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Electrostatics (Gauss’s law for sphere)

  1. Nov 1, 2012 #1
    1. The problem statement, all variables and given/known data

    I have some difficulty understanding a part of the following solved problem:

    Use Gauss's law to find the electric field inside a uniformly charged sphere (charge density ##\rho##).

    Answer:

    ##\oint E.da = E.4 \pi r^2 = \frac{1}{\epsilon_0} Q_{enc} = \frac{1}{\epsilon_0} \frac{4}{3} \pi r^3 \rho##

    ##\therefore \ E= \frac{1}{3 \epsilon_0} \rho r \hat{r}##

    I don't understand in the first step why the both the radius on the LHS and RHS are the same. I think the radius on the left should be that of the Gaussian surface and the one on the right should be that of the sphere itself... :confused:

    2. Relevant equations

    The integral form of Gauss’s law:

    ##\oint_S \ \vec{E} . \hat{n} da = \frac{q_{enc}}{\epsilon_0}##​

    3. The attempt at a solution

    Here is a diagram of the Gaussian surface of radius R around the sphere of radius r:

    http://img32.imageshack.us/img32/848/gaussian.jpg [Broken]​

    We have ##\rho=Q/Volume## for the entire sphere, so to get the Q needed for Gauss’s law we have to multiply it by the sphere's volume. So here is what I did:

    ##|E| \ \oint da= \frac{\rho.V}{\epsilon_0}##

    ##|E| 4 \pi R^2 = \frac{1}{\epsilon_0} \rho \frac{4}{3} \pi r^3##

    ##\therefore \ E= \frac{\rho r^3}{3 \epsilon_0 R^2}##

    I chose the Gaussian surface to be larger than the sphere itself, so that's why the two radiuses have to be different and so they don't cancel. Am I wrong or is this a mistake in the book? :confused:
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 1, 2012 #2

    alphysicist

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    Hi roam,

    The radii have to be the same because the problem asks for the field inside the sphere.

    For these types of problems, you make the Gaussian surface pass through the point you're looking at, and make it have the symmetry of the charges. So that's why the diagram you included has the Gaussian surface inside the sphere.

    Once you have that, on the RHS the enclosed charge is the charge inside just that Gaussian surface. So that's why the radius has to be the same. If you used the larger radius, you'll be including all the charge, not just the charge inside your Gaussian surface.




    Since you chose the radius to be larger than the sphere, you're finding the electric field outside the sphere. So your answer would be correct for that case (which is different from the original problem). But I did notice that you changed what r and R meant from what the picture has! In what you worked out, R is the radius of the Gaussian surface, and r is the radius of the charged sphere.
     
    Last edited by a moderator: May 6, 2017
  4. Nov 1, 2012 #3

    haruspex

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    What is the field at radius r from the centre of a uniformly charged shell of radius R > r?
    What is the field at radius r from the centre of a uniformly charged spherical annulus with internal radius R >= r?
    What is the field at radius r from the centre of a uniformly charged spherical annulus with internal radius r and external radius R > r?
     
  5. Nov 1, 2012 #4
    Hi alphysicist,

    Thanks a lot for the clarification. It makes perfect sense to me now. So the Gaussian surface has to be inside the sphere when we want the ##\vec{E}## inside it, and the charge is that of whatever is included in the surface. Also thanks haruspex for the response.
     
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