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

[mr_coffee]

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 \delta = Q/A; Then using E = \delta/Eo; Which was correct.

But for (b) I Saw Vf - Vi = -\int E.ds.
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?

 

[lightgrav]

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 !

 

[mr_coffee]

so my answer is right i just have to make it positive? 1198.905 V/>

 

[James R]

V=Ed only works if E is constant. Outside a sphere, E is not constant, but drops off as 1/r^2. The field outside a charged sphere is the same as the field of a point charge located at the centre of the sphere.

 

[mr_coffee]

Well a E field at the center of a sphere is 0 isn't it?

 

[whozum]

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.