At what distance has spheres decreased Electric potential by 700V?

[mr_coffee]

Hello everyone, I managed to get Parts (a) and (b) of this problem but can't get the last for some reason. A metal sphere of radius 15cm has a net charge of 2.0E-8C. THe question is, at what distance from the sphere's surface has the electric potential decrased by 700V? Well i found the electric field to be 7992.7 N/C at the sphere's surface. I also found If V = 0 at infinity, what is the electric potenital at the sphere's surface? I used Vsurface = kQ/(r). to find Vsurface being equal to 1198.9V. But I can't figure this part out. I pluged in 700V for Vsurface and solved for r, and got .2569m. r = kQ/Vsurface; This was wrong So i tried, $$Vf - Vi = -\int E.ds$$ Plugged 700V for Vf-Vi and solved for ds, because i know E from above and I know Vf-Vi, and i got .584m, also wrong any ideas? thanks

 

[Doc Al]

You know the potential at the surface from $V = k Q /r$. What's the potential at the point of interest? It's 1198.9V - 700V= 498.9V. Plug that into the formula and solve for the new distance.

 

[mr_coffee]

hm...that looked like it would work perfectly but its still wrong for some reason... I got .361m, r = [(9.0E9)(2.0E-8)]/(498.9)

 

[Doc Al]

That's the distance from the center; what they want is the distance from the surface.