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 the electric field between the plates. Explain the origin of the factor `1/2`.
Solution
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" = (1/2in_0"E"^2)"A"`
Electric intensity is given by,
`"E" = "V"/"d"`
∴ `"F" = 1/2 in_0("V"/"d")"EA"`
= `1/2(in_0"A" "V"/"d")"E"`
However, capacitance, C = `(in_0"A")/"d"`
∴ `"F" = 1/2("CV")"E"`
Charge on the capacitor is given by,
Q = CV
∴ `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, the field is E, and inside it is zero.
Hence, it is the average value, `"E"/2`, of the field that contributes to the force.
