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Electric Field of a single charge vs 2 charges

  1. Nov 1, 2007 #1
    1. The problem statement, all variables and given/known data

    So up till now we had problems where there was some sort of charge (line, surface, volume, point) and you pretty much had to find the electric field at some point (or as a function of some distance) or the same with voltage.

    Now, I have to calculate the capacitance between to things (irrelevant what, it's not my problem) and looking at the examples from the book they say something like:

    Two concentric spheres (one inside of the other, right?) have a charge of Q and -Q. Then they say that the electric field between them is Q/4pi(r^2).

    Okay... how come only Q contributes to the field and not -Q?

    Same thing with 2 plates, one Q and the other -Q, only Q is used for the electric field and -Q is ignored. I'm not understanding why...
     
  2. jcsd
  3. Nov 1, 2007 #2
    Never mind, it's all in the epsilons from Gauss' law...

    I must have misunderrepresented the question.
     
  4. Jan 25, 2008 #3
    your first question; why doesn't -Q contribute to the E field, the anwer is simple E field is force per coulomb first find the force it is Q^2/4pi R^2 then divide it by Q, second you asked the parallel plate capacitor, when you are trying to find the E field of an infinite sheet of charge you will find Q/2.(epsilon) each side of the plane but when you bring another plate with a -Q charge the charges will accumulate on one side of the conductors this is why you dont consider -Q's contribution, it would become the macroscopic example of coulomb's law for 2 particles
     
  5. Jan 25, 2008 #4

    Ben Niehoff

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    Gold Member

    Hmm...memow's explanation isn't quite correct.

    In the first case, simply apply Gauss's Law using a sphere in between the two spheres as your Gaussian surface. Note that the only contribution is from charges inside the Gaussian surface; therefore, the field between the spheres depends only on the inner sphere.

    In the second case, what actually happens is a neat cancellation. Remember that the field of an infinite conducting plane is

    [tex]E = \frac{\sigma}{2 \epsilon_0}[/tex]

    where the direction of E is away from the plane (or towards the plane, if [itex]\sigma < 0[/itex]). The direction of E is important, because what you're doing is putting two oppositely charged planes side by side:

    Code (Text):
          |      |
      E1  |  E1  |  E1
     <--- | ---> | --->
          |      |
      E2  |  E2  |  E2
     ---> | ---> | <---
          |      |
    Note that, outside the planes, E1 and E2 have opposite directions, and therefore they add to zero. Inside the planes, E1 and E2 have the same direction, so

    [tex]E = E_1 + E_2 = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}[/tex]

    so in this case, the charge -Q does affect the resultant field.
     
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