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.