Question

A parallel-plate capacitor has circular plates and no dielectric between the plates. Each plate has a radius equal to 2.2 cm and the plates are separated by 1.1 mm. Charge is flowing onto the upper plate (and off the lower plate) at a rate of 2.9 A. (a) Find the rate of change of the electric field strength in the region between the plates. V/m·s (b) Compute the displacement current between the plates and show that it equals 2.9 A. PLEASE EXPLAIN HOW YOU DERIVE THE EQUATIONS THAT YOU USE FROM THE ORIGINAL EQUATION WHILE YOU SOLVING THE PROBLEM. . .

Answer 1

We know,
Q = C * V & Q = I * t
So,
I * t = C* V
I/C = dV/dt

Area,
A = 3.14 * (2.2*10^-2)^2 m^2

C = ε*A/d
C = (8.85 * 10^-12) *  3.14 * (2.2*10^-2)^2 / (1.1 * 10^-3)
C = 1.22 * 10^-11 F

dE/dt = ( dV/dt ) / d
dE/dt = [2.9/(1.22 * 10^-11)]/(1.1*10^-3)
dE/dt = 2.16 * 10^14 V/m·s

Rate of change of the electric field , dE/dt = 2.16 * 10^14 V/m·s

(b)

Displacement Current is given by Id = ε * d(phi)/dt

Id = ε*d/dt(E.A)
Id = ε*A * dE/dt
Id = (8.85 * 10^-12) *  3.14 * (2.2*10^-2)^2 * 2.16 * 10^14
Id = 2.9 A

Hence Proved.