I know that a magnetic field exists when a capacitor is in the process of charging/discharging:
(a) But what if the capacitor is fully charged? Will the magnetic field still persist? Something like:
If there is no magnetic field around an electrostatic field, which of the Maxwell's laws prevents this? The electrostatic field is irrotational and a static (not varying in time), the Gauss law of electric field and magnetic field are satisfied. From the Maxwell-Faraday law (let's consider differential form):
$$
\nabla\times\mathbf E=-\frac{\partial\mathbf B}{\partial t}
$$
we have the LHS to be zero as the electrostatic field is irrotational. This means the time rate of change of magnetic field is zero too. But this doesn't mean the magnetic field itself is zero, it could have any constant value.
From the Maxwell-Ampere's law (again, let's consider the differential form):
$$
\nabla\times\mathbf B=\mu_0\left(\mathbf J+\epsilon_0\frac{\partial\mathbf E}{\partial t}\right)
$$
we have the RHS to be zero since this is an electrostatic field (not varying in time), and there is no current flow. This means the curl of this magnetic field is zero but that means the magnetic field can be constant and irrotational, and not that the magnetic field is zero.
(b) So, why can't there be a constant irrotational magnetic field around an electrostatic field between the fully charged capacitor plates?
Best Answer
You cannot forget Gauss’ law for magnetism. From that we have $$\nabla \cdot \vec B = 0$$ combined with $$\nabla \times \vec B =0$$ from the question, we have a Helmholtz decomposition of $\vec B$.
Now, the Helmholtz theorem says that if $\vec B$ goes to $0$ at infinity then this decomposition is unique. The only function which satisfies it is $$\vec B=0$$
The restriction that $\vec B$ goes to $0$ at infinity is reasonable since we are considering the circuit to be the only source of fields and so we know that at a distance greater than $d=(t-t_0)c$ the EM waves from the circuit will not have reached that location so the fields from the circuit will be zero. Since $\infty > d$ the field is known to go to zero, and therefore $\vec B=0$ is the unique solution.
(If we relax the condition at infinity then any $\vec B= const.$ would satisfy the Helmholtz decomposition, but clearly such fields could not be considered to be caused by the circuit)