Griffiths, in section 7.3 Maxwell's Equations, says:

There’s another way to see that Ampère’s law is bound to fail for nonsteady currents. Suppose we’re in the process of charging up a capacitor (Fig. 7.43). In integral form, Ampère’s law reads $$\iint \vec B \cdot d\vec l = \mu_{0} I_{\text{enc}}$$enter image description here I want to apply it to the Amperian loop shown in the diagram. How do I determine $I_{enc}$ ? Well, it’s the total current passing through the loop, or, more precisely, the current piercing a surface that has the loop for its boundary. In this case, the simplest surface lies in the plane of the loop—the wire punctures this surface, so $I_{enc} = I$ . Fine—but what if I draw instead the balloon-shaped surface in Fig. 7.43? No current passes through this surface, and I conclude that $I_{enc} = 0$!

A few pages later, he comes back to the same example to "show" that displacement current "fixes" this issue.

Let’s see now how displacement current resolves the paradox of the charging capacitor (Fig. 7.43). If the capacitor plates are very close together (I didn’t draw them that way, but the calculation is simpler if you assume this), then the electric field between them is $$E=\frac{\sigma}{\epsilon_{0}}=\frac{Q}{A\epsilon_{0}}$$ Where $Q$ is the charge, and $A$ is the area of the plate. $$\frac{\partial E}{\partial t}=\frac{1}{\epsilon_{0}A}\frac{\partial Q}{\partial t}=\frac{I}{A\epsilon_{0}}$$ Now equation 7.37 (4th maxwell's equation, with displacement current), reads in integral form: $$\int \vec B \cdot d\vec l = \mu_{0} I_{enc} +\mu_{0}\epsilon_{0} \iint \frac{\partial E}{\partial t} \cdot d\vec A$$ If we choose the flat surface, then $E = 0$ {I think this is a quasi-magnetostatic approximation} and $I_{enc} = I$ . If, on the other hand, we use the balloon-shaped surface, then $I_{enc} = 0$, but $\iint \frac{\partial E}{\partial t}\cdot d\vec A = \frac{I}{\epsilon_{0}}$. So we get the same answer for either surface, though in the first case it comes from the conduction current, and in the second from the displacement current.

Under the assertion that the plates are held extremely close together, we can assert that the field is uniform and is equal to what is given above, fine. I also understand why he is getting the same result: It is solely due to his choice of the surface the Amperian loop encloses. In this situation, consider the image below, which describes another surface, which the Amperian loop encloses (Labelled $S_2$ for convenience):

enter image description here

Here, the area that the surface in question "pierces" through the plate is not $A$, but some smaller area $A'$, therefore, the integral no longer evaluates to $\frac{I}{\epsilon_{0}}$, but some other, $\frac{I}{\epsilon_{0}A} A'$. I think this mistake is popping up because it is coming from the quasi-magnetostatic approximation. The question is: is this a mistake on Griffith's part, or what is actually happening?

Addendum: He has also, in considering the surface integral of $\frac{\partial E}{\partial t}$ through the balloon, ignored the contributions to the integral from the region between the conducting plate and the flat surface the Amperian loop encloses, only considering the part of the balloon that is between the two conducting plates. If this result were to be correct, this residual integral should exactly give 0. Now, considering my $S_2$ surface, under the same consideration, the residual surface integral should give me 0 (I don't agree with this, but lets just go with what Griffith's has done), then, clearly the integral contrubition from the surface between the two plates, is $\frac{I}{\epsilon_{0} A} A'$ which is giving me a different result from the flat surface.

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    $\begingroup$ No, you are just wrong; when the area is smaller than $A$ as you have drawn, there will be some current passing through the intersection between your drawn area and the capacitor plate, and thus you have to add that part in too. Then the agreement will be complete again. $\endgroup$ Sep 13, 2023 at 12:36
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    $\begingroup$ @naturallyInconsistent , OKk, I understand this mistake, I should have considered that current. but I have another question within this, regarding the applications of $\partial_t E=0$ approximation. When can we consider this approximation, and when can we not? Here, he ignores it in the flat surface, but considers it in between the plates. What allows us to pick and choose like that. see the addendum, it turns out the residual contribution is exactly equal to the surface integral over the flat surface, here taken to be 0. why is this true. $\endgroup$ Sep 13, 2023 at 12:46
  • $\begingroup$ If the plates are very close together they will make only a very small electric field outside of the region between the two plates. That's why it is ignored for the flat surface. So think of this argument as "the plates are arbitrarily close. if you aren't satisfied with the accuracy of ignoring that term, then make the plates closer." However, no matter how close together the plates are, the electric field between them gets no smaller, it remains $\sigma/\epsilon_0$, so it can never be ignored. The situation is set up to give you a clear conceptual example of how displacement current works. $\endgroup$
    – AXensen
    Sep 13, 2023 at 12:51
  • $\begingroup$ indeed in "real life" with plates that are not arbitrarily close together, there will be some small electric field going through the flat surface, and that will change the current+displacement current going through the loop, and that will change the magnetic field going around the edge of the loop. You could even calculate this in a toy model by replacing the square plates with two point charges. Find how the magnetic field going around the loop differs from the case with two planes. $\endgroup$
    – AXensen
    Sep 13, 2023 at 12:57
  • $\begingroup$ The increasing E field is almost completely confined between the capacitor plates. Only a very little bit of that is going out into the universe. On the flat surface, almost all of the E field that is there, is in the wires, where $\vec E=\sigma\vec J=$ const, and thus there $\partial_t\vec E$ is very small indeed, everywhere in the integral over the flat surface. $\endgroup$ Sep 13, 2023 at 12:58

1 Answer 1


Griffiths is right. Don't forget that that current flows outward on the plate to spread out the charge. The charge density on the plate is $\sigma=It/A$. In order to reach the area outside of $A'$ it has to come from somewhere. It flows through the gaussian surface, and the total surface current will be $I(A-A')/A$. Then the displacement current is the complementary fraction $IA'/A$ of the capacitors changing electric field.

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    $\begingroup$ I understand this mistake, I must have considered the current. But consider the addendum, another pressing issue I had was regarding the $\partial_t E=0$ approximation. When can we consider this and when can we not? The residual integral I am talking about in the addendum turns out to be the same as that through the flat surface, which is 0, but why is this true $\endgroup$ Sep 13, 2023 at 12:49
  • $\begingroup$ @nickbros123 answered as a comment on your question $\endgroup$
    – AXensen
    Sep 13, 2023 at 12:52

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