Electric potential for a sphere of charge

[peaceandlove]

Homework Statement

A sphere with radius 65 cm has its center at the origin. Equal charges of 2 µC are placed at 72 degree intervals along the equator of the sphere. The Coulomb constant is 8.99×10 N·m^2 / C^2. (a) What is the electric potential at the origin? Answer in units of kV. (b) What is the electric potential at the north pole? Answer in units of kV.

Homework Equations

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The Attempt at a Solution

The capacitance of an isolated sphere is C = 4*π*ε0*R = (1/K)*R where K = electric constant = 8.99*10^9 N-m²/C². C = 7.23*10^-11 F. The total charge on the sphere is 2*10^-6*(360/72) = 10*10^-6 C. It doesn't matter where the charges are, the conducting sphere will have the same potential at all points on the surface and inside the sphere. V = Q/C = 10*10^-6 / 7.23*10^-11 = 1.38*10^5 V = 138 kV.

The "north pole" would be at the "top" of the sphere, the equator being defined by the location of the charges. Of course, there is no distinction between "top" and "bottom", but it doesn't matter, since the potential is the same everywhere on the sphere. However, my answer got marked wrong for part (b), and I'm not sure why.

 

[turin]

I don't think that the sphere is conducting. In fact, I think that you whould basically ignore the sphere, and simply consider the ring of charges. The problem is worded strangely.

 

[peaceandlove]

Could you please explain how you would solve for (b)? Since there are 2 µC at 72 degree intervals along the equation, I tried dividing 2 µC by C, but that is wrong as well.

 

[diazona]

Are you familiar with this equation?
$$V = \frac{1}{4\pi\epsilon_0}\sum_i\frac{q_i}{r_i}$$
If not... you have some catching up to do