Gauss' Law and Infinite plane sheet of charge (1 Viewer)

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ximath

Hi All,

I am studying Gauss' law and have learned that using symmetry, we need to select a cyclinder in order to calculate electric field of an infinite plane sheet of charge.

$$2EA = \frac {\sigma A} {\epsilon}$$

That equation is written using Gauss' law and hence E field is found.

However, why wouldn't we use a cube instead of the cyclinder, for instance ?

Moreover, I am not able to understand why the cyclinder is suitable. (I know it is due to the symmetry arguments, but why ? ) I mean, the normal vector on the surface needs to be parallel to E field if we select the cyclinder. However, I can't see why would E field be parallel to the normal vector.

Doc Al

Mentor
Realize that the Gaussian cylinder is oriented so that its flat ends (not its curved surface) are parallel to the sheet of charge.

You are free to use any shape Gaussian surface you wish, as long as you can take advantage of symmetry. For an infinite sheet you can use a cube, cylinder, and many other kinds of surfaces to derive the electric field.

ximath

Hi Doc Al, thanks a lot!

As far as I know I am free to choose any closed surface in fact, however, symmetrical ones are preferable since they help us calculate the electric field.

If I used a cube instead of a cylinder, would it also have the symmetry property ? In other words, on the surface of that plane (which is parallel to the sheet), would electric field and normal vectors be parallel everywhere ? If this is the case, then why would all the textbooks mention about the cylinder but not about a cube as an instance ?

One more thing I am trying to realize is why E field is parallel to the normal vector everywhere on the surface that is parallel to the sheet... I think the E field components that are not perpendicular to the sheet must be canceled somehow, but I am having some problems imagining that. Do you have any visual Java applets or even images that explain this symmetry property ?

Doc Al

Mentor
As far as I know I am free to choose any closed surface in fact, however, symmetrical ones are preferable since they help us calculate the electric field.
Exactly.

If I used a cube instead of a cylinder, would it also have the symmetry property ? In other words, on the surface of that plane (which is parallel to the sheet), would electric field and normal vectors be parallel everywhere ?
Sure.
If this is the case, then why would all the textbooks mention about the cylinder but not about a cube as an instance ?
Beats me! (Maybe so they can use the same shape for a line of charge.)

One more thing I am trying to realize is why E field is parallel to the normal vector everywhere on the surface that is parallel to the sheet...
If the field did point in some other direction (not perpendicular to the surface) what would determine the direction? The sheet of charge is uniform in all directions, so no one direction can be chosen.
I think the E field components that are not perpendicular to the sheet must be canceled somehow, but I am having some problems imagining that. Do you have any visual Java applets or even images that explain this symmetry property ?
I don't have any off hand, but you might find something if you Google it.

mahela007

This thread look to be long dead... however, I did find it useful. I had the same question (more or less) as the OP and post #4 did a good job of explaining things. Thanks Doc Al.

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