1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Epotential of Point Charges - Really Hard Conceptual Problem

  1. Jan 26, 2009 #1
    A thick spherical shell of charge Q and uniform volume charge density p is bounded by radii r1 and r2, where r2 > r1. With V = 0 at infinity, find the electric potential V as a function of the distance r from the center of the distribution, considering the regions (a) r > r2, (b) r2 > r > r1

    a) I got V = (kQ)/r

    b) I found that p = 3Q/(4Pi(r2^3-r1^3))
    using that I found that qencl = Q*(r^3-r1^3)/(r2^3-r1^3)

    Then using Gauss' Law I found that E = Q/(4*PI*r^2*Eo) * ((r^3-r1^3)/(r2^3-r1^3))

    Change in E potential = -integral(E)*dr lower limit r^2, upper limit r

    Vr - Vr2 = -kQ/(r2^3-r1^3) * (r^2/2 - r1^3/r + r2^2/2 + r1^3/r2)

    I'm stuck at this part

    The answer is:

    I don't how they simplified the answer to that

    Attached Files:

    Last edited: Jan 26, 2009
  2. jcsd
  3. Jan 26, 2009 #2
    Should I move Vr2 to the other side?
  4. Jan 26, 2009 #3
    yea that's right I think, Vr2 would equal to kQ/r but I'm also stumped on how it is simplied to the final answer.
  5. Jan 26, 2009 #4


    User Avatar
    Homework Helper

    First of all check the signs in the integral.
    We have to find potential at distance r from the center not the potential difference between points at r and r2.
  6. Jan 26, 2009 #5


    User Avatar
    Homework Helper

    All looks well down to the integral. It seems to me the integral ought to go from infinity (where you know the potential) down to r. Of course you would have to do it in two parts, from infinity to r2 and then from r2 to r. Oh, I suppose you are already taking that first part into account as the Vr2.

    In the last step, Vr - Vr2, I'm having trouble with the signs on the terms in the last brackets. Surely there should be two negative terms - including the r^2/2.
  7. Jan 26, 2009 #6


    User Avatar
    Homework Helper

    I think you got it.
  8. Jan 26, 2009 #7
    Isn't that what I was doing? I subtract Epotential between r2 and r from the Vr2 to get Vr?

    I don't think so :(. I still don't know how to simplify it.

    By the way reuped the pic of the answer

    what I don't get is how they got (3r2^2)/2
  9. Jan 26, 2009 #8


    User Avatar
    Homework Helper

    Integrating field between limits r to r2 gives potential difference between the two distances. Potential at r is the potential difference between r and infinity.
  10. Jan 26, 2009 #9
    My prof just posted solutions:


    I still don't get how he sub it in and simplified it

    Attached Files:

    Last edited: Jan 26, 2009
  11. Jan 26, 2009 #10
    sorry i fixed the pics again.
  12. Jan 26, 2009 #11
    what happened to the r1^3/r2 term?
  13. Jan 26, 2009 #12


    User Avatar
    Homework Helper

    Put V0 = KQ/r2 and then take KQ/(r2^3 - r1^3) common.

    There are two terms r1^3/r2 with opposite sign and get canceled. The simplification is correct.
  14. Jan 26, 2009 #13
    I don't understand what you mean by take KQ/(r2^3 - r1^3) common.

    Do you mean:

    kQ/(r2^3-r1^3) * ((r2^3-r1^3)/(kQr2) - r^2/2 + r2^2/2 +r1^3/r + r1^3/r2)

    I get KQ on the bottom now.
  15. Jan 26, 2009 #14
    NVM i got it THANK YOU!!!!!!!!
  16. Jan 26, 2009 #15


    User Avatar
    Homework Helper

    V = [KQ/r2] - [KQ/(r2^3 - r1^3)] *[r^2/2 - r2^2/2 + r1^3/r - r1^3/r2]

    = [KQ/(r2^3 - r1^3)]*{[(r2^3 - r1^3)/r2 - [r^2/2 - r2^2/2 + r1^3/r - r1^3/r2]}

    = [KQ/(r2^3 - r1^3)]*{r2^3/r2 - r1^3/r2 - r^2/2 + r2^2/2 - r1^3/r + r1^3/r2}
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook