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What is the electric potential at the sphere's surface? having troubles!

  1. Sep 25, 2005 #1
    Hello everyone, I'm having troulbes figuring this one out. I found an equation in the book, tried it, but it was wrong. The problem is the following:
    A metal sphere of radius 15 cm has a net charge of 2.0 10-8 C.

    (a) What is the electric field at the sphere's surface?
    7992.7 N/C
    (b) If V = 0 at infinity, what is the electric potential at the sphere's surface?
    (c) At what distance from the sphere's surface has the electric potential decreased by 700 V?

    I found part (a) by finding [tex]\delta[/tex] = Q/A; Then using E = [tex]\delta[/tex]/Eo; Which was correct.

    But for (b) I Saw [tex]Vf - Vi = -\int E.ds.[/tex]
    So Vf- Vi is electric potential. So i tried:
    V = -E*d = -(7992.7N/C)(.15) = -1198.905 V but was wrong, any ideas why?
  2. jcsd
  3. Sep 25, 2005 #2


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    yea, it's a positive charge.

    Recall that the E outside the shell
    is the same as if it was a point Q,
    so V = kQ/r just as with a point charge.
    . . . V has same sign as the Q !
  4. Sep 25, 2005 #3
    so my answer is right i just have to make it positive? 1198.905 V/>
  5. Sep 26, 2005 #4

    James R

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    Gold Member

    V=Ed only works if E is constant. Outside a sphere, E is not constant, but drops off as [itex]1/r^2[/itex]. The field outside a charged sphere is the same as the field of a point charge located at the centre of the sphere.
  6. Sep 26, 2005 #5
    Well a E field at the center of a sphere is 0 isn't it?
  7. Sep 26, 2005 #6
    Right but for a closed sphere the E field (OUTSIDE the sphere) is the same as that of a point charge. inside the sphere its zero by gauss' laws.
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