Show that the force on each plate of a parallel plate capacitor has a magnitude equal to $(\frac{1}{2}) QE,$ 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.

Solution

Let,
Surface charge density of the capacitor= σ
Area of cross-section of the plate= A
Using the formula of charge on capacitor

Q = σA

and

E = σ / ε_{0} or ε_{0} = \frac{σ}{E}

...(i)

If, the separation of the capacitor plates is increased by a small distance

∆

x against the force F, then, work done by the external agency = F.Δx

Let u be the energy stored per unit volume or the energy density of capacitor, then, increase in the potential energy of the capacitor is,

= u \times increase in volume = u \times A . ∆ x

\Rightarrow

F . ∆ x = uA . ∆ x F = uA = (\frac{1}{2} ε_{0} E^{2}) A

F = \frac{1}{2} \frac{σ}{E} E^{2} A

[From (i)]

= \frac{1}{2} σAE = \frac{1}{2} QE

Therefore, the origin of the factor

\frac{1}{2}

lies in the fact that field is zero just outside the conductor and it is E inside. Hence, the average value E/2 contributes to the force.

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