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Charge distrubution in a sphere

  1. Dec 12, 2006 #1
    1. A sphere has a charge distrubution such that the charge density goes as P(r)=Ar^(1/2). Where A=0.200 C/ M^(7/2). If the sphere has a radius of 2.00m, determine the electric potential at a point 3.00m from the center of the sphere.



    I understand that P=Q/V (Im using P as Rho) but because the charge is not constant i must intergrate it. So P(r)= dQ/dV, then dQ=P(r)dV

    V of sphere= ((4pi)r^3)/3) So dV= (4pi)r^2 and therefore dQ=P(r)(4pi)r^2
    Now i beleive that A and R are constants so I removed them from the intergral. After intergrating both sides of the equation i get Q=(Ar^(1/2)) (((4pi)r^2)/3)
    I plugged in all the numbers and came out with 0.945 for Q. The answer for Q should be 0.812.

    Can anyone tell me where I went wrong?
     
  2. jcsd
  3. Dec 12, 2006 #2
    Is there anybody out there?
     
  4. Dec 12, 2006 #3

    Physics Monkey

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    The integral you're doing is over r so you can't just take all those r's out. The charge density is [tex] \rho(r) = A r^{1/2} [/tex] and so as you've said, the total charge is given by [tex] \int^R_0 \rho(r) 4 \pi r^2 dr [/tex] where [tex] R = 2 [/tex] m is the radius of the sphere. So again, little r is not constant, it is an integration variable. Do the integral and you should get the right answer.
     
  5. Dec 12, 2006 #4
    I did the intergral the correct way and came up to the correct answer. Now to find electric potential I plugged in the numbers to V=k (Q/R) this gave me a number for voltage of 3.65 X 10^9. I beleive this is correct but it seems to easy.
     
  6. Dec 12, 2006 #5

    Physics Monkey

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    What you have done is ok, but you should make sure you understand why. The reason why the potential is just kQ/r is because the charge distribution on the sphere is spherically symmetric and because you are looking outside the sphere. Remember that the potential of symmetric shell of charge is the same outside the shell as if all the charge were sitting at the center of the shell. The same basic principle is at work here. Similarly, if you were looking inside the sphere (i.e. at distances less than 2 m) then you would find that not all the charge contributes. Again this is because a symmetric shell of charge produces no field inside the shell.

    Make sense?
     
  7. Dec 12, 2006 #6
    I understand what you have said and I think I made a mistake. To find the potential at 3.00m. r is now 3.00 not 2.00? This time I came up with an answer of 2.4 X 10^9

    Thanks!
     
  8. Dec 12, 2006 #7

    Physics Monkey

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    Ah, ok, great. I didn't check your numbers so im glad you were paying attention.
     
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