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If you have two parallel planes with one density of 1 C/m^2 and the other with -1 C/m^2 density and you need to find the electric field at a point in the middle, how do you use Gauss's Law?

I drew a box enclosing one plane to find the electric field in the midpoint. I understand there's another plane exerting an additional force, but by Gauss's law it should only depend on the enclosed charge.

Is Gauss's Law being violated here?

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  • $\begingroup$ There are no infinite planes in nature and Gauss' Law in its usual form has a finite integration domain. It simply doesn't apply to the entire planes. Can it be extended with some non-trivial mathematical effort to infinite domains? Probably, but we don't teach it that way. That doesn't mean one can't use it for the problem... one can, by only integrating over a finite volume. $\endgroup$
    – CuriousOne
    Commented Jan 21, 2016 at 2:21
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    $\begingroup$ You can approximate a plane to be infinite by being sufficiently close to it. This is how it's taught and plenty of questions ask about infinite planes. I'm not sure what you mean by "doesn't apply to the entire planes" $\endgroup$
    – JobHunter69
    Commented Jan 21, 2016 at 2:23
  • $\begingroup$ Gauss' Law is a precise mathematical theorem. It is, in its original form, limited to a finite integration domain. Unfortunately many teachers still resolve to handing out "infinite whatever" questions, rather than to teach how to do proper approximations on finite objects. $\endgroup$
    – CuriousOne
    Commented Jan 21, 2016 at 3:34
  • $\begingroup$ Gauss's Law is not being violated, because when there are two parallel planes, then the electric field is zero at the side of the box outside of the two planes. When there's only one plane, it's equal on both sides. As a result the field is twice as much with two planes, which makes sense. $\endgroup$
    – knzhou
    Commented Jan 21, 2016 at 3:36
  • $\begingroup$ The answer to &"How does one apply Guass' law [...]?"* is "By taking advantage of symmetries." without caring much what goes in the [...]. If there are no symmetries than the answer "Don't bother". $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Jan 21, 2016 at 3:44

3 Answers 3

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you don't have to find it through gauss' law. you can simply calculate the resultant electric field by vector addition of the electric field of every plane as how you calculate resultant force while many forcers are applied on a particle.

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The electric field due to an infinite plane is constant in the entire space around it so if you have two planes with opposite charge density the electric field between the planes will be zero. And double on the out side of the planes

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  • $\begingroup$ I believe you mean the other way around? $\endgroup$
    – AccidentalFourierTransform
    Commented Jan 21, 2016 at 11:55
  • $\begingroup$ @AccidentalFourierTransform yes you are right, by mistake i have mixed the directions $\endgroup$
    – Ashok Sharma
    Commented Jan 27, 2016 at 8:08
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No, Gauss law is not at all violated here. enter image description here When you considered a single plate you get flux $\phi$ in box ABCD. Total field would be $\phi$ as sum on two sides AB and CD. Let here flux towards AB is $\frac{\phi}{2}$ and towards CD is $\frac{\phi}{2}$ as well.

Net flux = $\phi$. _________(1)

enter image description here When second plane of -ve charge is added. You again get an electric field. But this time you see that E is towards -ve plate. Flux $\phi$ entering the side CD is equal to flux exiting side AB .

Net flux due to -ve plate = +$\frac{\phi}{2}$ - $\frac{\phi}{2}$ = 0 ._______(2)

On adding (1) and (2) Total flux = $\phi$ +0 = $\phi$

As Gauss law stated Net flux from ABCD = $\frac{+q}{\epsilon}$ which is independent of -q on other plate as no net flux entered due to it.

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