Show that the force on each plate of a parallel plate capacitor has amagnitude equal to (½) QE, where Q is the charge on the capacitor,and E is the magnitude of electric field between the plates. Explainthe origin of the factor ½.
It is given that the charge on the parallel plate capacitor is Q and the magnitude of
electric field between the plates is E .
Let the force on each plate capacitor be F .
The work done by the force is,
W = F Δ d
The potential energy of the capacitor will increase by an amount of u A Δ d .
The work done should be equal to the increase in the potential energy.
F Δ d = u A Δ d F = u A = ( 1 2 ε E 2 ) A (1)
The magnitude of electrical energy is,
E = V d
Substitute the value of E in equation (1).
F = 1 2 ε ( V d ) E A = 1 2 ( ε A V d ) E
Substitute the value of capacitance, C = ε A d in the above equation.
F = 1 2 ( C V ) E
Substitute the value of Q in the above equation.
F = 1 2 Q E
It is given that the charge on the parallel plate capacitor is
electric field between the plates is
Let the force on each plate capacitor be
The work done by the force is,
The potential energy of the capacitor will increase by an amount of
The work done should be equal to the increase in the potential energy.
The magnitude of electrical energy is,
Substitute the value of
Substitute the value of capacitance,
Substitute the value of
