Question

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to $\left(1/2\right)QE,$ where $Q$ is the charge on the capacitor, and $E$ is the magnitude of the electric field between the plates. Explain the origin of the factor $1/2$.

Answer 1

Step 1: Find the relation between work and energy
Let, force applied to separate plates of parallel plate capacitor $=F$
Distance of separation of plates $=\Delta x$
$\because$ Work done = Force $\times$ displacement $W=F\Delta x$
Increase in potential energy $=uA\Delta x$
Where $A$ = Area of each plate.
$u=$ Energy density
This work done will be equal to the increase in the potential energy i.e.

$F\Delta x=uA\Delta x$...$\left(i\right)$

$F=uA=\frac{1}{2}{\epsilon }_0{E}^{2}A$...$\left(ii\right)$


Step 2: find the force in terms electric intensity
As we know that electric intensity is given by,
$E=\frac{V}{d}$;
$V=$ Potential difference
$d=$ distance between the plates
Substituting the values in equation $\left(ii\right)$, we get

$F=\frac{1}{2}{\epsilon }_0\left(\frac{V^2}{d^2}\right)A$

$F=\frac{1}{2}{\epsilon }_0\left(\frac{V}{d}\right)EA$...$\left(iii\right)$


Step 3: Find the force in terms of capacitance
As the capacitance is given by, $C=\frac{{\epsilon }_{0}A}{x}$
Therefore $F=\frac{1}{2}\left(CV\right)E$......$\left(iv\right)$ ,
$(C=\frac{Q}{V})$
$\therefore$ from equation $\left(iv\right)$

$F=\frac{1}{2}\frac{Q}{V}VE$

$F=\frac{1}{2}QE$