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E field outside a conducting body

  1. Apr 9, 2013 #1

    CAF123

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    Say we have a +Q charge located inside an isolated spherical conducting body and I want to know the E field at some point P outside the conducting body.

    I reliase that since this is a conducting body, the E field inside is zero. So by Gauss's Law, ##Q_{enclosed} = 0##. The only way for this to happen is if there is an induced charge -Q on the inner surface of the body.

    My question is: if I wanted to compute the E field at point P a distance r away from the surface of the body, why do I use R + r as my distance from the charge in the E field formula? (R the radius of the conducting body).
     
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  3. Apr 9, 2013 #2

    Simon Bridge

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    Try it for a simple geometry - say the conducting body is a spherical shell inner-radius a and outer radius b (so it is b-a thick).
     
  4. Apr 9, 2013 #3

    CAF123

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    I am not quite sure I understand what you mean - if I have +Q in the spherical shell of radius b-a, then to make the E field zero inside, there exists -Q/2 charge on one side of the shell and -Q\2 charge on the other. (both charges reside on the inside of the body). Does this help me?

    I think the reason you have to have R +r is because if I neglected the R term and just had r as the distance then I would be failing to consider the charge Q\2 on the other side. Is this what you were getting at?

    Many thanks.
     
  5. Apr 9, 2013 #4

    Simon Bridge

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    Start again:
    The shell has inner radius a and outer radius b, which is thickness b-a.
    The charge +Q is at the center.
     
  6. Apr 9, 2013 #5

    CAF123

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    Is the +Q charge at the common centre of the two spheres?
     
  7. Apr 9, 2013 #6

    Simon Bridge

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    Make a hollow sphere out of metal - keep the walls of constant thickness - you have one shell with two surfaces. But I do mean that the +Q charge is at the center of the shell, yes, which would also be the center of both surfaces. We'll look at what happens when the charge is off-center later - unless you figure it out first ;)

    The idea is to use an example with explicit and simple geometry that you can use to shed light on your question.
    I had hoped it would be faster but never mind.
     
  8. Apr 9, 2013 #7

    CAF123

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    I think the reason I am getting slightly confused is because I have attempted problems in which I need to calculate the E field outside of a conducting sphere and outside an uncharged sphere.

    For the conducting sphere, say we have some sphere of radius r containing charge +Q. Since it is conducting, E field inside is 0, so the charge inside is zero and so charge -Q is induced on the inner cavity. By Gauss's Law, it is clear why we take the distance from the pt we want the E field to the center of the sphere. In answer to your question above, it the charge was displaced from the center, I don't think this changes anything.

    However, what is not clear, is the physical reasoning. E = 0 inside so there is no charge. And yet we take the distance from the centre to the point we want to calculate the E field. In a conductor, charge resides on the outer surface, so why do we not just measure the distance between the E field position to the outside surface?
     
  9. Apr 9, 2013 #8

    Simon Bridge

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    OK - that's what I was trying to get you to think about. I know you'd rather I just told you but you won't understand it that way.

    Isolated charge +Q inside and isolated conducting shell inner radius a and outer radius b...
    Take it one step at a time.

    The presence of the enclosed charge exerts a force on the charges in the conductor. It will attract negative charges and repel positive ones, causing them to separate. The separation of charges produces it's own electric field inside the conductor which opposes the one from outside - cancelling it out.

    For r < a, what is the electric field?

    For a<r<b, what is it?

    Therefore, for r > b ...
     
  10. Apr 10, 2013 #9

    vanhees71

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    Why should the field inside the shell be 0? That's not true, because there is the charge inside. I guess one can solve this problem even analytically. For sure you can for a infinitesimally thin conducting shell with help of the method of image charges.
     
  11. Apr 10, 2013 #10

    CAF123

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  12. Apr 10, 2013 #11

    CAF123

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    I see what your saying and when I think about this it does make sense - we have charge enclosed in a dome in order to operate the Van de Graff generator. Hmm.. I have attempted a few problems where they say even with charge enclosed, the electric field is zero. The explanation was that the presence of +Q inside induces -Q on the inner layer. So what is right?
     
  13. Apr 10, 2013 #12

    Simon Bridge

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    The conductor occupies the spherical volume between r=a and r=b. The enclosed charge +Q is at r=0 - there is no conductor there.
    [edit: gah - I should read who the comments are from!!!]

    Recall, in the classical model you are using, charge is infinitely divisible.
    Each part of a clump of charge will repel every other part - so it will fly apart unless something holds it together. Inside a conducting volume, there is nothing to hold it together - by definition.

    Thus, any clump of charge inside a conducting volume will fly apart until it is spread over the surface of conducting volume. So there cannot be any charge inside the volume.

    If we embedded a small lump of charged insulator inside the metal - as a test charge - that charge experiences zero net force since the charges in the conductor are already arranged to cancel each other out... the total force on the insulator from the surface charge distribution is zero even though the force from each surface element may not be.

    Since the small test charge experiences zero net force, then, according to the definition of the electric field, there is zero electric field.

    This is purely abstract: it does not say anything about how a volume becomes conducting or what the charge is made up of.

    But you are jumping ahead of yourself - please take it one step at a time.
    The electric field for r<a would be:
    $$\vec{E}=\frac{kQ}{r^2}\hat{r}$$ ... there is nothing wrong with having an electric field in a volume of space enclosed by a conductor - otherwise Faraday cages would be lethal. You just can't have the electric field inside the conductor itself.
     
    Last edited: Apr 10, 2013
  14. Apr 10, 2013 #13

    CAF123

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    Ok, so in the conducting shell, E=0 (r between a and b). How does this conform with Gauss Law?
    Specifically, if I take a sphere of radius a<r<b, then the enclosed charge within that sphere is +Q.
     
  15. Apr 10, 2013 #14

    Simon Bridge

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    [edit] posts passed each other
    One way to imagine it is to sketch the two surfaces of the conductor in 2D - shade in the conducting region. Put a dot in the center for the positive charge - imagine it has a charge of +8 units - so you draw 8 electric field lines radiating away - arrows pointing away from the charge. Draw the field lines right up to the inner surface of the conductor - where they stop.

    They stop because the +8 charge has attracted 8 negative charges towards it. So draw them in - one negative charge for each field line. Now all the field lines start and end on a charge. They are completely used up.

    However, the conductor started out neutral.
    When the 8 negative charges got attracted to the inner surface, they left behind 8 positive charges free to move in the conductor. Those ones will repel each other ... where will they end up?

    One step at a time - you are getting ahead of yourself again!

    For a<r<b, E=0
    The only way this can happen is if there is a negative charge distribution on the inner surface at r=a given by: $$\sigma_a = -\frac{Q}{4\pi a^2}$$... so the total charge inside 0<r<b is zero.

    (it helps if you have the diagram).
     
    Last edited: Apr 10, 2013
  16. Apr 10, 2013 #15

    CAF123

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    On the other side of the conductor (inner surface of r=b). The E field at r >b would be 0 since the Efield has terminated at the surface r=a.

    So there is +Q in the centre and to make E =0 inside the conductor, then there has to be -Q on the inner surface. Hence any sphere drawn by Gauss's Law will give 0 electric field. Does this argument not assume we know the E field in the conductor to be zero already before we apply Gauss Law?
     
  17. Apr 10, 2013 #16

    Simon Bridge

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    Yep. Just like you need to know the enclosed charge already to apply Gausses law.

    You could apply Maxwells equations to show that the charges in the conductor will always arrange themselves to cancel out the field inside the conductor if you like - that won't follow from Gausses law alone. What you are doing here is electrostatics. Gausses law just lets you work out the field from the charge distribution.

    But you should be able to answer your question from post #1 now.
     
  18. Apr 10, 2013 #17

    CAF123

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    Are all my responses correct? In particular, for r <a, the e field is non zero, for a<r<∞ the E field is zero and
    Yes, I believe so - because of the distribution of charges in a conducting body. Going back to the point made by vanhees, if we consider a charge located inside a conducting sphere, then the E field inside would not be zero, right ?(because I can just invoke Gauss Law). This is just how the van de Graff works: charge is fed into the dome via a conveyor creating an E field inside.
     
  19. Apr 10, 2013 #18

    vanhees71

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    Ok, let's calculate the whole thing. For the charge in the center it's quite simple. Using Heaviside-Lorentz units, the task is to solve
    [tex]\Delta \phi=-Q \delta(\vec{x})[/tex]
    under the boundary conditions given by the spherical shell.

    Due to symmetry the potential is a function of [itex]r=|\vec{x}|[/itex] only. The general solution for the Laplace equation, which is valid everywhere except in the origin is
    [tex]\phi(r)=\frac{A}{r}+B.[/tex]
    The electric field is
    [tex]\vec{E}=-\vec{\nabla} \phi=\frac{A \vec{x}}{r^3}.[/tex]
    The constants [itex]A[/itex] and [itex]B[/itex] are different for the three regions [itex]I: \quad r<a[/itex], [itex]II: \quad a \leq r \leq b[/itex], and [itex]III: \quad r \geq b[/itex].

    The potential and the component of [itex]\vec{E}[/itex] tangent to the boundary surface should be continuous. The latter is fulfilled everywhere, because the field is radial and thus perpendicular to the spherical surfaces.

    Now for region I, we must have [itex]A_I=Q/(4 \pi)[/itex] in order to have the correct singularity due to the charge at the origin [itex]B_I[/itex] we leave open for the moment.

    In region II, we must have [itex]\phi=\text{const}[/itex], because there must be no field inside. This implies [itex]A_{II}=0[/itex], and continuity of the potential gives [itex]B_{II}=Q/(4 \pi a)+B_{I}.[/itex]

    Since the field vanishes inside the conductor, the induced surface charge at [itex]r=a[/itex] must be [itex]-Q[/itex] and due to spherical symmetry the surface-charge density there is [itex]\sigma_a=-Q/(4 \pi a^2).[/itex] Since the shell is isolated (and I assume it's total charge is 0), the charge on the outher shell must be [itex]+Q[/itex] again and the surface-charge density [itex]\sigma_b=+Q/(4 \pi b^2).[/itex]

    Finally in region III this implies there must be again [itex]A_{III}=Q/(4 \pi)[/itex]. Further we demand that [itex]\phi(r) \rightarrow 0[/itex] for [itex]r \rightarrow \infty[/itex], which implies [itex]B_{III}=0.[/itex]

    This implies, we must have [itex]Q/(4 \pi a)=B_{I}=B_{II}=\phi(b)=\frac{Q}{4 \pi b}[/itex], because the potential of the whole shell is constant.

    Finally we have
    [tex]\phi(r)=\begin{cases}
    \frac{Q}{4 \pi} \left (\frac{1}{r}-\frac{1}{a}+\frac{1}{b} \right ) & \text{for} \quad r<a,\\
    \frac{Q}{4 \pi b} \quad \text{for} \quad a \leq r<b,\\
    \frac{Q}{4 \pi r} \quad \text{for} \quad r \geq b.
    \end{cases}[/tex]
    The field is
    [tex]\vec{E}=E(r) \vec{e}_r[/tex]
    with
    [tex]E(r)=\begin{cases}
    \frac{Q}{4 \pi r^2} & \text{for} \quad r<a,\\
    0 & \text{for} \quad a<r<b, \\
    \frac{Q}{4 \pi r^2} & \text{for} \quad r>b.
    \end{cases}
    [/tex]
    The jumps of the normal component are precisely the surface charges, determined above
    [tex]E(a+0^+)-E(a-0^+)=-\frac{Q}{4 \pi a^2},[/tex]
    [tex]E(b+0^+)-E(b-0^+)=+\frac{Q}{4 \pi b^2}.[/tex]
    This completes the proof that the above given solution for the potential satisfies all boundary conditions.
    \end{cases}
     
  20. Apr 10, 2013 #19

    CAF123

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    Thanks, but what is the source of the E field for r>b? The charge at the center creates an outward E field but this terminates on the inner surface of the sphere of radius a. Furthermore, if I were to draw a Gaussian surface (say a sphere of radius r>b), then the charge enclosed is 0, and so this implies E=0. ( for a<r<b, E=0 as it has to be for a closed conductor => charge -Q induced on the inner surface of sphere r=a. So drawing a sphere of radius r>b contains this +Q and -Q => charge= 0 and so E=0). No?
     
  21. Apr 10, 2013 #20

    vanhees71

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    The charge enclosed in a Gaussian surface is [itex]Q[/itex] if the surface is either totally in the interior in region I at [itex]r<a[/itex] and outside of the shell in region III at [itex]r>b[/itex], if the shell carries no additional charge.

    If the Gaussian surface is completely inside the shell, the integral over [itex]\vec{E}[/itex] is of course 0, because [itex]\vec{E}=0[/itex]. My calculation also shows, how this works: There is an induced surface charge [itex]-Q[/itex] on the inner boundary of the shell at [itex]r=a[/itex]. Then I assumed the total shell is uncharged, so that there must be the opposite surface charge of [itex]+Q[/itex] at the outer surface at [itex]r=b[/itex].

    BTW: If the shell as a hole is charged, this additional charge must distribute uniformly over the outer shell at [itex]r=b[/itex], and this doesn't affect the field in the interior at [itex]r<b[/itex]!
     
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