3. (15 points) A parallel-plate capacitor with thin circular plates of radius a

Question

3. (15 points) A parallel-plate capacitor with thin circular plates of radius a is being charged with a constant current. (i) Show that the discontinuity of the tangential magnetic field across a capacitor plate is 0 K × n, where n is the unit vector perpendicular to the plates. Neglect fringe effects. (ii) K obviously is the surface current density that flows as the plate is being charged. Show that this surface current implies that the charge density remains uniform during the charging of the conductor. One way to do this is to calculate the rate of change of charge in an arbitrary annulus. If this is proportional to the area of the annulus, the charge density is constant.

Explanation / Answer

a.

at radius r

current included within the loop of radius r = I - i

i = Ir^2/a^2

hence

Ien = I - Ir^2/a^2

hence from ampere's kaw

B*2*pi*r = mu*(I - Ir^2/a^2)

B = mu*(I/2pi)(1/r - r/a^2) r'

(where r' is the radial unit vector)

hence

B = mu* k x n

where k = (I/2*pi)(1/r - r/a^2)

b. now, ien = (I - Ir^2/a^2)

ien = dq/dt

q = [integral](2*pi*r*dr*s(r))

hence

(I - Ir^2/a^2) = 2*pi*r*d(s(r))/dt

d(s(r))/dt = I(1 - r^2/a^2)/2*pi*r

now this means s(r) = kr^2

which means it is prportional to the area of the plate

hence the rate of chantge of charge on the plate is constant

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.