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Simple E&M questions

  1. Apr 23, 2005 #1

    quasar987

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    1. We have a grounded infinite conducting plane lying in the xy plane and a point charge held a distance d above it. Griffiths says that this implies that

    i) V = 0 at z = 0 (since the conducting plane is grounded

    ii) V --> 0 far from the charge (that is, for x² + y² + z² >> d²)

    My question is: How does grounded implies V = 0?

    And how can we have V = 0 at z = 0 and at infinity simultaneously? This would imply, according to [itex]W = Q\Delta V[/itex] that it takes no work to bring a charge Q from infinity to the plane along the z axis. Maybe it IS the case, but it's not obvious at all.


    2. This one is for people who own Griffith's book. Look at Figure 2.36 pp.88. How come E_above and E_below point in the same direction ?! I can think of at least one situation where this is not true: for an infinite charged plane, E points outward.
     
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  3. Apr 24, 2005 #2

    dextercioby

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    That's the convention.Earth's electric potential is 0,ergo all conductors tied to the surface of the Earth have 0 potential...


    Daniel.
     
  4. Apr 24, 2005 #3

    Galileo

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    1.
    Grounded means V=0 by definition. It doesn't matter what point you choose as your 'reference point' (point where V=0).
    It may look ambiguous since the the convention is to assign V=0 to the 'point at infinity'. What is meant is that the plate is kept at a potential equal to 0. So indeed it takes no work to bring in a charge from infinity and stick it on the plate.

    2.
    It's not the case that they should point in the same direction, but Griffiths drew the picture that way to get an idea.
    The point is that you have to pick a positive direction.
    The result: [itex]E^{\perp}_{\mbox{above}}-E^{\perp}_{\mbox{below}}=\frac{1}{\epsilon_0}\sigma[/itex] may change sign depending in which direction you choose to be "positive".
    For the infinitely large charged plane you have to take this sign into account. "Above" the plane the field is [itex]\frac{\sigma}{2\epsilon_0}[/itex] and "below" [itex]-\frac{\sigma}{2\epsilon_0}[/itex] (there's no such thing as "above" or "below", but there's a sign difference)
    So:
    [tex]E^{\perp}_{\mbox{above}}-E^{\perp}_{\mbox{below}}=\frac{\sigma}{2\epsilon_0}-(-\frac{\sigma}{2\epsilon_0})=\frac{\sigma}{\epsilon_0}[/tex]

    as required.
     
  5. Apr 24, 2005 #4
    quasar,

    Just imagine a big fat copper wire running between infinity and any point where you want the potential to be zero.
     
  6. Apr 24, 2005 #5

    quasar987

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    Ok, suppose that when I define my function potential by

    [tex]V(\vec{r}) = -\int_{\vec{a}}^{\vec{r}} \vec{E}\cdot d\vec{r}[/tex]

    I chose 'a' to be a point in the earth (or in the plane, or in the wire connecting them). Then this means setting a point where V = 0. And since there is no electric field in a conductor in the electrostatic régime, the potential difference between point 'a' and any point connected to the earth is 0, <==> the potential is 0 at any point connected to the earth.

    But if I do this, then there is no telling what the potential at infinity is! Most likely, it will be a constant that is not 0.

    Inversely, if I define my potential function to be 0 at infinity, and if I want it to be zero at the earth, I have to add a constant to my potential function, making it 0 no more at infinity.

    There is also the possibility that what you are suggesting is that we define the potential as being zero at infinity, and then since the earth-plate system is an equipolential (i.e. V = constant), let's set that constant to zero as well but without changing the potential at every other point outside the conductor. It doesn't matter anyway, since the gradient of any constant, zero included, is 0!

    But that would cause problems. Because as we transition from the earth-plane system to just outside that system, the potentiel will be discontinuous, giving out a gradient undefined, while the field certainly is.


    What does it mean to KEEP something at a potential zero? How would one go about doing this?

    But as I approach a test charge from the conductor, a charge of opposite sign will induce on its surface, creating a force of attraction on my test charge. And hence "working" on it.
     
  7. Apr 24, 2005 #6

    quasar987

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    You mean, he drew it like that at random? They could just as well have been drawn in opposite directions?

    If we take the perpendicular component of field to point outward (as it will be the case in the event that the field is entirely due to the sheet), we get the "opposite" relation: [itex]E^{\perp}_{\mbox{above}}+E^{\perp}_{\mbox{below}}=\frac{1}{\epsilon_0}\sigm a[/itex]
     
  8. Apr 26, 2005 #7
    Say a charge q is going from infinity to a neutral plate. At infinity q would not be
    attracted to the plate. When q comes closer to the plate, this plate is now not neutral anymore as seen from q. As is mentioned an opposite charge will be induced in the plate. This will cause a minor current flow in the plate. q and the induced charge can be seen as real and wiil exert forces on any other possible charges. When q is absorbed by the plate the potential of this plate is now changed depended on its size and whether or not things like wires or planets are connected to this plate.
    Eric
     
  9. Apr 26, 2005 #8

    quasar987

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    Thanks for replying Eric, I was starting to lose hope. But you could point out for me to which part (if any) of the topic discussed in the previous post are the notions you exposed related. Thx.
     
  10. Apr 27, 2005 #9

    Galileo

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    You're blowing up the problem to much (I wanted to say you're thinking too much about this, but that's actually a good thing :biggrin: ).
    The point at infinity has V=0 and the plate has that same potential.
    Don't worry about how this is done, it's simply a given for this somewhat artificial problem. It supplies you with the boundary conditions necessary to solve the Laplace equation.
    If it helps, think of a wire attached to the plate at stuck into the ground (hence grounded) far away (like jdavel suggested). That is your point at infinity which is zero by definition. Since the plate+wire is an equipotential you have V=0 everywhere.

    Calculating the potential by looking at the work done by bringing a charge from infinity is a virtual process, not an actual dynamical one. You are given an electrostatic potential and you look at the work you do when dragging it through that potential (which doesn't change!).
    I guess that was a little confusing. I mean the potential is zero at both points (infinity and plate). In actuality the charge you bring in will find resistance from the induced charge on the plate, but hopefully you see this is not relevant for the calculation of the potential.

    It's just a picture to get an idea. Yes they could have been drawn in opposite directions.

    We're not looking at magnitudes here, but at the component of the E-field in the direction perpendicular to the surface. This component can be negative. You are always looking at the difference so you get [itex]E^{\perp}_{\mbox{above}}-E^{\perp}_{\mbox{below}}[/itex] regardless.
    Ofcourse, if they point in different directions, one will be positive and one negative so their magnitudes add.
    You cannot choose the perpendicular component to always point outwards. You need to pick one direction to be the outward unit normal [itex]\hat n[/itex] (which we will call the 'above' direction). Then:

    [tex]\vec E \cdot \hat n = E^{\perp}_{\mbox{above}}[/tex]
    [tex]\vec E \cdot \hat n =E^{\perp}_{\mbox{below}}[/tex]

    where [itex]\vec E[/itex] is the field just above and below the plate respectively.
     
    Last edited: Apr 27, 2005
  11. Apr 27, 2005 #10
    Note to quasar. Say we reverse the process I described earlier and start from a plate with a 0 V potential which relates to your point 1. Now we introduce a test charge above the plate, say this charge comes somehow out of this plate (although it could come from anywhere). There's no way our plate has still a 0 V potential, no matter how much we ground to earth, because of the induced oposite charge.
    If this charge comes from our plate then we needed some energy to get it at this point. If it comes from outside we had a system with some positive energy to start with and the process I described above applies.
    Eric
     
  12. Apr 28, 2005 #11

    Meir Achuz

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    In order to keep the plane at V=0, it must be cononected to a battery or some voltage source. A disconnected plane would increase its voltage as a charge approached it. You cannot use simple conservation of energy as the charge moves.
    A charge released from rest at infinity would be drawn to the plane and hit it with large velocity. As the charge approaches the plane, current must flow into the plane from the battery. The charge ultimately would be getting its energy from the battery keeping the plane at ground.
     
  13. Apr 28, 2005 #12
    I am not so sure about the battery argument.

    But to go back to the original question (quasar's point 1): The force produced by a charge in a circular conducting plane (radius r) is:

    F=-(q^2)/4pi*E0*(2d)^2.

    E0 stands for epsilon 0. d is the distance between plane and charge parallel to centre of plane. 2d is used because the image of the charge is exactly a distance -d away from the surface of the plane. The - sign in -(q)^2 comes from the fact that we are dealing with a charge q and its image -q (or -q and q). Therefore the force is always attractive.

    To get a perfect mirror image d << r. When we let d > r the force falls away even faster than (2d)^2 until eventually F=0 as we all know. It becomes more interesting if we make d smaller. According to the formula F can get very big indeed. How big? In reality d cannot become smaller than the Bohr radius because of wave mechanics. But even long before it reaches this distance we have to deal with surface irregularities, crystals and +ve and -ve ions. What I can say is that the energy required to remove an electron of ideal surfaces (low work function) is about 4 to 5 volts.

    Anyway how do we get from this to a nice formula for E AND U ? I don't know.
    Eric
     
  14. Apr 29, 2005 #13

    quasar987

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    I'm just trying to figure out what Griffiths did. He says the statement of Gauss' law for a wafer-thing Gaussian pillbox is

    [tex]\int_S \vec{E}\cdot d\vec{a} = \frac{1}{\epsilon_0}\sigma A[/tex]

    and only the area of the pillbox lid contributes to the flux. So

    [tex]\int_S \vec{E}\cdot d\vec{a} = \int_{A_{above}} \vec{E}_{above}\cdot d\vec{a}_{above} + \int_{A_{below}} \vec{E}_{below}\cdot d\vec{a}_{below}[/tex]

    And here, it really matters whether the perpendicular component of E_above and E_below is in the same direction as the area vector of their respective pillbox lid or not. If E_above and E_below point in the same direction as indicated by Figure 2.36, then ok, the dot product [itex]\vec{E}_{above}\cdot d\vec{a}_{above}[/itex] is positive, and the dot product [itex]\vec{E}_{below}\cdot d\vec{a}_{below}[/itex] is negative, producing the result (2.31). But in other the case where the perpendicular component of E_above and E_below point in oposite direction, the integral yields [itex]E^{\perp}_{\mbox{above}}+E^{\perp}_{\mbox{below}}= \frac{1}{\epsilon_0}\sigma[/itex] out of Gauss' law.
     
  15. Apr 30, 2005 #14

    Galileo

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    But what you've just done is to create 2 different normal unit vectors for the surface! One for above the plane and one for below, so it will always point away from the plane. This is not possible (see my previous post).

    You chose: [itex]d\vec a_{\mbox{above}}[/itex] and [itex]d\vec a_{\mbox{below}}[/itex], but these normals are the same. First you choose a positive direction (the direction of the surface normal), its unit vector is [itex]\hat n[/itex], then [itex]d\vec a = \hat n da[/itex] (this is the definition of [itex]d\vec a[/itex]). So [itex]d\vec a_{\mbox{below}}=d\vec a_{\mbox{above}}=d\vec a=\hat n da[/itex].
     
  16. Apr 30, 2005 #15

    quasar987

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    Why would I use the same normal vector for both the pillbox lids? This is not what Gauss' law tells me to do. It tells me to chose an OUTWARD direction for by pillbox.
     
  17. Apr 30, 2005 #16

    cepheid

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    Yeah you're right! In fact if you look at example 2.4 on page 73, that's exactly what he does, and it makes way more sense. In the example we are currently discussing, on page 88, he says in the second sentence of the third blurb of text that we let upward be the positive direction "for consistency". But if you ask me, it seems totally inconsistent. I'd like this explained too. Why do it using this convention?
     
  18. Apr 30, 2005 #17

    cepheid

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    Actually, I figured out what he's doing. On page 73, his objective is to compute the electric field magnitude and direction due to an infinite plane of charge. In this example, he notes that we know a priori that the field always points away from the plane i.e. that the the field above is opposite the field below. He says, "By symmetry, E points away from the plane (upward for points above, and downward for points below." So then he invokes Gauss's law using normal vectors for the pillbox. He arrives at the result:

    [tex] \mathbf{E} = \frac{\sigma}{2\epsilon_0}\mathbf{\hat{n}} [/tex]...[1]

    Where n is a unit normal vector i.e. it always points away from the plane.

    On page 88, he is now trying to do something else. He wants to show the fact that the electric field is discontinuous at plane surfaces carrying charge density sigma and quantify that discontinuity without assuming anything a priori about the directions of the electric field. Instead, he does it systematically by invesigating the components of the above and below field that point in one specific direction. That direction he calls "perpendicular to the plane", and he symbolizes it using [itex] \perp [/itex]. I'm not going to do that, because it is leading to confusion between that and the normal vector, which does not have one specific direction. Instead, let's call this ONE SPECIFIC DIRECTION in which he chooses to systematically investigate the components of the electric field the z-direction. So all he is saying is that he has chosen the direction "upwards" (towards the top of the page) to be the positive z-direction. And downwards is the negative z-direction, by his convention. Fine. Now, let me see. Can we confirm his results using our result from above? I.e. can we figure out what the discontinuity should be? Sure. All we need to do is subtract the above and below electric field vectors. Why? Because if there were no discontinuity, that vector difference would be zero, i.e subtracting two equal vectors (same magnitude and direction) gives you zero. If in fact the above and below vectors differ in either magnitude or direction, then the the difference between them would be non zero. So that's how we check for a discontinuity. But tell me...how can I possibly subtract the above and below vectors? Equation 1 isn't helping me much as far as that goes. You can see how the changeable unit normal vector isn't of much assistance to us, because thanks to it, we have one single amalgamated expression for the the E field everywhere, instead of one for each of the above and below fields. Clearly we are only going to be able to do this by looking at the components of the E field in one specific direction. Using the z-direction we defined earlier, equation 1 becomes two eqns:

    [tex] \mathbf{E}^{\text{above}} = \frac{\sigma}{2\epsilon_0}\mathbf{\hat{z}} [/tex]

    [tex] \mathbf{E}^{\text{below}} = -\frac{\sigma}{2\epsilon_0}\mathbf{\hat{z}} [/tex]

    I want to emphasize that we have assumed this result a priori whereas Griffiths arrives at it without making any assumption about the directions of the fields and by simply looking at the components of E in the z (what he calls [itex] \perp [/itex]) direction.

    Now it's piece of cake for us. We know that:

    [tex] \mathbf{E}^{\text{below}} = -\mathbf{E}^{\text{above}} [/tex]

    So their vector sum is zero as it should be. What about their difference?

    [tex] \mathbf{E}^{\text{above}} - \mathbf{E}^{\text{below}} [/tex]

    [tex] = \mathbf{E}^{\text{above}} - (- \mathbf{E}^{\text{above}}) [/tex]

    [tex] = \frac{\sigma}{2\epsilon_0}\mathbf{\hat{z}} - \left(- \frac{\sigma}{2\epsilon_0}\mathbf{\hat{z}} \right) [/tex]

    [tex] = \frac{\sigma}{\epsilon_0}\mathbf{\hat{z}} [/tex]

    So that tells us by how much the field vectors are discontinuous above and below (ie. you'd have to add this result we got to the vector below in order to get the vector above).

    Almost there..phew. I just now need to point out that since all our fields are pointing in either the +ve or -ve z-direction, we can dispense with the vector notation, instead using z-components of the vectors. More specifcally, scalar components of the field in the z-direction. Positive or negative numbers that tell us the magnitude of the field, and whether it is pointing up or down, essentially. So rewriting this using the scalar notation:

    [tex] E_z^{\text{above}} = \frac{\sigma}{2\epsilon_0} [/tex]

    [tex] E_z^{\text{below}} = -\frac{\sigma}{2\epsilon_0} = -E_z^{\text{above}} [/tex]

    Then we have:

    [tex] E_z^{\text{above}} - E_z^{\text{below}} = \frac{\sigma}{2\epsilon_0} - \left(-\frac{\sigma}{2\epsilon_0}\right) = \frac{\sigma}{\epsilon_0} [/tex]

    This result is exactly Griffiths equation 2.31. Just replace your z's with [itex] \perp [/itex]'s. Whoops, I also interchanged my superscripts and subscripts. But that's okay. I'm used to E sub z meaning "the z component of E".

    I know that was long-winded, and apologies to those of you who already explained this more succinctly. But I wanted to go over it in detail because I never understood it the first time 'round either.
     
    Last edited: Apr 30, 2005
  19. Apr 30, 2005 #18

    Galileo

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    Ugh, okay I was talking about the normal of the charged surface, not the normal to the gaussian surface, these always point outwards.

    I`ll be more precise. Let [itex]\hat n[/itex] denote the unit normal to the charged surface (not the surface of the gaussian pillbox). Let the direction of [itex]\hat n[/itex] be the "above" direction. Then, by definition:

    [tex]\vec E^{\perp}_{\mbox{above}} = (\vec E \cdot \hat n)\hat n[/tex]
    where [itex]\vec E[/itex] is the field just above the surface and
    [tex]\vec E^{\perp}_{\mbox{below}} = (\vec E \cdot \hat n)\hat n[/tex]
    where here [itex]\vec E[/itex] is the field just below the surface.
    The components of these two vectors in the direction of the normal are respectively: [itex]E^{\perp}_{\mbox{above}}=\vec E \cdot \hat n[/itex] and [itex]E^{\perp}_{\mbox{below}}=\vec E \cdot \hat n[/itex].

    Now draw the pillbox, its normal vectors point outwards, so at the sides [itex]d\vec a[/itex] and [itex]\vec E^{\perp}[/itex] are perpendicular and contribute nothing to the flux. Above [itex]d \vec a[/itex] and [itex]\hat n[/itex] are parallel, but below the surface [itex]d\vec a[/itex] and [itex]\hat n[/itex] are anti-parallel, since [itex]d\vec a[/itex] and [itex]\hat n[/itex] point in the opposite directions here.

    So when [itex]\vec E^{\perp}_{\mbox{above}}[/itex] and [itex]\vec E^{\perp}_{\mbox{below}}[/itex] point in the same direction (say in the direction of [itex]\hat n[/itex]), then the components E_above and E_below are both positive.
    When [itex]\vec E^{\perp}_{\mbox{below}}[/itex] points in the other direction, E_below will be negative. Remember that this is the component of [itex]\vec E^{\perp}_{\mbox{below}}[/itex] and not its abosulte value. (E_above - E_below) will be bigger when E_below is negative (as it should when the fields point in opposite directions)
     
    Last edited: Apr 30, 2005
  20. Apr 30, 2005 #19

    OlderDan

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    You would not use the same vector for both lids. You are correct that the normal always points outward from the closed surface. A charge must accumulate on the surface of the conducting plate to keep the potential inside zero. This charge must arrange itself so that the electric field at the surface is perpendicular to the surface and the electric field inside the conductor is zero.

    This problem is often used to introduce the method of images. The problem can be replaced by a problem of two equal and opposite charges separated by twice the distance for the one charge to the plane. The added charge is thus the mirror image of the original charge with opposite sign. There is a nice applet at this site that will let you construct this mirror image represntation of this problem.

    http://webphysics.davidson.edu/mjb/DC_aapt_talk/images.html

    Click on both the start buttons at the top of the list. Add a minus charge to the right and move it unti you produce a filed that looks like the field on the left. If you get it right, the equipotential lines will be perfectly symmetrical.
     
  21. Apr 30, 2005 #20

    cepheid

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    I called the former "z hat", because it seemed to make things clearer. Is that such a sin?

    What you are saying makes perfect sense to me now, but it has to be made clear that in this context n has been defined as the "above" direction, as you stated.
     
    Last edited: Apr 30, 2005
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