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Gauss Law

  1. Aug 21, 2009 #1
    1. The problem statement, all variables and given/known data
    One ,non uniform but spherical in symmetry distribution of charge has charge density which is given by the following equations
    ρ=ρο(1-r/R) when r< R and ρ=0 when r>R, ρ0=3Q/πr^3 and it is constant
    a) show that the total charge included by this distribution is equal to Q
    b)show that for r>R the electric field is the same that occurs as if it comes from a point charge
    c)find an expression for the electric field for the area r<R


    2. Relevant equations
    the equations i used was gauss law and Qenc= integral pdV
    i hope u understand i don't to write yet in latex

    3. The attempt at a solution
    I solved a and b and i can't think about c
    i know that the desired expression is E=(p0r/12ε0)(4-3r/R)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 21, 2009 #2
    i tried the following path without success


    [tex]\int E dA[/tex]=Qenc/ε0 but Qenc=[tex]\int p Dv[/tex] therefore

    [tex]\int E dA[/tex] equals [tex]\int p Dv[/tex] i figure out how dV is so change it with the equal dr expression so i did for dA..i idi the math math but no result equal the solution i wrote above...

    any help is good cause i have tests in the university in few days :P
     
  4. Aug 21, 2009 #3
    Edit: Sorry I misread your post...you're asking for (c).
    The electric field inside a hollow thin shell of uniformly distributed charge is 0. For a position r from the center, note that it can be considered to be contained within a shell of inner radius r and outer radius R; this shell has a thickness and varying charge density, but you could also look at it as several infinitesimally thin shells of different but still uniform charge densities piled on top of each other. Also note that at the same time as being contained within a thick shell, it can also be considered to be right outside a charged sphere of radius r.
     
    Last edited: Aug 21, 2009
  5. Aug 22, 2009 #4
    Thanx for the reply but it doesn't help to much..
    The charge distribution is not uniform inside the volume and it behaves according to this equation
    ρ=ρ0(1-r/R) p= charge density R=radius of the main volume and r= radius of the gauss surface.. and ρ0=3Q/πR^3 and it is constant

    I need to find an expression for the electric field for r<R(inside the main volume) E(r)..I Looked at the solution and the desired equation is E=(p0r/12ε0)(4-3r/R)

    I can't get to this equation...i tried to do the following maths [tex]\int E dA[/tex] = Qenc/ε0 and Qenc= [tex]\int p dV[/tex] and compine those two
     
  6. Aug 22, 2009 #5

    kuruman

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    That's how you should do it. How far did you get doing each integral?
     
  7. Aug 22, 2009 #6
    i said dA=8πrdr and dV=4πr^2dr and r from 0 to R and did the maths..u can try it but u can't result to the desire equation...i took E as constant in the integral and solved by E

    can u think of anything different ?
     
  8. Aug 22, 2009 #7
    Try to visualize it like this. Take an arbitrary value r < R; this would imply the electric field at a spherical surface (surface G) with radius r inside the spherical charge distribution mentioned. So you have an electric field coming from the charge distribution beyond surface G and from the charge distribution within surface G. What my other post was stating was that the electric field from beyond surface G is 0 N/C; so the electric field at surface G (in other words, at a radius R) is due primarily to the charge distribution within surface G (meaning the charge distribution up to a radius r). So what you'd have to do is use Gauss's Law to calculate the electric field due to the charge distribution at a radius r --- The other main difference in (c) from (a) and (b) is that in (a) and (b) the charge distribution is Q...however, since you're only looking at the charge distribution up to a radius r, you'll have to evaluate a different value for the charge in part (c).
     
  9. Aug 22, 2009 #8

    kuruman

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    To begin with dA is not 8πr dr. Element dA is on the surface of the sphere where the radius is constant. Also, you say that the constant in the charge distribution is ρ0=3Qπ/r3. I think it should be ρ0=3Qπ/R3, otherwise it is not a constant. Maybe the r3 in the denominator confused you.

    Do each side of Gauss's Law separately,

    [tex]\int \vec{E} \cdot \vec{dA} = ?[/tex]
    [tex]\int \rho dV = ?[/tex]

    and tell us what you get for each. It would be easier to help if you show what you did.
     
  10. Aug 22, 2009 #9
    Ok boys i solved..i took ,like before, those two ,indefinite now, integrals and solved them to find E

    my mistake was that before i took those two definite integrals from 0 to R
     
  11. Aug 22, 2009 #10
    when i find the time i will post the entire problem with the solution so others can see...

    this was my first experience in this site and i found it very useful and nice :)

    excellent...and u guys rulez..
     
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