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

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?

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 $$C_1 = C_2$$.
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.