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Potential that has had me stuck for hours

  1. Feb 24, 2008 #1
    A total amount of positive charge Q is spread onto a nonconducting, flat, circular annulus of inner radius a and outer radius b. The charge is distributed so that the charge density (charge per unit area) is giver by o = k/r^3, where r is the distance from the centre of the annulus to any point on it. Show that (with V = 0 at infinity) the potential at the centre of the annulus is given by:

    V = (Q/8pie)((a+b)/ab)

    so I have

    dV = (1/4pie)(dQ/r)

    dQ = odA = o2pirdr = 2piQdr/r^2

    I'm really confused about the k in o = k/r^3
    I'm thinking it must either be the constant k = 1/4pie, but my professor never rights that as k, so I'm thinking that k=Q because otherwise I don't see a way to have Q in the expression.

    So I tried taking the integral of that expression from a to b with plugging dA and o in, but I'm not getting what I need.

    Any help is much appreciated!
    Thanks
     
  2. jcsd
  3. Feb 24, 2008 #2
    To find k: You are given the total charge (Q) and the charge density. Integrate the charge density over the annulus to get an expression for the total charge. Equate this with Q and solve for k.


     
  4. Feb 24, 2008 #3
    so if i integrate the charge density from a to b i get -2k/r^2
    Q = -2k/r^2
    k = -qr^2/2

    is this correct? how will i get rid of this negative?
     
  5. Feb 24, 2008 #4
    or do i do:
    odA = dQ
    Q = -4pik((1/b)-(1/a))
    Q = 4pik((b-a)/ab)
    k = Qab/4pi(b-a)
     
  6. Feb 24, 2008 #5
    Your second post is good, except you should check your work. Bringing the negative into the parentheses should change b-a to a-b, and from where did you get the 4?
     
  7. Feb 24, 2008 #6
    i used the - to make it 1/a - 1/b
    then changed it to b-a/ab so i was back at b-a

    the 4 i got from dA = 2pirdr
    then *-2 from the integral
    i used o=k/r^3 and odA = Q
    should i just use o=k/r^3 to get
    Q = -2k(1/b - 1/a)
    k = (Q/2)(ab/b-a)

    once i get k, what do i do with it?
     
  8. Feb 24, 2008 #7
    thats not right, without the dA i end up with
    Q = -2k(1/b^2 - 1/a^2)
     
  9. Feb 24, 2008 #8
    Sorry, that was my mistake. Your b-a is correct, but your k still isn't completely right. You are missing a [itex]\pi[/itex] in there.

    After you find k, you should then integrate to find the total potential. Remember, that

    [tex]V = \int \frac{k \cdot dq}{r}[/tex]

    EDIT:

    How are you finding Q? It should be the case that

    [tex]Q = \int dq = \int_a^b 2\pi r \sigma \cdot dr[/tex]
     
    Last edited: Feb 24, 2008
  10. Feb 24, 2008 #9
    i figure i must be integrating dQ = odA from a to b
    dq = int k/r^3 dA
    but i know dA = 2pirdr
    is that correct?
    i dont know how to integrate that
     
  11. Feb 24, 2008 #10
    Yes, that is correct, What you basically have is

    [tex]dq = \sigma dA = \frac{2\pi k \cdot dr}{r^2}[/tex]

    then

    [tex]Q = \int_a^b \frac{2\pi k \cdot dr}{r^2}[/tex]

    which you can do, because it is just using the power rule (take out the constants first). You almost had it right the first time.
     
  12. Feb 24, 2008 #11
    or will it be Q = 2pik * int dr/r^2
     
  13. Feb 24, 2008 #12
    that was what i did back when i got: k = Qab/4pi(b-a)
     
  14. Feb 24, 2008 #13
    Yes, you are right, but try integrating

    [tex]\int \frac{dr}{r^2}[/tex]

    again. If you do it correctly, then solve for k, you should have a 2, not a 4.
     
  15. Feb 24, 2008 #14
    ah yes, terrible mistake... brought over a -2 instead of a -1

    so when i intergrate the final part

    dV = kdQ/r
    do i want to have dQ = odA
    integrating that will give me what i got for Q again
    and what do i do with the r?
     
  16. Feb 24, 2008 #15
    Not quite, because you this time you are integrating dQ/(4[itex]\pi\epsilon r[/itex]), not just dQ, as you did when finding what k equals in terms of Q. Furthermore, you can replace the k in [itex]\sigma[/itex] with whatever you found k to be in terms of Q, a, and b.
     
  17. Feb 24, 2008 #16
    i can take out the 4pie as a constant so im left with integrating dQ/r which im not sure how to integrate..
    do I want to change dQ to odA then everything is in terms of r

    dV = odA/r4pie
    = k2pirdr/(r^4)(4pie)

    V = int dV = (Q/4pie)(ab/b-a) * int dr/r^3
    = (Q/4pie)(ab/b-a)(-2)(1/b^2 -1/a^2)

    is this correct so far?
     
  18. Feb 24, 2008 #17
    this leaves me with the correct answer, except with 2pie on the bottom instead of 8pie (I had a factor of 2 in the numerator instead of denominator)
     
  19. Feb 24, 2008 #18
    You're right, but you need a -1/2 instead of a -2 coming out of the integral.
     
  20. Feb 24, 2008 #19
    ah yes, i need to brush up on those integrals, thanks!
    i wouldn't have been able to solve this on my own
     
  21. Feb 24, 2008 #20
    wait, why would i get -1/2 from integrating 1/r^3
    shouldnt it be -2?
     
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