At t = 0 , a parallel-plate capacitor having plate-area A and plate separation d is connected to a battery of emf E and internal resistance R . Consider a plane surface of area A/2 , located symmetrically between the plates and parallel to them. Determine the displacement current through this surface as a function of time.

Step-by-step solution

The capacitance of the parallel-plate capacitor is C = ₀ A d . The conduction current in the circuit during charging is given by: i(t) = E R e^ - t RC The electric field between the plates is uniform and its magnitude is E = q(t) ₀ A , where q(t) is the charge on the capacitor. The electric flux through the given surface of area A 2 is: _E = E A 2 = q(t) 2 ₀ The displacement current through this surface is: i_d = ₀ d _E dt = ₀ d dt ( q(t) 2 ₀ ) = 1 2 dq dt = i(t) 2 Substituting the expression for i(t) and C , we ge