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Why V=max inside a charged sphere?

  1. Jan 15, 2016 #1
    Inside a charged sphere, the electric field strength is zero because there are electrostatic forces from all directions. I can't understand why electric potential inside the charged sphere is maximum ?
  2. jcsd
  3. Jan 15, 2016 #2


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    Hello hello :welcome:

    This is a conducting sphere I take it ?
    ##V## is constant if ##\vec E## is zero.
  4. Jan 17, 2016 #3
    E is zero inside a conducting shell, not sphere.
  5. Jan 17, 2016 #4
    Is this a uniformly charged sphere or or a sphere changed on the surface? For a uniformly charged sphere the e-field is not zero inside the sphere. For a sphere with only a charged surface the e-field is zero everywhere inside the sphere.

    The potential depends on what case you are talking about. If you can find the e-field you can find the potential or potential difference between two points. If the e-field is zero the potential is zero as well since potential is the path integral of the e-field in moving between two points.
  6. Jan 17, 2016 #5
    If E is zero V is not only constant but it is also zero.
  7. Jan 17, 2016 #6


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    That is not right (and BvU is correct). V is a potential so you can add an arbitrary constant to it everywhere and the point of zero potential is arbitrary. Using the most common convention for the hollow sphere, we'd say that V is zero at infinity, it is negative and obeys the ##-1/r## rule in the region outside the shell, and is constant and negative inside the shell.

    What you can say is that if E is the gradient of V so if E is zero then the V is constant at that point.
  8. Jan 17, 2016 #7
    Very true. I was thinking more along the lines of potential difference, but it is clear we are talking about potential...not potential difference. And potential difference is meaningless when the two points of interest are not specified. I should have known better.

    As mentioned, if the electric field is zero inside the shell then the dV/dr is zero, implying that V(r) is constant inside the shell. However, it is arbitrary what you choose the value of the potential to be.

    Thanks Nugatory.
  9. Jan 18, 2016 #8
    Thank you very much !

    V remains constant since E is zero and there's no change in E-field (dV/dx = 0) inside the shell ?
  10. Jan 18, 2016 #9


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    Again, for a conducting shell (or conducting sphere): yes.
    You will learn (or may have learned already) about the Gauss theorem that helps a lot with spheres that are non-conducting and uniformly charged.

    (And then you find there is a lot of difference between conducting and non-conducting. Your post #1 wasn't clear about which of the two, so some confusion occurred).
  11. Jan 18, 2016 #10
    Sphere is correct.
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