Question

Suppose two circular metallic plates of radius R and separation d forms a parallel plate capacitor. Let Q be the instantaneous value of charge on either plate and is changing with time. (a) Calculate the Poynting vectorS.(b) How is the net energy flow into the capacitor is related to the rate of change of capacitor energy?

Answer 1

given two circular metallic plates, radius R each
plate seperation = d

this forms a parallel plate capacitor
let Q be instantaneous charge on either plate, and its changing with time

a. now, cpaacitance C = A*epsilon/d
8.98*10^9 = 1/4*pi*epsilon
so,
C = pi*R^2*epsilon/d

now, since the capacitor is being charged
charge at any time = Q
hence current at any time, i = dQ/dt

now, electric field due to charge inside the capacitor is from +ve to -ve plate along the direction of current
and magnitude of this field is
|E| = 2*Q/pi*R^2*epsilon [ assuming d << R, the point inside capacitor is too close to the capacitor plate that the plate can be considered an infinite plate]

for magentic field, assume the current flows from south to north (along z direction)
then
magnetic field is tangential with magnitude given by
|B| = 2*mu*dQ/4*pi*r*dt , where r is the radial distance of the point from the center of the capacitor
and mu is permeability of free space

now poynting vector is S
S = E x H
since H is always tangential and E is along z, so S is always radially into the cylinderical axis
the magnitude is 2*Q/pi*r^2*epsilon * 2*mu*dQ/4*pi*r*dt
|S| = Q/pi^2*R^2*r*epsilon * mu*dQ/dt

|S| = (mu*Q/pi^2*R^2*r*epsilon) dQ/dt
radially inwards

b. from conservation of energy we can easily say that
net energy flow into capacitor = integral [ rate of change of capacitor energy * dt] from t = 0 to t = t