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Electric potential of two spherical shells

  1. Jan 25, 2012 #1
    1. The problem statement, all variables and given/known data[/b]
    Consider two thin, conducting, spherical shells as shown in cross-section in the figure below.
    http://capa.physics.mcmaster.ca/figures/sb/Graph25/sb-pic2565.png [Broken]
    The inner shell has a radius r1 = 17.2 cm and a charge of 10.3 nC. The outer shell has a radius r2 = 27.4 cm and a charge of -29.1 nC.
    Calculate the electric potential V at r = 10 cm, with V = 0 at r = infinity.

    2. Relevant equations

    3. The attempt at a solution
    I'm not sure if i need to use integrals or not since it said this is only a cross section. i've tried find the radius of the spheric shells in respect to the point at 0.1 m but that didn't work. Any help would be appreciated :D
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 25, 2012 #2
    The first thing is to think about what electrical potential means. The second thing is to remember or look up what is special about charges and electric fields inside, around and on a sphere.
    Last edited: Jan 25, 2012
  4. Jan 25, 2012 #3
    Well i know the electric field for the inner sphere is zero but for the outer sphere, the inner sphere should act as a charged particle sooo there should be some sort of field with reference to the outer ring? would V simply by V=Ke(-29.1e-9)/(0.27-0.1)?
  5. Jan 25, 2012 #4
    Ok so you have what act like point charges until you get inside each one so you can find the potential for the larger sphere to its surface but once you go inside it it takes no more energy or work to continue on toward the center of the entire system. The same thing is true of the inner sphere - it acts like a point charge until you get inside it and then no more work needs to be done to get anywhere inside it. See if you can go from there.
  6. Jan 25, 2012 #5
    Would the potential be 0V in that case?
  7. Jan 25, 2012 #6
    Remember your definition of potential - energy per unit charge to move from infinity to the point designated. So you find the potential at the surface of each sphere for the charge of that sphere using the formula (a very easy integral from infinity to the radius or just use the formula) and realize that once you go inside the smaller sphere it takes no more energy so no more potential.
  8. Jan 25, 2012 #7
    Why would the work be zero once your in the larger sphere?
  9. Jan 25, 2012 #8
    Only zero additional energy gained or lost once inside the larger sphere. You do them separately then add. So find the potential (I like to think energy because it is logical that it takes energy to move in toward the sphere or you gain energy depending on the charge) for the large one to its surface, then find the potential for the small one to its surface, there isn't any more inside since there's no electric field in there. So use your forumla for potential for a point charge but just stop at the surface of each sphere.

  10. Jan 25, 2012 #9
    for the radius, would it just the the given radius that you use in the equation of would it be

  11. Jan 25, 2012 #10
    No. Go to the link I have you and read it and look at the picture. There is no additional energy used to get inside the shell so you only use the radius of the big sphere for it's potential and of the small sphere for it's potential. The 10 cm is given just to see if you really understand the concept of electric fiend and potential of spheres. No matter where you are asked for the potential, if it's inside either sphere you have to calculate the potential for just the radius of that sphere you've gone inside. You can't use any additional distance inside it.

    Let's suppose you are being sucked into a giant spherical vacuum cleaner. The closer you get the harding it pulls on you but once you are sucked inside it it there is no more suction. You just float around in there. It's the same for a charged sphere acting on another charge. It will suck that charge in until it gets to the surface but if the charge is moved inside the sphere there won't be any more pull on it.

    so the 10 cm is a TRICK value to see if you can be confused. Or if you understand the principle well enough to know once you're inside both spheres there is no increase in potential so you just use the given radii of the two sphere for each of their individual potentials, then add them.
  12. Jan 26, 2012 #11
    How would you know that the system is in electrostatic equilibria because the electric potential anywhere inside is equal to that of the surface only when that happens.
  13. Jan 26, 2012 #12
    ignore the last comment :D
  14. Jan 26, 2012 #13
    thank you for the help :) It's hard to get my head around E and M LOOOL
  15. Jan 26, 2012 #14
    Good point - because there's no wires connecting them and it says nothing about the spheres floating in a conducting liquid or anything like that and it gives you a specific charge on each which means that's the charge. It says nothing about that charge changing. There's another homework problem on here where a wire connected the inner and outer sphere in which case all the charge goes to the outer sphere but these spheres are NOT touching or it would say so.
  16. Jan 26, 2012 #15
    That was a very sneaky problem
  17. Jan 26, 2012 #16
    so if we consider it in equilibrium, I'm assuming the significance of a charge inside a sphere(in this case created by the smaller sphere) wouldn't matter? That the electric potential inside is still equal to that of the surface.
  18. Jan 26, 2012 #17
    Yes but you have to do each one independently. Find the potential of the larger one using its radius and charge. Then find the potential of the smaller one using its radius and charge, then add them. That will give you the total potential at the surface of the smaller sphere. Once you inside the smaller sphere there is no change at all in the potential so you've already got your answer.

    What threw me off on your question is because there are two spheres, the potential inside the big one is the same as the potential at its surface but the small ones potential is still changing until you get to its surface. So hopefully you have the right idea.

    Try to think of the vacuum cleaner analogy. If there are two vacuum cleaners pulling on you, once you're pulled inside the big one its stops pulling on you so you can caculate its potential using its radius. But the smaller vacuum cleaner is still pulling you in so more work will be done on you until you get to its surface. Once you're there you can now calculate the potential of the smaller vacuum cleaner because it can't pull any more. You'll go inside both now and there is no more pull. You're finished with the problem once you reach the surface of the smaller sphere.
    Last edited: Jan 26, 2012
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