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

  • Thread starter mr_coffee
  • Start date
  • #1
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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?
V
(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?
 

Answers and Replies

  • #2
lightgrav
Homework Helper
1,248
30
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 !
 
  • #3
1,629
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so my answer is right i just have to make it positive? 1198.905 V/>
 
  • #4
James R
Science Advisor
Homework Helper
Gold Member
600
15
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.
 
  • #5
1,629
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Well a E field at the center of a sphere is 0 isn't it?
 
  • #6
2,209
1
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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