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Electric Potential inbetween Conducting Metal Ball and Thick Shell

  1. Feb 11, 2014 #1
    1. The problem statement, all variables and given/known data
    A conducting metal ball of radius 2m with a charge of 3μC is surrounded by a concentric spherical shell of inner radius 4m and outer radius of 5m with a total charge of 4μC. Determine the electric potential in volts between the ball and shell at a radius of 3m.



    2. Relevant equations
    V = k∑(q/r)
    V = k∫(dq/r)
    V = -∫E dA


    3. The attempt at a solution

    The charge outside of a conductor treats the conductor like a point charge, so...

    V = (k(3E-6))/(3) = 9000V

    I'm not really sure how to find the voltage inside a spherical shell like this. I think for conductors the potential inside is equal to the potential at the surface, so...

    V = (k(4E-6))/(5) = 7200

    Vtot = V1 + V2 = 9000V + 7200V = 16200V

    But this question came from a practice exam from my professor, and apparently, the answer is actually 14,850V. Where did I go wrong?
     
  2. jcsd
  3. Feb 11, 2014 #2
    I would look at it in three steps. Assuming V = 0 at infinity, you want to find the change in voltage from infinity to that point. So that means integrating the field from infinity to the shell, then through the interior of the shell, then through the inside space to the point in question. It can be evaluated without actual integration, but that's what's going on behind the scenes.
     
  4. Feb 11, 2014 #3
    So if I use V = ∫kdq/r then...

    V = k (∫5 q/r + ∫54 q/r + ∫43 q/r )
    = kq (((1/∞) - (1/5)) + ((1/5) - (1/4)) + ((1/4) - (1/3)))
    = kq (-1/3)
    = (9E9)(4E-6)(-1/3)
    = 12000

    which still doesn't add to 9000 to equal 14850. Am I using the wrong equation? I'm confused about what I should be integrating.
     
  5. Feb 11, 2014 #4
    You have the right idea, but the field you want to integrate depends on your limits, because the field varies depending on the location. For example, outside both the sphere and shell, both contribute to the field, and when you're inside the shell, there's no field since its a conductor. Try the calculation again with this idea in mind.
     
  6. Feb 11, 2014 #5
    Okay so the integral from 4 to 3 just equals zero because the electric field is zero inside a conductor...

    Am I taking the other two integrals as kq∫dr/r2 using V = -∫E dr ? Does Eoutside of shell = kq/r2?

    V = -kq ( ∫5 dr/r2 + ∫45 dr/r2 )

    V = -kq (((1/∞) - (1/5)) + ((1/5) - (1/4)))

    V = -(9E9)(4E-6)((-1/5) + (-.05))
    V = 9000

    Do I have the wrong formula for E?
     
  7. Feb 11, 2014 #6
    No, it's the same formula for E: [itex]E = k \cdot \frac{q}{r^{2}}[/itex], but it's a matter of the charge that contributes to the field. Outside of the shell, both the charge on the sphere and on the shell contribute to q for the calculation. However, inside of the shell (as in between the sphere and the shell), the contribution from the shell disappears, so the charge on the sphere is the only thing that contributes to q.
     
  8. Feb 11, 2014 #7
    So from infinity to 5 and from 5 to 4, I add the charges to get 7E-6, then add to the potential inside the shell...

    V = -k(7E-6) ( ∫5 dr/r2 + ∫45 dr/r2 ) + k(3E-6)/3

    V = -k(7E-6) (((1/∞) - (1/5)) + ((1/5) - (1/4))) + k(3E-6)/3

    V = -(9E9)(7E-6)((-1/5) + (-.05)) + (9E9)(3E-6)/3
    V = 24750V

    I must still be doing something wrong...?
     
  9. Feb 11, 2014 #8
    You're nearly there. But remember that the integral from 4 to 5 is zero, since that's the part that's entirely within the shell -- as in between the two radii of the shell -- 4 and 5 meters. And then when you calculate k(3E-6)/3, that would be the potential from the center of the sphere out to 3 meters, but it's the opposite -- you're coming from infinity, so it has to be the potential from the inside radius of the shell (4) to 3, which involves another integral.
     
  10. Feb 11, 2014 #9
    Okay, I finally got the right answer, but can you tell me if I'm understanding the integrals correctly...? Can I think of the charge that is considered in each integral qenclosed for a Gaussian surface at the closest distance?

    V = -k (∫5 E dr + ∫45 E dr + ∫34 E dr)

    Ebetween 4 and 5l = 0 because everything between 4 and 5 is a conductor, and E = 0 inside a conductor.
    V = -k (∫5 qencl/r2 + (∫45 0 + ∫34qencl/r2 )

    V = -k (∫5 (7E-6)/r2 + (∫45 0 dr + ∫34 (3E-6)/r2 )

    How come when I take the integral of 0dr it doesn't equal r?

    V = -k ((7E-6)((0-1/5)) + (0) + ((3E-6)(.25-.3333333))

    V= 14850V
     
  11. Feb 11, 2014 #10
    Glad you came to the right answer. And you can think of figuring out the value for q that way, since in this case you're dealing with spherical shapes. I would double check that just by reasoning though, and if you have different shapes, go through the whole reasoning with Gauss's law to calculate field, since it will end up being different.

    And when you take the integral of 0dr, you get zero since essentially you're integrating nothing. If you think of integration as a summation, then an integral of zero just gives you a sum of zeros, which is zero. You can also think of zero as a constant that you can slip out of the integral. Then you integrate dr and do indeed get r, but then no matter what your limits are, there's still that zero out in front that makes the whole thing zero.

    One final note. If you don't like dealing with all the integration, you can think of it too in terms of potential differences. So you can find V = kq/r at each limit -- each change in the field -- and then add up all the differences. It's the same math as with what you did with the field and integration, but essentially with this case you're just using a formula that has already been integrated.
     
  12. Feb 11, 2014 #11
    Thank you so much for all of your help! I really appreciate it, and I understand the concept of potential a lot better now!
     
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