question_answer 28)

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 electric field between the plates. Explain the origin of the factor 1/2. 

Answer:

Let the distance between the plates be increased by .                 Work done by external agency = Fx                 If U is energy density, the increase in P.E. is given by                 U ()                 But                         U =                    and force             F = Ua                 i. e.                         F =                                                                   Outside the conductor, the field is E but inside it is zero. Thus, average value                 i.e. decides the force.