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How can Gauss's Law be Correct if We Consider Charges Not on The Gaussian Surface?

by modulus
Tags: charges, correct, gauss, gaussian, surface
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modulus
#1
Jan15-10, 04:17 AM
P: 103
Gauss' Law only figures out the the total charge enclosed by the Gaussian surface and on it, I treid this in a few cases where charges were placed only inside the surface, and my answers were correct.

But, when I tried imagining how to do this if there were charges present outside the Gaussian surface , I figured the flux of the electric field would change, because charges outside the surface would vary the electric field lines of charges present inside the surface.

So we won't be getting the flux due to charges only in the surface, but outside it too. That would give you the total charge of all charges around the surface, not only tose enclosed by the Gaussian surface, right?
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Pengwuino
#2
Jan15-10, 04:27 AM
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You get a flux from charges outside the surface, but it will all cancel out if there are no charges inside the surface.
vela
#3
Jan15-10, 04:51 AM
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Any field lines from an outside charge that enter the volume enclosed by the surface will also exit, so the net flux from that charge will be zero.

modulus
#4
Jan18-10, 03:56 AM
P: 103
How can Gauss's Law be Correct if We Consider Charges Not on The Gaussian Surface?

Sorry for replying so late. I had gone out of town.

OK, I understand that the electric field lines entering the surface will exit it. But, it will exit from the other side, correct?

Now this brings up two problems:

Suppose the Gaussian surface is irregular. Then, it wouldn't be neccessary that the electric field lines will exit in the same manner that they entered, so as to cancel out. Therefore, the total flux should be affected by the external field lines



Also, if there are charges present inside the Gaussian surface, they should not allow the electric field lines due to the external charges to exit normally.

Suppose there is a positive charge inside a cylindrical (symmetrical) surface and there is a positive charge outside of it too. Well then, the net electrical flux due to that charge would cancel out only if it exits fom the surface opposite to the one from where it entered. But, the positive charge inside would deflect the field lines of the positive charge outside, so as to prevent it from exiting from the opposite surface. This means the net electric flux would *not* be zero.
uart
#5
Jan18-10, 05:22 AM
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Quote Quote by modulus View Post
Now this brings up two problems:

Suppose the Gaussian surface is irregular. Then, it wouldn't be neccessary that the electric field lines will exit in the same manner that they entered, so as to cancel out. Therefore, the total flux should be affected by the external field lines
Hi Molulus, in electrostatics there is one thing that an electric field can never do, and that is curl into a closed loop. Now the only way that a field line could enter an enclosed surface without subsequently leaving it would be if it formed a closed loop inside the surface (which of course it can not do).
modulus
#6
Jan19-10, 04:29 AM
P: 103
But, I do not believe it is neccessary for the electric field line (of the external charge) to make a loop in the enclosed surface. Can't it just diverge slightly after entering the lopp and then exit out non-symmetrically?
And, won't the fact that it is an irregular surface prevent any firld line from exiting symmetrically?
vela
#7
Jan19-10, 04:54 AM
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The symmetry you are assuming is required is, in fact, not required. Consider an incompressible fluid and defining a Gaussian surface within the fluid. The amount of fluid entering the volume has to equal the amount exiting if there are no sinks or sources inside the volume. It doesn't matter how a piece of the fluid enters the volume or how it exits; it all works out so that the total amount entering and exiting the volume are equal. It's the same way with the electric field lines from external charges.

The only way you can get a disparity is if some of the field lines terminate inside the volume. Because field lines only start and end on charges, the disparity is proportional to the amount of charge inside the volume.
uart
#8
Jan19-10, 05:21 AM
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Yes I think you have some misconception of Gauss's Law modulus.

Gauss's Law simply states that the total flux leaving a closed surface is equal to the charge enclosed. The total flux is the surface integral (over the entire surface) of,

[tex]\vec{D} \cdot d\vec{s} [/tex],

it doesn't have to be symmetrical or uniform. ([itex]D = \epsilon E[/itex] btw).
modulus
#9
Jan20-10, 02:18 AM
P: 103
All right, so the symmetry has nothing to do with it. The field lines will exit out anyways. They change when there is some charge inside the surface which will cause them to diverge. That divergence indicates the net charge in the surface. Thanks a lot., that cleared up my doubts. (gosh, you guys should write a textbook on physics or something! Your language was much clearer than the language of my book. And the analogy was great, Vela, thanks!)

But, I still have a teeny little problem, having to do with the mathematical part of it. The value we finally get after integrating over the Gaussian surface; is it equal to the net charge inside the surface, or is it the net charge inside and on the surface, or is it the net electric field on the entire surface?

Also, when we do the integration, we find the net electric flux on the total surface. But I'm still having some trouble with the concept of what flux exactly is (and with dipole moments too). I mean, what is the physical meaning of a value which is equal to the product of the net electric field on a surface and the surface's area....doesn't it make more sense to figure out the net electric field on the surface divided by the area?


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