Question

A parallel plate capacitor of area $60cm^2$ and separation 3 mm is charged initially to $90 \mu C$. If the medium between the plates gets slightly conducting and the plate loses the charge initially at a rate of $2.5 \times 10^{-8} C/s$, then what is the magnetic field between the plates?
A. $2.5 \times 10^{-8}$T
B. $2.0 \times 10^{-8}$T
C. $1.63 \times 10^{-11}$T
D. Zero

Answer 1

Hint: Changing electric fields are responsible for producing changing magnetic fields. Electric field between the plates of a capacitor can be known (general from gauss’s law). We are given the rate of change of the charge or leakage current of the capacitor.

Formula used: Ampere’s law: The line integral of magnetic field passing from a given surface (closed loop) is equal to $\mu_0$ times the total current enclosed in that loop, with a correction for displacement currents is given as:
$\oint \vec{B}. \vec{dl} = \mu_0. I + \epsilon_0 \mu_0 \dfrac{d\vec{E}}{dt}$

Complete step by step answer:
For a capacitor, the electric field is independent of the separation distance. It just directly depends on the charge and inversely on the dimensions of the two plates. So, we need not bother about the separation of 3 mm (redundant information).
There is a changing electric field due to changing charge.
Electric field for a capacitor is given as:
$\dfrac{q}{\epsilon_0 A}$
We could just place this in the second term of the Ampere’s law here as we are given the changing value of q.
But, consider the left hand side of the expression. It requires us to find the magnetic field inside the capacitor. The capacitor is acting as a conductor, and we know that even in a conducting wire, all charge flows on the surface, so we expect no current flowing on the inside of the capacitor. It flows only on the surface.
So, due to the Ampere’s law, we can say that the magnetic field is zero.

So, the correct answer is “Option D”.

Note: It is a very tricky question. A simple concept from the Ampere’s law had to be applied here. Had it been the case, that we were not having our loop inside the capacitor, but outside it, we would have followed the analysis by inserting the rate of change of electric field here. Displacement current would have been the decay current in such cases.