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Electric potential

  1. Jun 14, 2012 #1
    A non conducting spherical shell is uniformly charged. The electrostatic potential at the centre is 200v and the electrostatic potential at the distance r=50cm from the centre is 40v. Find the radius of sphere a.

    I am not sure where go with this.

    I know that the potential difference is equal to the work done in moving from one point to another.

    Any help or direction would be appreciated,

    Thanks
     
  2. jcsd
  3. Jun 14, 2012 #2
    Do you know the shell theorem?

    If so, what does it tell you about the potential inside a uniformly charged shell? And what is the potential outside the shell?

    Frame equations based on answers to the above questions, and you should be able to solve for radius.
     
  4. Jun 14, 2012 #3
    But the question states that the potential inside the shell is 200v
     
  5. Jun 14, 2012 #4
    Yes it does. How can you express the potential inside the shell mathematically? Say the shell has a charge Q, radius R.
     
  6. Jun 14, 2012 #5
    Q/r^2
     
  7. Jun 14, 2012 #6
    Noo!!

    You need to add in the constant of proportionality [itex]k=1/4\pi \epsilon[/itex] before you can use that. And even then, kQ/r2 is the electric field outside the shell. You need the potential inside the shell...
     
  8. Jun 14, 2012 #7
    Is the potential energy constant through the shell?
     
  9. Jun 14, 2012 #8
    Yes it is. Or equivalently, the electric potential is constant throughout the shell. That's indirectly what the shell theorem states....
     
  10. Jun 14, 2012 #9
    So obviously r has to be less that 50cm.

    Can we say Q/r^2= 200v and Q/(r^2 - 50^2) = 40v?

    Then we can find Q and substitute it in to one of the equations so find r?
     
  11. Jun 14, 2012 #10
    Yes!! :approve:

    You didn't pay attention to this post..

     
  12. Jun 14, 2012 #11
    Yes my mistake. Thanks for your help. Was slightly confused at first but now makes sense.

    Very much appreciated
     
  13. Jun 14, 2012 #12
    Glad you figured it out. :smile:
     
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