Class 12 Physics, CBSE and UP Board, Electrostatic Potential and Capacitance, Additional Exercise,Chapter-2 Exercise 2.28, NCERT Solutions for Class 12 Physics
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.
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Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of Δx. Hence, work done by the force to do so = FΔx
As a result, the potential energy of the capacitor increases by an amount given as uAΔx.
Where,
u = Energy density,
A = Area of each plate,
d = Distance between the plates
V = Potential difference across the plates
The work done will be equal to the increase in the potential energy i.e.,
FΔx = uAΔx
F= uA= (1/2 ε0 E²) A
Electric intensity is given by,
E=V/d
Therefore F = 1/2 ε0 (V/d)EA =1/2 ( ε0 A V/d)E
However ,capacitance , C = ε0 A/d
Therefore F = 1/2 (CV)E
Charge on capacitor is given by , Q=CV
Therefore F =1/2 QE
The physical origin of the factor, 1/2, in the force formula lies in the fact that just outside the conductor, field is E and inside it is zero. Hence, it is the average value, E/2, of the field that contributes to the force.