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Homework Help: Physics olympiad - electrostatics

  1. Oct 26, 2008 #1
    (Any number of options may be correct.)

    Three concentric spherical shells have radii r, 2r and 3r with charges q1, q2 and q3 respectively. Innermost and outermost shells are earthed. Then,

    (a) q1+q3=-q2
    (b) q1=-q2/4
    (c) q3/q1=3
    (d) q3/q2=-1/3

    The answer key tells me that (a), (b), (c) are correct. Could anyone please help me solve this?
    Last edited: Oct 27, 2008
  2. jcsd
  3. Oct 26, 2008 #2
    I know that I must write the net potentials of the innermost and the outermost shells as zero.

    Could someone please tell me how to write the net potential of a particular shell? Does it depend on the other shells as well?

    Thanks in advance.
  4. Oct 26, 2008 #3
    Show your work...which laws do you think you'll need?
    Last edited: Oct 26, 2008
  5. Oct 26, 2008 #4
  6. Oct 26, 2008 #5
    That equation is correct for a single point charge (or a charge q distributed in a spherically symmetric way) once you choose your potential equal to zero at infinity.

    Do you know Gauss' law?

    Do you know how to calculate a potential once you know the electric field?
    Last edited: Oct 26, 2008
  7. Oct 26, 2008 #6
    Yes I know Gauss' law.
  8. Oct 26, 2008 #7
    E=-dV/dr ?
  9. Oct 26, 2008 #8
    Apply Gauss' law to find E, then use E to find the potentials: not by integrating from infinity, but starting at the origin r=0.
  10. Oct 26, 2008 #9
    Yes, kind of: you need the inverse relation since you can calculate E from Gauss' law.
  11. Oct 26, 2008 #10
    I'll try your method tomorrow and tell you if I was able to solve it or not. Anyway, thanks for your help. :)
  12. Oct 26, 2008 #11
    Okay: you should find that exactly *one* of the answers is correct.
  13. Oct 27, 2008 #12
    Actually, the answer key tells me that (a), (b), (c) are correct.
  14. Oct 27, 2008 #13
    One condition follows from the given that the potentials on shells 3 and 1 are the same. The other two are really the same conditions then, and follow if you insist on having a potential zero at infinity (I didn't think you need that, but okay....you do).

    And, did you use the approach I suggested?
    Last edited: Oct 27, 2008
  15. Oct 27, 2008 #14
    Yes, but now you have to get the potential from the fields, start integrating from either r=0, or r=\infty.
    Actually, in this problem r is a constant for some reason, so use a different variable for the variable distance.
    (I think you mean your answer only for the constant distances given in the problem, but that is not enough).
  16. Oct 27, 2008 #15
    Using Gauss' law I get

    [itex]E_14 \pi r^2=\frac{q_1}{\epsilon}[/itex]
    [itex]E_24 \pi (2r)^2=\frac{q_1+q_2}{\epsilon}[/itex]
    [itex]E_34 \pi (3r)^2=\frac{q_1+q_2+q_3}{\epsilon}[/itex]
  17. Oct 27, 2008 #16
    Could you please complete the solution for me? I'm really stuck! :(
  18. Oct 27, 2008 #17
    No, I'm not gonna complete it for you.

    For a variable distance r between R and 2R (I'm using R now for what is called r in the problem), the field is E(r)=q1/4\pi \epsilon r^2. You have to integrate this from R to 2R to get the potential difference between shells 2 and 1. Then you do the same for r between 2R and 3R: that gives you the potential difference between shells 3 and 1, which should be zero. That actually gives you one of the conditions a,b,c,d, but I won't tell you which one.

    If you can't integrate, then you can't solve the problem.
  19. Oct 27, 2008 #18
    yeah..keep going
  20. Oct 27, 2008 #19
    [itex]E=\frac{q}{4\pi\epsilon R^2}[/itex]
    [itex]V_2-V_1=\int^{2r}_{r}\frac{q}{4\pi\epsilon R^2} dR[/itex]
    [itex]V_2-V_1=\frac{q}{4\pi\epsilon R} (r\ to\ 2r)[/itex]
    [itex]V_2-V_1=\frac{q}{4\pi\epsilon}(1/r-1/2r) [/itex]
  21. Oct 27, 2008 #20
    Now you have to put in the correct values of q...and then you're basically there!
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