Area of reference to find magnetic field inside a (dis)charging capacitor

Question

Faraday's law of induction can be expanded:$$V=-N\frac{d(BA)}{dt}$$ and for a circuit which creates an area between the wires if that area doesn't change then it becomes $$V=-N\frac{dB}{dt}A$$ The area of the formula is well defined. Suppose a capacitor is being (dis)charged during the (dis)charge a magnetic field is created inside the capacitor. What I don't get is what does the electric flux mean in the formula, of which area do we find $d(EA)/dt$ in order to get the magnetic field?

Answer 1

It looks like you're asking about the Ampere-Maxwell Law, $$ \oint_{\Gamma}{\mathbf{B}\cdot\,d\mathbf{l}} = \mu_0\int_\mathcal{S}{\mathbf{J}\cdot\,d\mathbf{a}} + \mu_0\epsilon_0\,\frac{d}{dt}\int_\mathcal{S}{\mathbf{E}\cdot\,d\mathbf{a}}. \tag{1}$$ Here, $\Gamma$ is a fixed closed loop in space and $\mathcal{S}$ is any surface bounded by $\Gamma$. So, yes, we do have some freedom in choosing the surface through which to calculate the flux of $\mathbf{J}$ and $\mathbf{E}$.

The point is that the RHS of Eq. $(1)$ is the same, regardless of which surface we choose (so long as it is still bounded by the loop $\Gamma$). This is because the combination $\mathbf{J} + \epsilon_0\frac{\partial \mathbf{E}}{\partial t}$ is divergenceless (see end of post), as we can quickly show using the continuity equation and Gauss's Law: $$ \begin{split} \nabla\cdot\left(\mathbf{J} + \epsilon_0\frac{\partial \mathbf{E}}{\partial t}\right) &= \nabla\cdot\mathbf{J} + \epsilon_0\frac{\partial}{\partial t}\left(\nabla\cdot\mathbf{E}\right) \\ &= -\frac{\partial \rho}{\partial t} + \frac{\partial \rho}{\partial t} = 0. \end{split} $$ (Alternatively, we could have noted that $\mathbf{J} + \epsilon_0\frac{\partial \mathbf{E}}{\partial t}$ is the curl of $\mathbf{B}/\mu_0$, and the divergence of a curl is always zero.)

Theorem: Let a vector field $\mathbf{A}$ be divergenceless. Then for a given loop $\Gamma$, the flux of $\mathbf{A}$ is the same through any surface bounded by $\Gamma$.

Proof: Call $C_1$ the flux of $\mathbf{A}$ out of a surface $\mathcal{S}_1$ bounded by $\Gamma$ and $C_2$ its flux through a second surface $\mathcal{S}_2$, also bounded by $\Gamma$. The flux of $\mathbf{A}$ out of the closed surface formed by $\mathcal{S}_1$ and $\mathcal{S}_2$ is zero (by the Divergence Theorem), so $C_1 = C_2$.

Answer 2

From your description I suspect your confusion arises because until now you have encountered Faraday's law always in conjunction with a wire loop or with a coil (made up of several wire loops). But this isn't necessary. Faraday's law is independent of any wire. Actually Faraday's law is just a relation between the electric field $\mathbf E$ and the magnetic field $\mathbf B$. $$\oint_C \mathbf E\ d\mathbf l = - \frac{d}{dt} \int_A \mathbf B\ d\mathbf A$$ where $C$ is a closed curve enclosing an area $A$. The integral on the right side is the magnetic flux (or the number of magnetic field lines) passing through the area $A$. Notice that this formulation does not mention any wire at all. But of course, for technical applications with wires it makes sense to choose the curve $C$ along the wire loop.

And now for your problem with a (dis)charging capacitor. We have a similar situation like above for the Ampere-Maxwell law: $$\frac{1}{\mu_0}\oint_C \mathbf B\ d\mathbf l = I_\text{enclosed} + \epsilon_0 \frac{d}{dt} \int_A \mathbf E\ d\mathbf A$$ The integral on the right side is the electric flux (or the number of electric field lines) passing through the area $A$. You can choose any closed curve $C$ you want. But for practical purpose (calculating the magnetic field inside the capacitor) it will make the most sense to use a circle or a square parallel to the plates of the capacitor.