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Confusion On Electric Field Inside Capacitors

  1. Mar 10, 2014 #1
    When deriving capacity of a parallel plate capacitor, we consider the electric field (sigma/epsilon), i.e. the field due to one of the plates while both the plates are charged with Q and -Q respectively...so should not be the field due to both the plates, i.e. (2sigma/epsilon) since the direction of both fields are same??
    And when doing for two parallel wire of equal and opposite charges, my book says the field between these wires is the addition of fields due to each wire.....Why so??
    Please someone clarify this.
  2. jcsd
  3. Mar 10, 2014 #2

    Simon Bridge

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  4. Mar 10, 2014 #3
    Thanks for the link. But one thing I'm not getting clearly is E=(sigma/epsilon) when fiele is in one side of the plate and E=(sigma/2epsilon) when field is in both side....for the parallel plates field is in one side of each...so why the resultant field is (sigma/epsilon) and not (2sigma/epsilon) ??
    And the problems I posted are already worked out and derived correct result...the capacy of a parallel plate capacitor of area A is (epsilon*A/d).
  5. Mar 10, 2014 #4
    The field of each plate is E = σ/(2ε). Both plates together produce a field twice as big E = σ/ε
  6. Mar 10, 2014 #5

    Philip Wood

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    And outside the capacitor the fields cancel!
  7. Mar 10, 2014 #6

    Simon Bridge

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    You didn't do that math did you? :)

    put two infinite plates equal charge distribution, opposite charge, parallel to x-y plane, with the +q plate at z=+d/2 and the -q plate at z=-d/2 and remember that the electric field is a vector.

    Field due to +q plate is
    $$\vec E_+ = \left\{\begin{align} -\frac{\sigma}{2\epsilon_0}\hat{k} &:\; z>d/2\\ 0 &:\; z=d/2\\\frac{\sigma}{2\epsilon_0}\hat{k} &:\; z<d/2\end{align}\right.$$
    Field due to -q plate is
    $$\vec E_- = \left\{\begin{align} \frac{\sigma}{2\epsilon_0}\hat{k} &:\; z>-d/2\\ 0 &:\; z=-d/2\\-\frac{\sigma}{2\epsilon_0}\hat{k} &:\; z<-d/2\end{align}\right.$$ (caution: don't take my word for it: check!)

    Now you do the math: the total field is $$\vec E = \vec E_+ + \vec E_-$$
    ... you should be able to confirm dauto's result:
    Because the electric field is a vector:
    ... but that is only for infinite plates. Real life capacitors have some field outside and the above is just an approximation.
    Taking the approximation too literally can lead to odd results.
    i.e. http://www.lhup.edu/~dsimanek/museum/advanced.htm
    (3rd one down: 2nd of the "two puzzles" - try it before looking at the answer).

    The path through this sort of confusion is to take things one step at a time checking as you go.
  8. Mar 10, 2014 #7
    Thanks to all for clearing my confusion about parallel plates.

    One last thing, what will be the electric field just outside a charged spherical shell? Will it be (sigma/epsilon) or (sigma/2epsilon)?
  9. Mar 10, 2014 #8

    Simon Bridge

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    Why not work it out - or look it up - for yourself and see.
    But why would you expect a closed curved surface to have a similar field to an infinite plane surface?
    The geometry is different.
    Perhaps check?

    For a sphere radius ##R## carrying surface charge density ##\sigma##
    Work through it
    - what is Gauss' Law?
    - where would you need to put the gaussian surface?
    - what is the total charge on the sphere?
  10. Mar 11, 2014 #9
    It is already worked out in my book by considering a Gaussian pillbox through the charged surface. But it is titled ''Electric Field on the surface of a charged onductor''. It didn't say whether the surface is closed or not.
    That's why I'm confused. And I myself imagined this many times but couldn't get clearly.
  11. Mar 11, 2014 #10

    Simon Bridge

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    For the sphere - try drawing the Gaussian surface around the whole sphere.
    For r>R, the sphere acts as a point charge. You should be able to work it out from there.

    If we keep feeding you potted answers, you'll never get this stuff.
  12. Mar 11, 2014 #11
    You couldn't understand my problem...
    If we consider a small region of a charged spherical shell then there will be field outward, I agree.
    But what about inward??
    The confusions are..
    1. There are fields inward but they cancel out. So the net field inside is zero.
    2. There is no inward field.

    So the results will be different.
    Now my question is which path should I follow? 1 or 2?
  13. Mar 11, 2014 #12
    Both 1 and 2 are correct. 2 is correct because 1 is correct. For a spherically symmetric shell by the shell theorem there is no field inside the shell. For a metallic object of any shape assuming you give the charges enough time to reach an equilibrium position (electrostatics) there is no field inside.
  14. Mar 11, 2014 #13


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    1 and 2 resulted in the SAME outcome! So why would this matter? The gaussian sphere inside enclosed NO CHARGE, so it is already obvious that there will be no E-field (net or otherwise) crossing the gaussian surface!

    It really appears that you are making this more difficult than it seems, and you seem to mix the case for unbounded surface (when you were dealing with a capacitor plate), with a bounded surface, as in the sphere. They are not the same!

  15. Mar 11, 2014 #14

    Simon Bridge

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    No - I understood your problem all right.
    If you won't follow suggestions I cannot help you.

    You are considering only the field due to the small part you isolated?
    What is the field due to an isolated bit of charge?

    Both approaches should yield the same result though.
    Did you do the rest of the working?

    Note: it is easier to put the gaussian surface about the whole sphere.
  16. Mar 11, 2014 #15
    Sorry for mixing. But the source of these confusions was capacitor.
    Well I'm coming to a conclusion that there is no inward field in a closed surface. The term 'cancellation' is meaningless until there are fields. So no field, no chance of cancellation.
    If there were inward fields (though cancelled out), the resultant outward field would be (sigma/2epsilon) (just like charged plane sheet). But actually it is (sigma/epsilon).
    Thanks to all.
  17. Mar 11, 2014 #16

    Simon Bridge

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    If you won't do the maths you won't understand it.
    Good luck anyway.
  18. Mar 11, 2014 #17
    No, it wouldn't. It would be (sigma/epsilon). The fields inside the shell cancel each other out and the fields outside the shell add to (sigma/epsilon). Half of it (sigma/2epsilon) comes from near by charges and the other half (sigma/2epsilon) comes from distant charges adding to (sigma/epsilon).
  19. Mar 11, 2014 #18


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    Indeed, one should do the math. Let's take the most simple analytically solvable example, a capacitor consisting of a sphere (or spherical shell) of radius [itex]a[/itex] and a spherical shell with (inner) radius [itex]b[/itex] connected to a battery of voltage [itex]V[/itex].

    First let's think what happens qualitatively: The voltage causes that electrons are moving such that the inner surface carries a total surface charge [itex]+Q[/itex] and the spherical shell of [itex]-Q[/itex]. Due to symmetry both charges are distributed homogeneously over the surfaces, i.e., you have charge densities [itex]Q/(4 \pi a^2)[/itex] and [itex]-Q/(4 \pi b^2)[/itex]. Inside the inner sphere, obviously the electric field is [itex]0[/itex], because that's the only spherically symmetric solution which is not singular at the origin. Outside of the outher shell the electric field is also 0 due to Gauß's Law.

    So there is only an electric field for [itex]a<r<b[/itex], and the spherically symmetric solution is
    [tex]\vec{E}=\frac{A}{4 \pi r^2} \vec{e}_r.[/tex]
    Now we have to determine the integration constant [itex]A[/itex]: From Gauß's Law,
    [tex]\vec{\nabla} \cdot \vec{E}=\rho/\epsilon,[/tex]
    applied to a little cylinder with one surface inside the inner shell and one surface outside at [itex]r=a+0^+[/itex] gives
    [tex]\left . \vec{e}_r \cdot \vec{E} \right |_{r=a+0^+}=\sigma/\epsilon=\frac{Q}{4 \pi a^2 \epsilon}.[/tex]
    Here, [itex]\epsilon[/itex] is the dielectric constant of the medium. This fixes the the integration constant to [itex]A=Q/\epsilon[/itex] and thus
    [tex]\vec{E}=\frac{Q}{4 \pi \epsilon r^2} \vec{e}_r \quad \text{for} \quad a<r<b.[/tex]
    The voltage is
    [tex]V=\int_{a}^{b} \mathrm{d} r E_r=\frac{Q}{4 \pi \epsilon} \left (\frac{1}{a}-\frac{1}{b} \right ),[/tex]
    and thus the capacitance is
    [tex]C=\frac{Q}{V}=4 \pi \epsilon \frac{ab}{b-a}.[/tex]
    If you need more details of how to derive the expressions from Maxwell's equations (electrostatics), let me know.
  20. Mar 11, 2014 #19

    But in the mathematical derivation total flux cosidered outward...no inward flux considered in Gaussian pill box shaped surface....so the field is (sigma/epsilon).
    Can you give me your mathematics?
    I think my book derived it wrongly.
  21. Mar 11, 2014 #20

    I'm doing math also. If you think I would be clear by mathematics then show me your mathematics so that I can easily understand.
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