1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook