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Gauss' law

  1. Nov 16, 2009 #1
    1. The problem statement, all variables and given/known data

    A sphere of radius R=10 cm made of insulating material has a uniform charge density p=100 C/m-3

    a)Find the magnitude of the electrical field at a distance r<R, using Gauss' Law.What is the value of the electric field at r=5 cm?
    b)What is the total charge on this sphere?
    c)Find the magnitude of the electric field at a distance r>R using Gauss' Law.What is the value of the electrical field at r=10cm? At r=20 cm?
    e)Take the voltage at r=0 to be 0 Volts.Find the Voltage V(r) at any r by integrating the Electrical field E(r) along the radial direction.

    2. Relevant equations

    Gauss' Law= [tex]\oint[/tex]E.ds=Qinside/E0 (permittivity constant)
    3. The attempt at a solution

    I managed to do b) since the density is provided to us we only have to find the Q from d= Q/V, V is 4/3Pi R3.... but i certainly do need help with other questions.
  2. jcsd
  3. Nov 16, 2009 #2
    I will start by helping you with part (a).

    on the left hand side of Gauss' equation we are able to move the E field through the surface integral. why?

    side note: (are you in calculus based physics or algebra based?)
  4. Nov 16, 2009 #3
    We are able to move the E field through the surface integral because the electric field is parallel with the surface vector of that area am i right?

    (Calculus based physics)
  5. Nov 16, 2009 #4

    now, we can modify our equation a little.


    what is [tex]\oint{dA}[/tex] equal to?
  6. Nov 16, 2009 #5
    [tex]\oint{dA}[/tex] equals to surface area i assume?
  7. Nov 16, 2009 #6
    yes, now how would one express the charge in term of the charge density?

    (remember that we desire an express in terms of the radius)
  8. Nov 16, 2009 #7
    well, charge density should be equal to total charge/Volume,,, and volume can be expressed as a function of the radius.
  9. Nov 16, 2009 #8
    right, so this is our expression:

    [tex]E4\pi r^2=\frac{\rho 4 \pi r^3}{3\epsilon_0}[/tex]

    does this make sense to you?
    Last edited: Nov 16, 2009
  10. Nov 16, 2009 #9
    To be honest i haven't seen this around before, would be glad if you can tell me the way that you calculated such a formula?
  11. Nov 16, 2009 #10
    We just calculated it together:

    [tex]E[/tex]= electric field
    [tex]4 \pi r^2[/tex]= surface area of a sphere =[tex]\oint dA[/tex]
    [tex]\rho=\frac{Q_{en}}{V_s}[/tex] (where V is the volume of the sphere.)

    so [tex]\frac{\rho 4 \pi r^3}{3}=Q_{en}[/tex]

    what doesn't make sense?
  12. Nov 16, 2009 #11
    right! i got it now. its all clear now :)
  13. Nov 16, 2009 #12
    So how are we going to plug this to the question. It feels like there is still things that im missing ?
  14. Nov 16, 2009 #13
    (sorry I went out to eat, I am back now for 15mins then I have class just f.y.i)

    nothing is missing just plug in what is known and solve for the E field.

    Note that this is the E field for only 0<r<R
  15. Nov 16, 2009 #14
    thats ok =). I have a question about that.What changes when we take a point that is greater than our radius, and when we take a point that is smaller than the radius, which is actually within the sphere.
  16. Nov 16, 2009 #15
    The charge enclosed changes from linear to constant.

    When: 0<r<R

    the charge enclosed is a function of the radius. This is quite intuitive because the closer we get to R the more charge we have enclosed.

    When: R<r<infinity

    the charge enclosed is a constant function. This is also quite intuitive because there is no charge outside the sphere which would add to the enclosed charge.
  17. Nov 16, 2009 #16
    Ok that makes sense except 2 points;

    First how are we going to find the formula for the other situation, what do we need to change?

    Second one;

    For the situation, R<r<infinity , this actually means the point we are considering is somewhere out of the sphere.Are you saying that the E field is constant for all the points that is left out of the sphere?, shouldn't it be decreasing with proportional to 1/d2
  18. Nov 16, 2009 #17
    Your first point:
    we are going to find the formula for the other situations in the same manner. We need to change the charges dependence on r.

    point two:
    I am not saying the E field is constant outside of the sphere. I am saying that the enclosed charge is constant outside of the sphere. You should be able to differentiate between these two things.

    (Note: the E field will fall off inversely proportional to the radius squared)
    Last edited: Nov 16, 2009
  19. Nov 17, 2009 #18
    Alright i understood.Thanks for all the help
  20. Nov 17, 2009 #19
    My pleasure ^^
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