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Charged spherical shell

  1. Sep 7, 2015 #1
    1. The problem statement, all variables and given/known data
    A charged spherical insulating shell has inner radius a and outer radius b. The charge density on the shell is ρ.
    insulator.gif
    What is the magnitude of the E-field at a distance r away from the center of the shell where r < a?

    2. Relevant equations
    Gauss' Law

    3. The attempt at a solution
    I read through the Gauss' law chapter in my textbook, and it stated that the internal electric field of such a charged insulator is zero. Therefore, I know the answer is 0 but I do not know why. Any explanation as to why this answer is correct would be appreciated. Thanks.
     
  2. jcsd
  3. Sep 7, 2015 #2
    Well, Gauss's law states that: [tex]\oint \mathbf{E} \cdot \ d \mathbf{A} = \frac{Q_{enc}}{\epsilon_0}[/tex] What is the enclosed charge if you use a Gaussian surface inside the cavity?
     
  4. Sep 7, 2015 #3
    0. Is it because there is no charge in the cavity?
     
  5. Sep 7, 2015 #4
    Correct. If you imagine your Gaussian surface to be a sphere inside the cavity centered at the center of the larger sphere, no charge is inside (so Qenc = 0). All the charge is in the insulated sphere in this problem.

    Now, the integral being equal to 0 is not a necessary condition to declare that the electric field is 0. For instance, [itex]\int _{-1} ^1 x \ dx = 0[/itex], so this is not enough to declare that there is no electric field inside.
    Edit: changing the argument because I saw a flaw in it. Simply put: by symmetry, the electric field should cancel out everywhere inside the sphere.

    If it didn't cancel out, which direction would get precedence? Why should any direction of a perfectly symmetrical sphere get the privilege of having the electric field not cancel out?
     
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