I know that a magnetic field exists when a capacitor is in the process of charging/discharging:

enter image description here

(a) But what if the capacitor is fully charged? Will the magnetic field still persist? Something like: enter image description here

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?

  • 1
    $\begingroup$ Recall that $\nabla\cdot\vec B=0$ as well $\endgroup$
    – Toffomat
    Mar 22, 2022 at 11:07
  • $\begingroup$ At the moment I cannot show it mathematically stringent, but I also think that an irrotational magnetic field would imply $\nabla \cdot \vec B \neq 0$ somewhere, which is not allowed. $\endgroup$
    – Chris
    Mar 22, 2022 at 11:12
  • 2
    $\begingroup$ You can use the same logic for a discharged capacitor or for no capacitor at all. Is there a magnetic field around a neutral piece of wood? It satisfies the same conditions. $\endgroup$
    – nasu
    Mar 22, 2022 at 11:18
  • $\begingroup$ related: do note that (outside of perfect capacitor theory) in reality no capacitor is perfect, and will self-discharge at some rate even if unconnected to electrical circuit, so in practice capacitor will never be in "fully charged" state but will always be discharging (or charging) $\endgroup$ Mar 23, 2022 at 18:09
  • 1
    $\begingroup$ @MatijaNalis There is the answer: The self-discharge is associated with a current hence a magnetic field. Another answer would be: "Yes, the earth's magnetic field." I also think it depends on the observer. If the capacitor moves with respect to the observer, both separated charges each constitute a current and in the eye of the observer will cause a magnetic field. $\endgroup$
    – HarryH
    Mar 23, 2022 at 18:32

5 Answers 5


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)

  • 2
    $\begingroup$ For OP: if you are only considering the fields produced by the setup described then its logical B=0, but in general the condition that B goes to zero at infinity is purely forced by human will, and in real life, a background solution could theoretically exist $\endgroup$ Mar 22, 2022 at 11:45
  • 3
    $\begingroup$ I agree, which is why I think it is important to explicitly state that assumption, justify why it is reasonable to assume, and also be aware of what we would observe if that assumption were violated $\endgroup$
    – Dale
    Mar 22, 2022 at 14:02
  • 3
    $\begingroup$ Without vanishing at infinity, not only $\mathbf{B} = \mathrm{const}$ would be possible, but also many growing solutions such as $\mathbf{B} \propto (y, x, 0)$. $\endgroup$
    – nanoman
    Mar 23, 2022 at 15:19

I believe I understand your question and infact it is a very valid question and gives a very extraordinary answer.


$\nabla \cdot \vec{B} = 0$

if $\vec{B} = \hat i$

then in the presence of no charges and currents this field would satisfy maxwells equations.

As taking the divergence would be 0, and the curl would be zero also.

I believe you have stumbled on the intrinsic non uniqueness of maxwells equations. The constant field that you have described is a freedom on maxwells equations that we can add to any field "produced" by charges.

As you would be 100% correct that the normal field produced by charges and currents + this constant field WOULD also satisfying maxwells equations.

What you may not have realised, is that for this constant field to actually satisfy maxwells equations, it must be present in ALL SPACE, not just the field surrounding a capacitor since this field isn't "caused" by the capacitor. This field is the background B field that permeates all of space!

Consider maxwells equations in freespace:

$\nabla \cdot \vec{E} = 0$

$\nabla × \vec{E} = -\frac{\partial \vec{B}}{\partial t}$

$\nabla \cdot \vec{B} = 0$

$\nabla × \vec{B} = \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}$

What does "freespace" actually mean? Well normally we input some charge distribution and solve maxwells equations to find the field, but what if the universe DIDNT have any charges or currents in? What would the fields be? Let's solve!

Manipulating these equations we get

$\nabla^2 \vec{E} = \mu_0 \epsilon_0 \frac{\partial^2 \vec{E}}{\partial t^2}$

$\nabla^2 \vec{B} = \mu_0 \epsilon_0 \frac{\partial^2 \vec{B}}{\partial t^2}$

The solution of this equation is any function in the form

$\vec{F}(\vec{k} \cdot (\vec{r} - ct\hat k))$

Which is not 0, meaning a universe that has nothing in it, can still have a EM field present!

This is a homogenous solution, which can be added onto any solution of maxwells equations. So to answer your question, there could be a static field, there is nothing wrong with there being a static field, there could also be a wave like solution there aswell. But the true nature of the homogenous solution is unknown to us, since it is not determined by charges and currents

  • $\begingroup$ Edit: fixed the solution to homogenous wave eq $\endgroup$ Mar 23, 2022 at 9:26

By Gauss' law $\vec \nabla \cdot \vec B = 0$, $\vec B$ is divergence-free. If it is also curl-free, then $\vec B$ is a constant field in time and space. Maxwell's equations don't disallow such a solution, but we don't detect such a "background" magnetic field in experiments. Maxwell's equations similarly permit a "background" electric field, even in the absence of any charges.

Even if a background magnetic field does exist, it exists in all space, so it can't be associated with or viewed as being caused by whatever is going on in the capacitor. It would be akin to Earth's magnetic field: this is an "external" magnetic field "sourced" by currents far away from the capacitor, and is not associated with it.

EDIT: As nanoman's comment points out, there are more general solutions for a divergence-free and curl-free vector field, if one doesn't require that the field magnitude remain bounded. Maxwell's equations permit static "background" solutions of the form $$\vec B=a_x(b_y+a_yy)(b_z+a_zz)\hat x + a_y(b_x+a_xx)(b_z+a_zz)\hat y + a_z(b_x+a_xx)(b_y+a_yy)\hat z. $$

  • $\begingroup$ "If it is also curl-free, then $\vec{B}$ is a constant field in time and space" -- see this comment. $\endgroup$
    – nanoman
    Mar 23, 2022 at 15:22
  • $\begingroup$ @nanoman Thanks for pointing that out. I've edited my answer to include more general static solutions, if one allows the field to grow as $r\rightarrow\infty$. $\endgroup$
    – Puk
    Mar 23, 2022 at 18:50

There could be, but such a magnetic field would not be produced by that capacitor.

The Maxwell equations state that the only producers of magnetic field are either electric currents, or else the coupling between electric and magnetic fields when the two vary in time. In fact, in a static capacitor situation, both these terms are zero.

But that doesn't stop you from putting the capacitor in a magnetic field generated by something else. Maxwell's equations are linear, so you can do that and the magnetic field will permeate the inside of the capacitor with no problem. And indeed, every capacitor you encounter on Earth has some magnetic field permeating it, thanks to at least Earth's own magnetic field.

Still, though, such fields do not come from the capacitor.


But what if the Capacitor is fully charged? Will the magnetic field still persist?

I understand that the discharged capacitor you're describing is equivalent to two charged disks with charges of equal value and opposite sign. No, in a system with only stationary electric fields (no dependence with time) and no electric currents, there cannot be magnetic fields.

Which of the max law prevents this?

The Maxwell equations give you the curl and the divergence for both the electric and magnetic fields. The reason that this is interesting is because according to Helmholtz's theorem, any fields (that behaves well) can be understood as a results of scalar sources and vector sources. The scalar sources are the divergence of the field and the vector sources are the curl of the field. With this knowledge you can understand the Maxwell equations as follow:

  • $\nabla\cdot\vec{E} = \frac{\rho}{\epsilon_0}$: The scalar sources of electric fields are the electric charges.

  • $\nabla\times\vec{E} = -\frac{\partial\vec{B}}{\partial t}$: The vector sources of electric fields are time-changing magnetic fields.

  • $\nabla\cdot\vec{B} = 0$: Magnetic fields don't have scalar sources.

  • $\nabla\times\vec{B} = \mu_0\left(\vec{J}+\epsilon_0\frac{\partial\vec{E}}{\partial t}\right)$: The vector sources of magnetic fields are time-changing electric fields AND electric currents.

Once you know the mathematical expression of the sources, you can obtained the expressions of the fields by applying Helmholtz's theorem.

Going back to your example, the scalar sources of magnetic fields are always $0$ ($\nabla\cdot\vec{B} = 0$) and the vector sources are also $\vec{0}$, since both $\vec{J}$ and $\frac{\partial\vec{E}}{\partial t}$ are $\vec{0}$. Since there are neither scalar nor vector sources then the magnetic field must be $\vec{0}$ (again, according to Helmholtz's theorem).

  • 2
    $\begingroup$ The conclusions you've drawn are only correct if we as humans impose the condition that B goes to 0 at infinity. The mathematics of maxwells equations allow for such a field to theoretically exist $\endgroup$ Mar 22, 2022 at 11:42
  • 1
    $\begingroup$ "No, in a system with only stationary electric fields (no dependence with time) and no electric currents, there cannot be magnetic fields." By E fields with no dependant on time, I assume you mean no moving charges, even in a universe with no charges, a B field COULD theoretically exist. $\endgroup$ Mar 22, 2022 at 11:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.