r$  is? [ p & q are smallest integers and V is potential across the battery]

 

Let in new equilibrium the upper plate is displaced by x units.
If we get real x less than L then only new equilibrium is achieved.
Here, force between plates of capacitor = spring force
$Kx=\frac{Q^2}{2A{\epsilon }_0},$   where, x = extension in spring and A is area of the plate.
We know that, after connecting the switch,  $Q=C\text{'}V$
 So,  $Kx=\frac{1}{2A{\epsilon }_0}{\left[\frac{A{\epsilon }_0}{L-z}V\right]}^2$
Maximum value of  $K_x{(L-x)}^2$
We get  $K{(L-x)}^2+2K(L-x)(-1)=0$
 $x=L/3$
Maximum value of    $Kx{(L-x)}^2=K\frac{4L^3}{27}$
So for real x,
  $K\frac{4L^3}{27}\ge \frac{A{\epsilon }_0}{2}{V}^2$  $\Rightarrow V\le \sqrt{{\left(\frac{2}{3}\right)}^3\frac{L^{3}K}{A{\epsilon }_0}}$  
Hence, the value of  $p+q+r=8$