1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Non-Uniform charged sphere

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data
    A non-uniformly charged sphere of radius R has a charge density p = p_o(r/R) where p_o is constant and r is the distrance from the center of the spere.

    a) find the total charge inside the sphere

    b) find the electric field everywhere (inside & outside sphere)

    c) find the electric potential difference between teh center of the sphere and a distance r away from the center for both r<R and r>R.

    d)Graph the potential as afunction of r

    2. Relevant equations



    3. The attempt at a solution

    a) Find the total charge in sphere.

    p = dQ/dV

    Q = integral p dV where dV = 4(pi)R^2dR and it is integrated from 0 to R

    Q = integral from 0 to R p_o*(r/R)*4*pi*(R^2)*dR = 2*pi*p_o*r*(R^2)

    b) Find electric Field inside & out of sphere

    Outisde r>R

    Q = 2*pi*p_o*r*(R^2)

    [tex]\oint[/tex]EdA = Q_encl / E_o = E(4*pi*(R^2)) = (2*pi*p_o*r*(R^2))/E_o

    E = (p_o*r)/2E_o

    Inside... r<R

    If the rest is correct, can i get a hint on finding the inside as to the outside. this confuses me. it seems the only difference is r<R, but how would that effect the equation.
     
  2. jcsd
  3. Apr 26, 2010 #2

    [tex]dQ = \rho dv = (\frac{\rho_0}{R})r 4\pi r^2 dr[/tex]

    So your integral for (a) is wrong because your dv is wrong. To find the E field inside the sphere draw your Gaussian surface at some arbitrary r inside the sphere. What is [tex]Q_{enc}[/tex]? You know that [tex] dQ = \rho dv [/tex] but what are your limits now?
     
  4. Apr 26, 2010 #3
    ok so for part A i integrated and got Q = (4[tex]\pi[/tex]por4)/R and wouldnt that be the same as Qencl because it was the charge in the sphere

    Einside = (1/4[tex]\pi[/tex]r2)(Qencl/Eo)

    what differes when i find it outside
     
  5. Apr 26, 2010 #4
    [tex] Q_{total} = \frac{4\pi\rho_0}{R}\int_{0}^{R}r^3dr =
    \frac{4\pi\rho_0}{R}\frac{R^4}{4} = \pi\rho_0R^3 [/tex]

    To find charge inside all you do is change your limits from 0 -> R to 0 -> r where r is an arbitrary radius inside the sphere.

    p.s learn latex it's not that hard https://www.physicsforums.com/showthread.php?t=8997.
     
  6. Apr 26, 2010 #5
    So if you want the E field outside the sphere, [tex] Q_{enc} = Q_{total} [/tex] since the whole sphere is enclosed with your Gaussian surface.
     
  7. Apr 26, 2010 #6
    ok so for part a i wanted the total charge inside sphere which would be Qenc.. correct?

    Part A)

    Qencl = [tex]\pi[/tex]por4/R

    Then for part B i need the Electric field inside and outside

    Inside..

    E(4[tex]\pi[/tex]r2) = Qencl/Eo

    and then outside i would solve for Qtotal as you did earlier and use that instead of Qencl in the same equation for the inside


    why does my pi always end up like an exponent
     
  8. Apr 26, 2010 #7
    Use latex and that link I gave you, it is simple.

    (a) Your answer is wrong (I showed you the answer in a previous post) but you are right that [tex] Q_{enc}(r) = \frac{\pi\rho_0}{R}r^4 [/tex] --> This equation is the enclosed charge as a function of your radius with r between 0 and R. It is asking for the total charge of the sphere though. So what is the total enclosed charge if the whole sphere is included? It is just that equation but with r = R.

    (b) For inside the sphere: Looks good so far and you know what [tex] Q_{enc} [/tex] is :
    [tex] Q_{enc}(r) = \frac{\pi\rho_0}{R}r^4 [/tex].

    For outside the sphere: You know what [tex] Q_{enc} = Q_{total} [/tex] or in otherwords your answer from (a).
     
  9. Apr 26, 2010 #8
    ok sorr i was confused.. i thought that the charge inside would not include the total sphere

    ok now i need to find the potential difference for r<R and r>R
    Im confused on the integration limits

    r<R do i integrate from 0 to R

    r>R i integrate from R to r
     
  10. Apr 26, 2010 #9
    It asks for the potential difference FROM THE CENTER OF THE SPHERE and a distance r away from the center

    r<R is that a distance r from the center?

    r>R is that from the center of the sphere?
     
  11. Apr 26, 2010 #10
    the exact wording is

    Find the electric potential difference between the center of the sphere and a distance r away fr the center for both r<R and r>R. Be sure t indicate where you have chsen yur zero reference potential.

    the example in the book finds the potential for r>R and they integrate from ra to rb, setting rb = infinity

    there is not really an exaple for r<R so im lost

    is there a rule for doing this, the book is not clear
     
  12. Apr 26, 2010 #11
    Well if you are setting your zero potential to be at r = infinity then for r>R you would do the same integration. For r<R you still have to integrate from r = infinity but to a point r inside the sphere. This requires two different integrals because there are two different E fields (one for r>R and one for r<R).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Non-Uniform charged sphere
  1. Uniform Charge on Sphere (Replies: 22)

Loading...