Question

The capacitors referred to in this problem have only empty space between the plates.

(a) Show that a parallel-plate capacitor has a displacement current in the region between its plates that is given by Id = C dV/dt, where C is the capacitance and V is the potential difference between the plates. (Do this on paper. Your instructor may ask you to turn in this work.)

(b) A 5.00-nF parallel-plate capacitor is connected to an ideal ac generator so the potential difference between the plates is given by V = V0 cos ωt, where V0 = 3.00 V and ω = 500π rad/s. Find the displacement current in the region between the plates as a function of time. (Use the following as necessary: t and π.) Id = _____________µA

Answer 1

let d be the distance between the two plates of the capacitor and let A be the area of cross section of plate.

we know that capacitance of a parallel plate capacitor is given by C = ₀A/d

using Gauss law, we can say total electric flux through plate is _E = Q/₀   where Q = total charge on plate

   also Q = CV   ............................(1)     where V= potential difference between the plates.

now displacement current is given by I_d = ₀ (d/dt)(Q/₀)

                                                           = dQ/dt

                                                           = dCV/dt                ....................[since(1)]

                                                     I_d   = C dV/dt    ...................(2)         [C being constant comes out]    proved.

now C = 5 nF

       V= V₀cost ;     V₀ = 3 V , = 500(pi) rad/s

putting all these values in (2), we get

       I_d= C d(V₀cost)/dt

          = -CV₀(sint)                 [since derivative of cosx= -sinx]

           = - 5 * 10^-9 * 3* 500 (3.14) (sin500(pi)t)

           = 23.5 sin(500t) A