In an ideal charged capacitor (with infinitely large parallel plates), the electric field outside the area between the plates is zero.

Will be there any current flowing through the red wire from plate 1 to plate 2 if I attached it just like on the image below?

image originally posted as oi63.tinypic.com/35bw1zt.jpg

I think the answer is no. Because the electric field in the points where I attached the wire to the plates is zero, just like it is all along through the wire.

What I'm a bit confused about is the fact that you can actually say that there IS a path connecting those two points (marked with red dots, connected by the wire) with non zero voltage.

Just look at the definition of voltage:

As long as the integral over the path we take (where $E$ is the electric field) from point 1 to 2 isn't zero, the voltage won't be zero = there will be current. Like in this case (forget about the capacitor plates here, just assume the electric field exists in the area between the virtual plates as in this picture):

originally posted as http://oi64.tinypic.com/wclb21.jpg

The integral over the red path won't be zero, so the current should be able to flow between those two points once they are connected with a wire, right? But there's nothing to push the charge from the first to the second point, because the electric field is zero at those points. Could anyone explain that to me?

  • $\begingroup$ If you connect 2 oppositely charged objects with a wire, why wouldn't a current flow through the wire? There will be a PD between the 2 points of contact until the charge starts flowing, then it will quickly reduce to zero. $\endgroup$ Mar 1, 2017 at 6:19
  • $\begingroup$ You seem to be engaging in fallacious reasoning with limits. As the size of the capacitor goes to infinity, the electric field (i.e., voltage per unit length) goes to zero. You're arguing that, therefore an "infinite" capacitor would have no total voltage. But as the size of the capacitor goes to infinity, the length of a wire to go around it goes to infinity, so the total voltage = electric field*length = 0 * infinity, which is undefined. $\endgroup$ May 7, 2018 at 21:36

4 Answers 4


Your question demonstrates the difference between electrostatics in "normal" Euclidean space and in periodic space. In all cases however, a current will flow through the wire.

In your problem you sketch how the wire goes from one side of the infinitely large capacitor to the other without passing through the field inside the capacitor. However, in our everyday Euclidean space this not possible.

This leaves three options:

Option 1: The wire goes through a hole somewhere in the capacitor. In this case the wire goes through the electric field, which will make a current flow.

Option 2(inspired by John's answer): The capacitor is not truly infinitely large. In this case the wire can go around the capacitor. However, due to edge effects caused by the finite size, the field at the sides of the capacitor won't be zero. Therefore the wire will pass through an electric field and a current will flow.

Option 3: The capacitor does not live in Euclidean space, but in a space with periodic boundary conditions. This means that when you move out of the image on the left side, you will enter the image again on the right side. It is similar to a rolled up piece of paper, where you can move in circles around the cylinder. In such a space it is possible to connect the two plates without crossing the capacitor and your problem seems to persist.

However, in such a periodic space your assumption that there is no field outside of the capacitor fails. In fact, it is not longer trivial what to consider as the inside and outside of the capacitor. You can draw Gaussian surfaces to see that there will be an electric field everywhere, except for inside the conductors. The field will point from the positively charged plate, to the negatively charged plate along two different routes. Any wire connecting the two plates will thus experience an electric field and a current will flow.

  • $\begingroup$ Thanks, that makes sense. Let's forget about capacitors for a second - what would happen in the following case? imgur.com/Ap3phEC There's an electric field and points A and B nearby the field. As you can see, the field is zero at those points. Will there be a current flow from point A to B once I connect them with a wire? According to the definition of voltage, it's a line integral of the electric field over the path from A to B, so in this case the voltage across A and B won't be zero. But I think that only electrons in the area of non-zero electric field will move... $\endgroup$ Mar 1, 2017 at 18:21
  • $\begingroup$ ... and because A is not in a non-zero electric field, no electrons will flow from that point towards B. If I'm wrong, then is it because the electrons located in the middle part of the wire will move towards B, and that will create a shortage of electrons "on the left edge of the non-zero electric field" (looking at the image) and the electrons will be pulled from the area near A? $\endgroup$ Mar 1, 2017 at 18:25

I think much of the confusion arises because you are neglecting edge effects. You are assuming that there is no electric field outside the plates, which is equivalent to taking those plates to be infinite. What this means is that you can't actually take the loops that you've drawn, for they will intersect the plates somewhere.


Without going into a particular case of a capacitor with infinite plates, which has already been nicely covered by Crimson, I'll directly address your question involving points A and B, lying on the two sides of the capacitor.

This case, in principle, has already been addressed by John, when he mentioned the edge effects, but I'd like to make it a little more explicit.

Basically, the assumption that the electrical field at points A and B is zero is not correct. The field near A and B is going to look more like this:

enter image description here

We can say that, if a capacitor does not have infinite plates and is charged to some non-zero voltage, the field from the capacitor will extend to infinity in all directions - it won't be limited to the area near the edges.

So, an electron at point A will be surrounded by the non-zero field and will move toward point be.

In general, if there is an electrostatic field anywhere, it will be everywhere unless it is fully shielded.


In any real capacitor, the moment you connect the two plates with a wire, the charge will instantly flow from the left side to the right side through the wire, and the negative and positive charges will cancel each other and go to zero.

In the absence of the wire, the positive and negative charges are attracted to each other across the space between the plates. Those are the "closest" charges of the opposite type that can be "seen."

Connecting the wire changes this situation instantly. The charges are free to move throughout the contiguous metal regions. With the wire in place, the positive and negative charges can "find" each other through the wire. Essentially you have reduced the space between the opposite charges from "d" (the plate spacing) to zero. There is a strong electric field through the wire that instantly draws the positive and negative charges together.

  • $\begingroup$ If that is your video (as seems likely by the view count), you should disclose it as yours. Otherwise it may get marked as spam. $\endgroup$
    – Chris
    May 6, 2018 at 0:42
  • 2
    $\begingroup$ I've removed the link, unlocked this post and undeleted it. SVTechWr, you're welcome to edit to add the link back in, as long as you do so in a way that clearly says if it's your video. I'd also recommend explaining what the video shows, so that people will understand why you chose to include that link in this answer. $\endgroup$
    – David Z
    May 7, 2018 at 20:32
  • $\begingroup$ Thank you for moderating the site. I appreciate that you are keeping up the high quality of the posts. I'll be sure to follow your guidelines in future posts. $\endgroup$
    – SVTechWr
    May 9, 2018 at 3:11

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