A circular parallel plate capacitor with radius a = 2.5 cm and plate separation

Question

A circular parallel plate capacitor with radius a = 2.5 cm and plate separation d = 1.0 mm is connected to an infinitely long wire as pictured. A switch is closed at t = 0 s, and the capacitor is charged with a constant current of I(t) = 0.5 A. Assume that the capacitor is ideal (i.e., there is no electric field outside the plates). Consider a location halfway between the capacitor plates, but at a radius of r > a (i.e. outside the capacitor itself.) Is the magnetic field strength at this point larger, smaller, or the same as the magnetic field the same distance r from the wire, but away from the capacitor? Explain your answer. Now consider a location halfway between the capacitor plates, but inside the capacitor at at a radius of r

Explanation / Answer

From maxwells equations

Displacement current = I(d) = epsilon dE/dt

E is electric field whic is related to V in a parallel plate capacitor as E = V/d

So dE/dt = 1/d (dV/dt) = 1/dC * dQ/dt = 1/(Cd) I(wire)

By amperes law intergral ( B.dl) = mu ( I in ) = mu * I(d)

So B* 2*pi*r = mu * I(d)

Since displacement current is constant, magnetic fieldvaries inversely with r. So closer to the plates, more the magneticfield.

Electric field is increasing with time because dE/dt = (1/Cd) * I(wire) whih is a constant.

So E must increase linearly with time.

Magnetic field at a particular distance r will be constant with time as B = mu * I(d) / 2*pi*r

Magnetic field due to wire away from capacitor = mu * I(wire) / 2*pi*r

For capacitor we have mu * I(d) / 2*pi*r

Sincedisplacement current is related to current in wire . I(d) = (epsilon) i(wire)/ Cd.   

Since capacitance will be epsilon(A)/d for a pure parallel plate and for circular also it is related to area,

I(d) will be inversely proportional to area of capacitor. So large area capacitors have small I(d)

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