A parallel plate capacitor consists of two circular plates of area S with vacuum

Question

A parallel plate capacitor consists of two circular plates of area S with vacuum between them. It is connected to a battery of constant emf E. The plates are then slowly oscillated so that they remain parallel but the separation between them is varied as d=d0+d1Sint.

(a) Find the magnetic field between the plates produced by the displacement current if the battery remains connected.

(b) ) Similarly, find the magnetic field between the plates if the capacitor is charged up to E, is then disconnected from the battery and then the plates are oscillated in the same manner.

Explanation / Answer

Given
Parallel plate capacitor, Plate area = S
Materisal between the plates is vaccum

a. Connected with battery of EMF = E
   distance between plates, d = do + d1*sin(wt)
   so charge on thee plates be q(t)
   then q(t) = C(t)*E
   where C is time dependent capacitance
   for a parallel plate capacitor
   C(t) = S*epsilon/(do + d1*sin(wt)) [ epsilon is permiottivity of free space]
   so, q(t) = S*epsilon*E/(do + d1*sin(wt))
   displacement current = d*q(t)/dt = -S*epsilon*E*(d1*w*cos(wt))/(do + d1*sin(wt))
   Magnetic field due to diplacement current at distance x , B = 2ki/x [ k = mu/4*pi where mu is permeability of free space ]
   B(t) = -2*k*S*epsilon*E*(d1*w*cos(wt))/x*(do + d1*sin(wt))
b. Connected to battery, charged then removed
   so initial capacitance, Co = Ae/do
   initial EMF = E
   initial charge = q = CoE = Ae*E/do

   now when the osscilations start, q does not change
       so, dq/dt = 0 = displacement current
       Magnetic field, B = 0

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