A parallel plate, vacuum capacitor with circular plates, each of area S, is conn

Question

A parallel plate, vacuum capacitor with circular plates, each of area S, is connected to a battery of constant emf, epsilon. The plates are then slowly oscillated so they remain parallel, but the gap d between them is given by d = d_0 + d_1 sin wt. Find the magnetic field, H, between the plates produced by the displacement current. Similarly find H if the charged capacitor is first disconnected from the battery and then the plates are oscillated as before. Within the region 0 lessthanorequalto x lessthanorequalto a of the vacuum, the electric and magnetic fields are given by

Explanation / Answer

Problem B3:

One of Maxwell's equations is...

integral (B ds) = o  o  dE /dt + oIenc

The current enclosed is zero, so you have

integral (B ds) = o  o  dE /dt .

The electric flux between theplates can be written as EA, where A is constant (the area of theplate) and E changes. E changes because the charge on the plateschange, because the capacitance changes, as the plates oscillate.So you can rewrite the expression as:

integral (B ds) = o  o  A dE/dt          and

     E = / d = / (d0+d1sin(t) )

Notice that you can now take dE/dt, which is...

      E = - d1 cos(t)[d0+d1sin(t)]-2

For any valueof r, or (since that's what we're using for "r"), theintegral becomes    B (2). Also note that on the rightside of the equation, A is just 2

Putting all this together:

B ( 2 ) =o  o  2 (- d1 cos(t)[d0+d1sin(t)]-2)     

By simplifying we get,

    B = - (1/2)   o  o   d1 cos(t)[d0+d1sin(t)]-2

There is an additional factor of o here and the o comes right from Maxwell's equation.

b) Since the capacitoris disconnected from the battery, the charge on the plates is constant, so the E field is constant. So consider the same as above, but this time dE/dt = 0 , so the whole rightside of the equation is zero, so the B field is zero.

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