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Electric Field of a Conducting Surface

  1. Jan 15, 2014 #1

    kmm

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    I understand that taking the boundary condition that the electric field changes [itex] \frac{\sigma}{\epsilon_{0}} [/itex] when you cross a charged surface and the fact that the electric field is zero inside a conductor that the electric field outside the conductor should be [itex] \mathbf{E} = \frac{\sigma}{\epsilon_{0}} \hat{\mathbf{n}} [/itex].

    What's troubling me is considering electric fields other than the field due to some patch. In the case of a conductor, the electric field [itex] \mathbf{E} = \frac{\sigma}{\epsilon_{0}} \hat{\mathbf{n}} [/itex] is due to the patch and other areas, even external ones. This was because of the boundary condition which was derived for a single patch of a charged surface. The field must come from other areas, not just the patch.

    Although, when deriving the boundary condition, the Gaussian pillbox used over the surface was made small enough as not to include areas with different values of [itex] \sigma [/itex] and not to include any curving of the surface. (Wouldn't that mean the boundary condition would be different if we did include these other regions?) Yet, by invoking the boundary condition for a conductor, we now have to include these other fields.

    Assuming the "other" fields are the same on top and bottom and using a positive charge distribution, I took [itex] \mathbf{E}_{above} = \mathbf{E}_{other} + \mathbf{E}_{patch} [/itex] and [itex] \mathbf{E}_{below} = \mathbf{E}_{other} + \mathbf{E}_{patch} [/itex]. Above, [itex] \mathbf{E}_{patch} = \frac{\sigma}{2 \epsilon_{0}} \hat{\mathbf{n}} [/itex] and below [itex] \mathbf{E}_{patch} = - \frac{\sigma}{2 \epsilon_{0}} \hat{\mathbf{n}} [/itex] so when we take [itex] E_{above} - E_{below} [/itex] the "other" fields cancel and we get [itex] \frac{\sigma}{\epsilon_{0}} [/itex], as expected. It's not surprising to me that the external fields cancelled though because [itex] E_{above} - E_{below} [/itex] was obtained from gauss's law and any external fields through a Gaussian surface won't contribute to the flux. This makes me wonder if the other fields could change the boundary condition. It also makes me question applying the boundary condition to a conducting surface. I don't think I'm right, I'm just not sure what I'm thinking wrong.
     
    Last edited: Jan 15, 2014
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  3. Jan 15, 2014 #2

    kmm

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    To be more concise, the boundary conditions for a charged surface are found only by considering the electric field due to some patch. However, when invoking the boundary conditions to determine the electric field outside a conductor, we must now include the electric field due to spots other than the patch. Why is this? This seems to be a contradiction.
     
  4. Jan 15, 2014 #3
    This statement is where your mistake is. It's not quite true.

    Gauss's law gives us a relationship between the charges inside some volume and the electric field on the surface of that volume--the whole electric field, created by all charges in the system, not just the ones inside the volume. That's one of the remarkable things about Gauss's law: it says that charges outside of a volume may affect the electric field at different points on the surface of that volume, but they won't affect the total integral of the electric field.

    Now, in the conducting surface case, we look at a small Gaussian pillbox on some small patch of the surface. The electric field on that pillbox is created by all the charges in the system, not just the ones in the pillbox, so we don't know what the electric field is, though we know that it will be pretty much constant if we pick a small enough pillbox. However, Gauss's law tells us that whatever that electric field is and however it was created, the integral of the electric field coming out of the surface of the pillbox must be proportional to the charge inside the pillbox. From that, we get a relationship. E = σ/ε0

    It's a bit subtle, but we're not actually finding the electric field created by the charges in the pillbox. Rather, we're finding a relationship between the charges inside the pillbox and the electric field created by all charges in the system. It's a little surprising, maybe, but because of Gauss's law we actually can find such a relationship.

    If we were to, say, use Coulomb's law to figure out the electric field due to that small patch of charges, we would get the wrong answer. If you think about it, that approach wouldn't explain at all why the electric field is zero on one side of the patch (inside the conductor) and non-zero on the other side (outside the conductor). You would expect the E field to point away from the surface charge on both sides, not just on one side. So clearly, with the Gauss's law approach, we are somehow including the contribution from the other charges implicitly. (Actually, we're pretty much doing that when we invoke the fact that E=0 inside the conductor. We're saying that all the charges everywhere must conspire to make E=0 in the conductor, and then we use Gauss's law to figure out what it must be outside the conductor.)

    Maybe you're still skeptical, though, about how we found the electric field due to all charges while only considering a small portion of the charges. It seems too good to be true, right? Well, it kind of is. Gauss's law does indeed give us a relationship between the charges inside a volume and the total electric field (due to all charges) on the surface of that volume, but it's actually rare that we can figure out the electric field on the surface of a volume by applying Gauss's law to that volume. Gauss's law only gives us a little bit of information about the electric field (the surface integral), so if we don't have some other information to add (e.g. symmetry, or in this case, the fact that the pillbox is infinitesimally small), we can't actually figure out what the electric field is.

    NOTE: I've spoken a bit loosely here so that I could focus on the key aspect of your question, but I should add that even in this case, Gauss's law alone doesn't actually give us the full electric field. It only tells us the perpendicular component of the electric field; there still could be a tangential component. However, in this case, using Ampere's law, we can show that that tangential component is zero, completing the boundary condition. In other cases, symmetry might tell us that the tangential component is zero, but Gauss's law only ever tells us about the E field perpendicular to a surface.

    Edit: I realize that was kind of a long, rambly response. Let me know if it needs some clarification...
     
    Last edited: Jan 15, 2014
  5. Jan 15, 2014 #4

    kmm

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    If charges outside the system don't effect the integral, then the integral should only tell us about the electric field due to the charges. If I have 2 electrons floating in space and I surround only one of them with a Gaussian surface, we could only determine the electric field from the charge we surrounded. Obviously, the electric field Gauss's law would give us would be incorrect because the other electron we ignored would have an effect on the total field.

    Right, so here, Gauss's law would give us the electric field due to the patch.

    Again, I don't think Gauss's law tells us anything about the electric field due to what's outside the pill box.

    I would think Coulomb's law could tell us that. Using coulombs law, we would fine the electric field due to the charge on the outside of the conductor, but take into account the charges that cancel the field on the inside of the conductor.

    Yes, it's because a conductor could have some non symmetric surface structure or charge density and that's what is making it hard for me to see how we can use Gauss's law to come up with a specific boundary condition for a charged surface, then apply it to a conductor.

    But in the integral, the dot product [itex] \mathbf{E} \cdot d \mathbf{a} [/itex] picks out the component of [itex] d \mathbf{a} [/itex] along [itex] \mathbf{E} [/itex]. It wouldn't just be the component of the E field perpendicular to the surface. You want the E field to be parallel to [itex] d \mathbf{a} [/itex] to make the integral simpler, otherwise you would have to take that into account and couldn't take [itex] \mathbf{E} [/itex] out of the integral. Symmetry just makes it easier.
     
    Last edited: Jan 15, 2014
  6. Jan 15, 2014 #5

    kmm

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    If surrounded a charge with a sphere or cube, we would get the same value for the flux and could determine [itex] \mathbf{E} [/itex], using a cube would just be harder.

    *Edit:

    I suppose your right that we have used Gauss's law to determine only the perpendicular component of E in this case. But yes, combining that result with the fact that the tangential component is zero, we know the total field due to the patch.
     
    Last edited: Jan 15, 2014
  7. Jan 16, 2014 #6

    Meir Achuz

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    Above, [itex] \mathbf{E}_{patch} = \frac{\sigma}{2 \epsilon_{0}} \hat{\mathbf{n}} [/itex] and below [itex] \mathbf{E}_{patch} = - \frac{\sigma}{2 \epsilon_{0}} \hat{\mathbf{n}} [/itex] so when we take [itex] E_{above} - E_{below} [/itex] the "other" fields cancel and we get [itex] \frac{\sigma}{\epsilon_{0}} [/itex], as expected. [/QUOTE]
    That is often given as a homework problem.
    "Prove that if a hole is punched in the surface of a conducting shell, the E field just inside the hole is one half the E field just outside the shell before opening the hole."
    The proof is that you have just shown that half the field comes from the 'patch;, and half from the rest of the conductor.
     
  8. Jan 16, 2014 #7

    kmm

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    Yes. Griffith's get's this field from the electrostatic boundary condition of crossing a charged surface. It seems to me that the boundary condition only considers the field due to the patch and not the other fields. So it seems invalid when applying the boundary condition to a conductor since we have to consider the fields from the rest of the conductor.
     
  9. Jan 16, 2014 #8
    First, you probably know this, but I just want to make sure Gauss's law is crystal clear.

    [tex]\iint_{\substack{\text{closed}\\\text{surface}}} \mathbf{E} \cdot d \mathbf{l} = \frac{Q_{\text{enclosed}}}{\varepsilon_0}[/tex]

    E is the electric field. It is the entire electric field created by all the charges in the entire system. Qenclosed is only the charges in the volume.

    So Gauss's law is a relationship between the charges inside a volume and the electric field created by all charges everywhere--including the charges outside the volume.

    They don't affect the integral, but they certainly affect all the individual contributions to the integral. If we have a surface which doesn't have any charge enclosed, but I tell you that the integral of the electric field over 99% of the surface is +10, then you know the integral over the other 1% must be -10. If that small chunk of surface is small enough, you can be pretty confident that the E field points into the surface on that chunk. So, we've used Gauss's law to get information about the electric field due to outside charges.

    It's the same thing with the conductor problem. I tell you that the integral over half the pillbox is zero because E=0 in a conductor. From that, you know that the integral over the other half must equal the charge enclosed by the pillbox. However, now we're dealing with an integral over half a surface, not over an enclosed surface, so the fields due to outside charges will not, in general, integrate to zero on that half-surface. Thus, when we perform that integral, we are indeed accounting for the outside charges.

    The key here is that we're looking at integrals over small parts of the surface rather than the whole surface. The electric field from outside charges will, in general, affect these small parts.

    This statement seems to indicate a fundamental misunderstanding of Gauss's law...

    In the case you describe, Gauss's law is perfectly true, but it does not allow you to determine what the electric field is. There's no symmetry or additional information to exploit to allow you to bring E outside of the integral, so you're just stuck. You can get an answer by ignoring the outside charge, but it's the wrong answer because all you've done is applied Gauss's law incorrectly. If you apply Gauss's law correctly over a single surface, you can't get much info about the electric field in this case, because of the outside charge.

    No. Gauss's law gives us a relationship between the whole field and the charge in the patch.

    Think of it this way: the charges on the rest of the conductor are actually quite dependent on the charge in the patch. All the other charges have to align to make sure that the E field created by the patch gets cancelled out inside the conductor. I think that was Meir Achuz's point.

    Yes it does, because we're looking at individual contributions to the integral, not just the whole integral. Fields from outside charges cancel out over the whole integral, but not in the individual parts of the integral. You can't just pretend outside charges don't exist when applying Gauss's law.

    You're right, of course, but that wasn't my point. My point was that if we were indeed only finding the field due to the charges in the patch, we would find a non-zero field inside the conductor. We only have a zero field inside the conductor because we are properly accounting for all the other charges.

    Just to drive it home... we really are actually accounting for all the other charges. You can't just have any arbitrary charge distribution on the surface of a conductor. They have to have a very specific distribution which makes the E field zero in the conductor.

    [itex]d\mathbf{a}[/itex] is perpendicular to the surface...

    Symmetry doesn't just make the problem easier, symmetry is what allows you to solve for the electric field at all. If we can't make statements like "the electric field has constant magnitude over the surface," then we can't find the electric field using a single application of Gauss's law. In the example you gave above with two charges, the existence of an outside charge means we can't make statements like that, and it's impossible to find the electric field.

    In fact, even if both charges were inside the surface, you still wouldn't be able to find the field because there's no symmetry. You can't say things like "the E field is constant over the surface," so how could you bring E out of the integral?

    That's actually not true. You can't find the electric field using a single cubic surface and one application of Gauss's law. It's not that it's just harder to do; you actually can't do it. Gauss's law doesn't give you enough information. Symmetry isn't just "nice" in Gauss's law, it's critical. Gauss's law is always true, but if there's no symmetry, then applying Gauss's law to a single surface will rarely give you enough information to solve for the electric field.
     
  10. Jan 16, 2014 #9

    kmm

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    OK, I feel that I might be getting somewhere now. I still have some things I want to clear up and am not totally sure about though.

    I think this is probably the root cause of my misunderstandings. The way I've been looking at it is, the E field that we pull out of the integral if we are applying Gauss's law correctly (ie to a symmetric system of charges), and the value we get for [itex] \mathbf{E} [/itex] would be the electric field due to the charges within the surface, not the E field of the entire system. I assume this because,

    [tex]\iint_{\substack{\text{closed}\\\text{surface}}} \mathbf{E} \cdot d \mathbf{a} [/tex]

    is the flux of the E field through the surface. Any charges outside of this surface contributes nothing to the flux because those charges E field lines go in and then right back out. This is why I'm not seeing how [itex] \mathbf{E} [/itex] is the electric field created by all the charges in the system, in addition to those charges inside the surface.


    I'm not sure how the outside charges can not effect the integral but have an effect on the individual contributions to the integral.

    In this case, I can understand how we could infer what's going on with the rest of the E field. But we can only know what the electric field due to outside charges because we knew the total field prior to the calculation. Maybe this is what you're getting at?

    Here seems to be where the boundary condition for a charged surface is applied to a conductor. I see how because the boundary condition says that the E field has a discontinuity of [itex] \frac{\sigma}{\epsilon_{0}} [/itex] and since we know that [itex] \mathbf{E} = 0 [/itex] inside the conductor, we're inferring that the E field must be [itex] \frac{\sigma}{\epsilon_{0}} \hat{\mathbf{n}} [/itex] above the surface. Again, this is where I'm hung up. I don't see how just because the electric field has been made zero within the conductor that the electric field above the conductor must also change. I thought I could get away with saying that the charge density on the surface of a conductor changed to make the E field change from [itex] \frac{\sigma}{2 \epsilon_{0}} \hat{\mathbf{n}} [/itex] to [itex] \frac{\sigma}{\epsilon_{0}} \hat{\mathbf{n}} [/itex], but that doesn't work because that only changes [itex] \sigma [/itex] and doesn't get rid of the [itex] \frac{1}{2} [/itex]. That's where the "other fields" come in because the field due to the patch is [itex] \frac{\sigma}{2 \epsilon_{0}} \hat{\mathbf{n}} [/itex]. Sorry for repeating myself, but here I'm still not sure why it is now that we are considering other fields, but not when we created the boundary condition in the first place. Maybe I'm being circular here.

    I actually agree with you here. This was actually my point of the example was to show that I couldn't determine what the electric field is and that the answer I would get is wrong.

    Again, here I assume you mean we are inferring from the individual integrals what the total integral would be if we integrated over the entire surface.

    I see now that symmetry is crucial in determining what the electric field is. Without symmetry, we can't pull E out of the integral.
     
  11. Jan 16, 2014 #10
    Let me try to state more precisely what I said about the field due to outside charges contributing to individual parts of the integral, but not the whole thing.

    Suppose we have a closed surface, S, and we break that surface up into N different non-closed surfaces, S1, S2, ... SN, so that S = S1 + S2 + ... + SN.

    Let's also break up the charge, so that qin is the charge inside the surface S, and qout is the charge outside the surface S. The total charge in the whole system is q = qin + qout.

    On a similar line, we'll break up the electric field: E = Ein + Eout, where Ein is the E field created by qin and Eout is the E field created by qout

    Now, Gauss's law says:

    [tex] \iint_S \mathbf{E}\cdot d\mathbf{l} = \frac{q_{in}}{\epsilon_0} [/tex]

    My point was, it's true that

    [tex] \iint_S \mathbf{E}_{out}\cdot d\mathbf{l} = 0 [/tex]

    and

    [tex] \iint_S \mathbf{E}_{in}\cdot d\mathbf{l} = \iint_S \mathbf{E}\cdot d\mathbf{l} [/tex]

    but it is not necessarily true that

    [tex] \iint_{S_i} \mathbf{E}_{out}\cdot d\mathbf{l} = 0 [/tex]

    and thus we can't say that

    [tex] \iint_{S_i} \mathbf{E}_{in}\cdot d\mathbf{l} = \iint_{S_i} \mathbf{E}\cdot d\mathbf{l} [/tex]

    for any given i.

    Now let's look at the conductor boundary condition problem agani. We have two surfaces which are important to the pillbox: S1 (inside the conductor) and S2 (outside the conductor).

    We know that

    [tex]\iint_{S_1} \mathbf{E}\cdot d\mathbf{l} + \iint_{S_2} \mathbf{E}\cdot d\mathbf{l} = \frac{q_{in}}{\epsilon_0}[/tex]

    At this point (and I suspect this is actually a big part of your confusion), we could remove the Eout part if we wanted to. A true statement is

    [tex]\iint_{S_1} \mathbf{E}_{in}\cdot d\mathbf{l} + \iint_{S_2} \mathbf{E}_{in}\cdot d\mathbf{l} = \frac{q_{in}}{\epsilon_0}[/tex]

    But the next step of the derivation doesn't actually work if we do this. The next step is to say that E=0 on S1 (inside the conductor). We cannot say that Ein or Eout is zero, only that E = Ein + Eout is zero inside the conductor. So we have some information about Eout which tells us that

    [tex]\iint_{S_1} \mathbf{E}\cdot d\mathbf{l} = 0[/tex]

    Obviously, this is more useful in the equation involving the whole field, and not the equation involving just Ein, the field due to the patch. Using this, we get

    [tex]\iint_{S_2} \mathbf{E}\cdot d\mathbf{l} = \frac{q_{in}}{\epsilon_0}[/tex]

    which pretty much completes the boundary condition. It would be wrong to say that

    [tex]\iint_{S_2} \mathbf{E}_{in}\cdot d\mathbf{l} = \frac{q_{in}}{\epsilon_0}[/tex]

    because we didn't prove that [itex]\iint_{S_1}\mathbf{E}_{in}\cdot d\mathbf{l} = 0[/itex], we proved that [itex]\iint_{S_1}\mathbf{E}\cdot d\mathbf{l} = 0[/itex]

    So, my point is, we're always working with the whole field. I think what's maybe getting you is the fact that we could eliminate Eout in Gauss's law if we wanted to. If we did that, then you are correct, we would only get information about Ein and we would have to find Eout some other way. However, that doesn't mean we can't get information about Eout through Gauss's law. We don't have to eliminate Eout from Gauss's law, and in fact, Gauss's law tells us a very important characteristic of Eout: it integrates to zero over the closed surface.

    So, if we can inject some information about Eout over part of the surface, Gauss's law tells us what must happen to Eout over the rest of the surface. In this case, we inject some information about Eout on the part of the surface that's inside the conductor (we know that Eout + Ein = 0) and Gauss's law tells us about Eout on the rest of the surface (it tells us that Eout + Ein = σ/ε0)

    I hope that makes a little more sense.
     
  12. Jan 16, 2014 #11
    Like Meir Achuz said. The patch generates an electric field of [itex] \frac{\sigma}{2 \epsilon_{0}}[/itex] on both sides of it. Inside the conductor, the electric field generated by the patch is [itex] \frac{\sigma}{2 \epsilon_{0}} [/itex] pointing inward. Outside the conductor, the electric field generated by the patch is [itex] \frac{\sigma}{2 \epsilon_{0}}[/itex] pointing outward. Note that there's an abrupt change in the direction of the field when we pass through the patch.

    Now, inside the conductor, the total E field has to be zero. So if, inside the conductor, the field generated by the patch is [itex] \frac{\sigma}{2 \epsilon_{0}} [/itex] pointing inward, then the field generated by all the other charges must be [itex] \frac{\sigma}{2 \epsilon_{0}} [/itex] pointing outward to cancel it out. However, while the field from the patch abruptly changes direction at the surface, the field from the other charges won't do that: it will still point outward. So while it cancels the field from the patch when you're inside the conductor (giving a total field of zero), it adds to the field from the patch when you're outside the conductor, giving a total field of [itex] \frac{\sigma}{2 \epsilon_{0}} + \frac{\sigma}{2 \epsilon_{0}} = \frac{\sigma}{ \epsilon_{0}}[/itex].

    This is probably an easier way to understand the boundary condition compared to my last post... The last post was trying to help you understand that Gauss's law really does tell us about the whole field, not just the field from charges inside the volume. That's just something that's important to understand in general.
     
  13. Jan 16, 2014 #12

    kmm

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    Everything will come together for me (I think) if I understand your point in post #10. I was able to follow you until here:

    It seems like a situation where we could have an infinite regress. Over some total surface, you said that it's true that [tex] \oint_{S} \mathbf{E}_{out} \cdot d \mathbf{a} = 0 [/tex]

    But that [tex] \oint_{S_{i}} \mathbf{E}_{out} \cdot d \mathbf{a} [/tex]
    doesn't necessarily equal zero. But couldn't we look at [itex] S_{i} [/itex] as some new surface S in which [tex] \oint_{S} \mathbf{E}_{out} \cdot d \mathbf{a} = 0 [/tex]

    ?
     
  14. Jan 16, 2014 #13
    There's a critical difference between S and Si, though: S is a closed surface, but Si is not. Gauss's law only tells us that the integral of Eout over a closed surface is zero. Over a non-closed surface, the integral could be anything. Each individual integral [itex]\int_{S_i} \mathbf{E}_{out} \cdot d \mathbf{a}[/itex] can be anything, as long as they all add up to zero:

    [tex]\int_{S_1} \mathbf{E}_{out} \cdot d \mathbf{a} + \cdots + \int_{S_N} \mathbf{E}_{out} \cdot d \mathbf{a} = \oint_{S} \mathbf{E}_{out} \cdot d \mathbf{a} = 0[/tex]

    (Note my use of the \int vs \oint symbols. "\oint" shouldn't be used with Si since it isn't a closed surface.)
     
  15. Jan 16, 2014 #14

    Meir Achuz

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    How could so much be written about a UG homework problem?
     
  16. Jan 16, 2014 #15

    Brevity isn't a strength of mine...

    Though to be fair most of the discussion has centred around key misconceptions about Gauss's law, not just deriving the boundary condition.
     
  17. Jan 16, 2014 #16

    kmm

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    Ah yes this is true. I think I have it now. This means that if I take an arbitrary non-conducting charged surface, and build my pill box around a patch, since we know the total flux is not effected by outside E-fields, any E-fields due to charge outside the pill box will cause the e-field to fluctuate throughout the pill box, but in a way to insure that the flux remains constant. In situations like the conductor where the electric fields have fluctuated in a special way, that is to make E=0 inside the conductor, since the flux must be constant we can infer what the other fields are. Therefore, Gauss's law actually can tell us about the electric field from all sources. Thanks for the help!
     
  18. Jan 16, 2014 #17

    kmm

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    I appreciated the detail.
     
  19. Jan 16, 2014 #18

    I have no regrets about my uber-long posts then. :) You seem to have a pretty good handle on this now; glad I could help!
     
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