Question

# 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.

Open in App
Solution

## 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=FΔd The potential energy of the capacitor will increase by an amount of uAΔd. The work done should be equal to the increase in the potential energy. FΔd=uAΔd F=uA =( 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 )EA = 1 2 ( εA V d )E Substitute the value of capacitance, C= εA d in the above equation. F= 1 2 ( CV )E Substitute the value of Q in the above equation. F= 1 2 QE

Suggest Corrections
0