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Electric potential: point charge in a hollow charged conductor

  1. Mar 8, 2009 #1
    [solved] electric potential: point charge in a hollow charged conductor

    1. The problem statement, all variables and given/known data
    A hollow spherical conductor, carrying a net charge +Q, has inner radius r1 and outer radius r2 = 2r1. At the center of the sphere is a point charge +Q/2.

    GIANCOLIch23p22.jpg



    d) Determine the potential as a function of r for 0 < r < r1.

    2. Relevant equations

    (π = pi)
    For r > r2, the electric field is (3Q)/(8πε0r2).
    For r1 < r < r2, the electric field is 0 (ie, field inside conductor is zero in static situations).
    For 0 < r < r1, the electric field is Q/(8πε0r2).


    The potential as a function of r for r > r2 (where voltage is taken to be 0 when r is infinite) is (3Q)/(8πε0r).

    The potential as a function of r for r1 < r < r2 is (3Q)/(16πε0r1).


    3. The attempt at a solution

    My first instinct was to add the potential (3Q)/(16πε0r1) to Q/(8πε0r), which is the potential from infinity to r if the shell wasn't present. However, the answer is wrong. I also made many other fruitless attempts at this problem, but none of them very logical.

    Can someone tell me what I'm doing wrong, and how to find this potential when r is within the cavity of the conductor?

    I would greatly appreciate your help with this problem (and thank you in advance)!
     
    Last edited: Mar 8, 2009
  2. jcsd
  3. Mar 8, 2009 #2

    gabbagabbahey

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    When in doubt, go back to the definition of the potential at a point [itex]\vec{r}[/itex]:

    [tex]V(\vec{r})=-\int_{\infty}^{\vec{r}}\vec{E}\cdot\vec{dl}[/tex]

    The electric field has different values in different regions, so you break the integral into pieces:

    [tex]V(\vec{r})=-\int_{\infty}^{\vec{r}}\vec{E}\cdot\vec{dl}=-\int_{\infty}^{r_2}\vec{E}_{\text{outside}}\cdot\vec{dl}+-\int_{r_2}^{r_1}\vec{E}_{\text{between}}\cdot\vec{dl}+-\int_{r_1}^{\vec{r}}\vec{E}_{\text{inside}}\cdot\vec{dl}[/tex]
     
  4. Mar 8, 2009 #3
    So are you saying that I should add all three potentials like this?
    ie, (3Q)/(16πε0r1) + [Q/(8πε0)](1/r - 1/r1)

    -I realize I derived the potential from r to r1 incorrectly.

    -Also, I didn't include the potential in the region from r2 to infinity when I first solved the problem because I thought that (3Q)/(16πε0r1) basically came from the total potential from infinity to r1 (ie, potential is constant in that region aka the integral representing the change in potential is zero, so potential anywhere in r2 to r1 is just the potential as we approach r2). Am I right in assuming this or do I need to add (3Q)/(16πε0r) to the sum of the integrals too?
     
  5. Mar 8, 2009 #4

    gabbagabbahey

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    Well, what does the middle integral actually add to the potential inside r<r1 ?
     
  6. Mar 8, 2009 #5
    No, it doesn't add to the integral because E is 0 in that region.

    Okay, thank you for your help. :)
     
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