Potential difference inside a spherical shell

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The discussion focuses on calculating the potential difference between the north pole and the center of a uniformly charged spherical shell. The user attempts to apply Gauss's law to find the electric field (E) by integrating the surface charge density (sigma) over the shell's surface area. They express confusion about the left-hand side of their equation and whether to integrate E or the potential directly. The user concludes that E is uniform along the surface, leading to a simplified expression for E as sigma/epsilon0. The thread highlights the complexities involved in deriving the potential difference using both electric field and direct potential integration methods.
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Homework Statement



A top half of a spherical shell has radius R and uniform charge density sigma. Find the potential difference V(b)-V(a) between point b at the north pole, and point a at the center of the sphere.

Homework Equations





The Attempt at a Solution



\oint E ds = \frac{\int \sigma ds}{\epsilon_{0}}

I integrated sigma ds from 0 to 2pi and 0 to pi/2, and got the total surface area as 2pi r^2. Now I want to solve for E, and then use -\int E dl to give me the potential difference, but what does the left hand side of my original equation equal?

E is uniform along the surface, so I can pull it out of the integral and be left with the integral of ds. But wouldn't that be 2pir^2 as well? I'd be left with E=sigma/epsilon0
 
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Would it be easier to integrate the potential directly?

V(r) = C * ∫ σ(r')*da'/[r - r']
 
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