Gauss' law and parallel plates

1. Mar 8, 2015

hitemup

1. The problem statement, all variables and given/known data

Two large, flat metal plates are separated by a distance that is very small compared to their height and width. The conductors are given equal but opposite uniform surface charge densities +- $\sigma$. Ignore edge effects and use Gauss's law to show

a) that for points far from the edges, the electric field between the plates is $E = \frac{\sigma }{\varepsilon_0}$.

b) that outside the plates on either side the field is zero.

c) How would your results be altered if two plates were nonconductors?

2. Relevant equations

$\phi = \oint E *dA = \frac{Q_(enc)}{\epsilon_0}$

3. The attempt at a solution

I've searched a lot to find a solution to this problem.

http://aerostudents.com/files/physics/solutionsManualPhysics/PSE4_ISM_Ch22.pdf (solution number 24)
http://www.phys.utk.edu/courses/Spring 2007/Physics231/chapter22.pdf (page 21)

In both of these links, the approach to find electrical field between the plates is

1- create a cylindrical gaussian surface
2- put one end of the cylinder to one of the plates where the area is uncharged(due to attraction between two plates)
3- put other end to be between the plates.

Since the flux will pass through only one end of this cylinder

$$EA = \frac{\sigma A}{\epsilon_0}$$
$$E = \frac{\sigma }{\epsilon_0}$$

My question is, why didn't we do the same thing for the other plate, and then use superposition principle?
Or simply, why didn't we multiply what we found by 2 because of superposition?

2. Mar 8, 2015

BvU

The simple answer is that Gauss law really holds.
It is not so that the opposite plate has no influence on the electric field ! The influence is in establishing the $\sigma$.

3. Mar 8, 2015

hitemup

You have two point charges along the same axis, and you want to calculate the field between them.
Creating a gaussian sphere would lead to Coulomb's law, kq/r^2, where q_enclosed in the law is q, and r is the distance from that charge.
We just found an electrical field equation, but is that all? Shouldn't we consider the other charge?

4. Mar 8, 2015

hitemup

I wanted to double it because I thought that the calculation lacks the contribution of the negative charged plate. How can I know that the negative charged plate is involved in calculation?

5. Mar 8, 2015

rude man

Really good question! I deleted my previous answer because it was inadequate. Need to mull this one over myself, including the the 2-point charge problem. In the 2-point charge problem, the second charge clearly adds to the E field produced by the 1st charge, whereas the charges on the second plate do not add to the E field produced by the 1st plate.
EDIT: the charges on the 2nd plate are there because of the charges on the 1st plate, whereas your 2 point charges are independent of each other.
Even if you started with one plate charged and the second plate with zero net charge, the second plate would still have induced surface charges such that σ is the same for both plates.
Gauss' theorem states that ε∫E⋅dA = Qfree. But the charges on the second plate are not free: they are there because of the charges on the 1st plate. Or vice-versa. The surface charges are not independent, i.e. not free. I'm not 100% satisfied with this 'explanation' myself but maybe it'll help you think about it. Hopefully others will pitch in. TSny?

Last edited: Mar 8, 2015
6. Mar 8, 2015

hitemup

Is it because of the renewed charge distributions?

For example, a simple plate would be charged like this

+| |+

But when you have a system with two plates, they are charged like this.

|+ -|

We are doing the calculation with respect to this charge distribution, does that guarantee the field will be contributed from both plates?

7. Mar 8, 2015

rude man

Yes. You can't have one without the other. Put a gaussian surface running from the inside of one plate to the inside of the other. There has to be net zero charge inside the surface since the E field inside both plates is zero. So σ1right = -σ2left.
You can have 2 plates with different net charge on each, but the surface charge density on the inside surfaces would still have to be equal and opposite. The excess charges will move to the two outside surfaces.
The one question that eludes me at the moment is what the charge distribution between unequally charged plates (Q1 and Q2) would be. I can get 3 equations but there are 4 unknowns: σ1left, σ1right , σ2left and σ2right.
Assuming unit plate area,
Q1 = σ1left + σ1right ,
Q2 = σ2left + σ2right and
σ1right = -σ2left.

8. Mar 9, 2015

BvU

Anxious to hear from you how you find the field that way!

Because you no longer have spherical symmetry as you seem to think; there is only symmetry under rotation around the axis joining the two charges.

The flux through the gauss sphere is still $q\over \epsilon_0$. But between the charges the field strength is a lot more (case $+-\$; for $++$ a lot less) than on the far away side !

9. Mar 9, 2015

rude man

For each of the two charges individually there is clearly spherical symmetry of the E field. So superposition can be used to determine the total E field at any point where that point is common with the intersection of the two spherical surfaces.

The OP has brought up a very interesting question IMO.

10. Mar 9, 2015

BvU

We agree. I'm definitely not arguing with superposition, just saying that Gauss' law holds and that by itself it's not enough to determine the $\vec E$ field in the case as posed in post #3.

And yes, the problem statement and the further posts challenge what many take for granted, but have difficulty explaining clearly (at least I do...).

Assume the lower plate has the positive charge and is not connected to anything. When the upper plate is removed, the charge spreads evenly over the entire surface. If we can ignore the edge effects, this halves the field strength close to the surface !

11. Mar 9, 2015

rude man

It is enough to determine the field of post 3.
W/l/o/g, place the two charges q1 and q2 on the x axis at q1(x1,0,0) and q2(x2,0,0). Pick any point P(x,y,z).
Wrap a gaussian spherical surface about q1, touching P. The E1 field at P due to q1 is by Gauss magnitude kq1/r12 where r1 is the distance from x1 to P. The direction of E1 is the direction of x1 to P.
Do same for q2 to get E2.
Then the total E field is the vector addition of E1 and E2. We have used Gauss exclusively, obviously getting the same E magnitudes as with Coulomb:
ε∫∫E⋅dA = ∫∫∫ρdV = qfree (Gauss + Maxwell)
or 4πr2εE = q or
E = q/4πεr2 = kq/r2 = Coulomb's law.
Coulomb's law is thus directly derivable from Gauss's law with Maxwell's ∇⋅D = ρ.
The OP is simply wondering why q1 and q2 both contribute to the total E field whereas only one plate's charges do. A very good point, so far not explained to my satisfaction.

12. Mar 9, 2015

BvU

Yeah, I forgot to state I don't mean two trivial Gausses plus one superposition. I mean one Gauss, e.g. around one of the two charges in the presence of another charge. But you're right, I didn't say that.

And it is not so that only one plate contributes to the total E field. Witness posts #2 ("The influence is in establishing the σ ") and #10 (remove a plate E $\rightarrow$ E/2) !

My gratitude will go to whoever brings up a clearer explanation !

13. Mar 9, 2015

SammyS

Staff Emeritus
With these two charges present, Gauss's Law is not of any use in determining the E field at any place on this sphere. Gauss simply tells us the total flux leaving the surface of the sphere. The field is not of uniform magnitude on the surface of the sphere and its direction has varying orientation relative to vectors normal to the surface. So Gauss is not useful in this setting for finding E field at a particular point.

As BvU points out; Yes, it is possible to use Gauss's Law to determine the electric field due to either of these charges if the charge is isolated. Then superposition can be used for the combination.

14. Feb 8, 2016

rude man

This is REALLY ancient history but IMO a good answer was never produced.

OK, point #1 is that Gauss's theorem as applied to Maxwell's equations is ALWAYS correct. You can set up any gaussian surface between or inside the plates and you will always get the same E field. This has been pointed out already above.

Point #2 is this: superposition in general applies ONLY to independent sources.
If you recall from circuits theory , to use superposition you have to include all dependent sources concomitant with their respective independent sources. For example, suppose you have independent source A and dependent (on A) source B. You cannot compute voltages and currents (v and i) by separately (mentally) exciting source A and B. B must always tag along with A. But if you have a third source C, assumed independent of A, then you can compute v and i due to A (incl. dependent sources), then separately compute v and i using B (along with any of its dependent sources), then you can add all the voltages and currents when all sources are excited.

And so it is with this situation: you can not separately compute the E field using the + and - induced charges, then apply superposition, since they are not independent; in fact, charge on plate 1 = - charge on plate B. Note that, irrespective of the initial charge of each plate, the inner surface charges will always be equal & opposite providing the distance between them is arbitrarily small (or their areas are arbitrarily large). Again, gauss's theorem proves it.