Potential difference inside a spherical shell

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SUMMARY

The discussion focuses on calculating the potential difference \( V(b) - V(a) \) between the north pole and the center of a spherical shell with radius \( R \) and uniform charge density \( \sigma \). The user applies Gauss's Law, represented by the equation \( \oint E \, ds = \frac{\int \sigma \, ds}{\epsilon_{0}} \), to derive the electric field \( E \). The user concludes that \( E \) can be expressed as \( E = \frac{\sigma}{\epsilon_{0}} \) due to the uniformity of the electric field along the surface, and considers integrating the potential directly using the equation \( V(r) = C \int \frac{\sigma(r') \, da'}{r - r'} \).

PREREQUISITES
  • Understanding of Gauss's Law and its application in electrostatics
  • Familiarity with electric field concepts and uniform charge distributions
  • Knowledge of potential energy and potential difference in electrostatics
  • Basic calculus for evaluating integrals
NEXT STEPS
  • Study the application of Gauss's Law in various geometries, particularly spherical shells
  • Learn about electric field calculations for uniformly charged surfaces
  • Explore the derivation and application of electric potential from electric fields
  • Investigate the implications of potential differences in electrostatic systems
USEFUL FOR

Students studying electrostatics, physics educators, and anyone interested in understanding electric fields and potentials in spherical geometries.

warfreak131
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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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