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Why does a charge distribution with cylindrical symmetry have to be infinitely long when applying Gauss's law? It seems unnecessary especially since the electric field emits radially.

Anna
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1 Answers1

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It emits radially only if it is infintely long. Otherwise it has an axial component. For finite cylinder the field is purely radial only along tha radius that starts in the middle of the cylinder's height. There is another question treating this problem and it has an image that shows the field. Why we cannot use Gauss's Law to find the Electric Field of a finite-length charged wire?

Ifyou look at images of the field lines for finite cylinder you can see that if you are not too far from the middle line the field is close to being radial. If you move towards the ends the axial component increases and is directed towards the closest end of the cylinder. Now, if the cylinder is infinite, then

  1. You are never close to the end so you can consider that you are always in the "middle" and
  2. You cannot say that you are closer to one end than the other end so which way will the axial componets points? As you cannot pick a direction the component should be zero. Note that those are just some intuitive ways to help understanding. These are symmetry considerations and I know that it takes a little experience to become familiar with using symmetry to draw conclusions.

On a more practical approach, if you look at points very close to the cylinder (as compared to its length) the field can be approximated by the field of an infinite wire. You will see the same situation for many other models, like parallel plate capacitor (with infinite plates) and so on.

nasu
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  • Why, you can't have an axial radiation if it is infinitely long since the ends cannot be reached? – Anna Apr 25 '22 at 01:21
  • Does this also apply to a flat plate? There would be electric field lines on the edges unless infintely long? – Anna Apr 25 '22 at 01:30
  • I added some details. And yes, for a finite plate the field curves when you get close to the edge. For capacitors this is called the edge effect or the fringe field. But if the distance between the plates is small compared with the dimensions of the plates the field is aproximately uniform. – nasu Apr 25 '22 at 01:37
  • If we place this cylindrical charge inside a cylindrical Gaussian surface, isn't there going to be an electric field out from the circular tops and bottoms of this cylindrical Gaussian surface? – Anna Apr 25 '22 at 02:30
  • For finite cylindrical charge, it will be, yes. – nasu Apr 25 '22 at 10:57
  • Then how come when we do calculations with a cylindrical Gaussian surface with definite L surrounding an infinitely long wire we don't calculate the electric field on the faces? – Anna Apr 25 '22 at 17:55
  • on the circular faces that is. L is height of cylinder. – Anna Apr 25 '22 at 18:30
  • I said "for finite". And you ask how come that for a different situation (infinite wire) you have a different behaviour? What is the logic in this? – nasu Apr 25 '22 at 18:44
  • Let us continue this discussion in chat. – Anna Apr 25 '22 at 20:19