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What kind of a surface can we use for Ampere's circuital law? I was taught that any enclosed surface can be used for Gauss's Law(something like a cube,a sphere)-essentially 3-D enclosed surfaces.

For Ampere's law, I have used a circular ring for calculation of Magnetic field due to wire(and also rectangular shapes for solenoids)-essentially 2-D enclosed surfaces like a ring,rectangle etc.

Now when I came to know about Maxwell's Displacement Current, the book uses a bucket kind of a shape like in the figure.enter image description here

I am confused.One side is open and the other is closed.(I apologize if this sounds dumb.I only have a basic idea about the law.I don't really understand the 'surface integral' part and how it's supposed to be used.)

I've seen this link.I don't understand what is said there.

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  • $\begingroup$ I'm not sure what you mean by "one side is open and the other is closed". Both S1 and S2 are open surfaces. S2 is the brown, paraballoidal surface only. S1 is not a subset of S2. If S1 were made out of rubber, you could form S2 by stretching it over the end of an (American) football. $\endgroup$
    – The Photon
    Mar 1, 2016 at 18:45

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Here is an annotated picture of a butterfly net which shows that all you need is a loop and an open surface which adjoins the loop.
So the closed loop can be any shape you like as can the open surface linked to it.

enter image description here

You can choose the shape of your surface to suit your problem and so for the ideal capacitor which has no edge effects choose a surface part of which is at right angles to the electric field. It makes the integration a lot easier.
This is because the electric field will only be present between the plates and if you choose the surface to be at right angles to the electric field, and hence the rate of change of electric field, then the integral will become $\mu_o \epsilon_o \frac {dE}{dt}A$ where A is the area of a capacitor plate.

$E=\frac {q}{\epsilon_o A}$ so the integral is $\mu_o I$ because $\frac{dq}{dt}=I$ which makes it the "more familiar" right hand side of Ampere's law

enter image description here

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Neither S1 nor S2 is meant to be a closed surface.

S1 is just a disk, the shape of a dinner plate.

S2 is a truncated paraboloid. A bowl shape. The blue area in the drawing is not part of S2. Taken away from the rest of the drawing, it would look something like this:

enter image description here

(Image source: Wikemedia user Krishnavedala)

The whole point of this argument for displacement current is that it doesn't matter what shape the surface is, as long is its boundary (P1) is kept the same. You can take your initial flat disk that passes through the wire, and stretch it out so that it goes between the capacitor plates instead, and you should still obtain the same surface integral. Therefore there must be some kind of current passing between the plates, and we call this current displacement current.

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  • $\begingroup$ A disk? Or do you mean a ring? I used to think that it was a ring. $\endgroup$ Mar 2, 2016 at 15:42
  • $\begingroup$ A disk. If it were a ring, then the current wouldn't necessarily passing through its surface. $\endgroup$
    – The Photon
    Mar 2, 2016 at 15:44
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It's an application of Stoke's theorem: https://en.wikipedia.org/wiki/Stokes%27_theorem

I find "Div, Grad, Curl and All That", by H.M. Schey, to be a great introduction to how and why vector calculus works; it uses the electric field as the example throughout.

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