Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1/2) QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor 1/2.
Let the force applied to separate the plates of a parallel plate capacitor by a distance of 
x be F. Hence, work done = F
x.
The increase in potential energy of the capacitor = u
v = u
A
x
Where, u = Energy density, A = area of each plate and v = volume between plates of capacitors.
Since, work done will be equal to the increase in the potential energy i.e.,
work done = increase in potential energy of the capacitor
F
x = u
A
x
F = u
A
Since, u = 
and 
= E (Electric intensity or Electric field)
and 
Therefore, F = 1/2 (CV)E and Q = CV
Thus, F = 
QE
The physical origin of the factor in the force formula lies in the fact that just outside the conductor, electric field is E and inside it is zero. Hence, it is the average value 
of the field that contributes to the force.