Finding electric potential between two concentric spheres.

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SUMMARY

The discussion focuses on calculating the electric potential difference between the center of a spherical shell with inner radius A and outer radius 3A, which has a uniform charge density (ρ0). The correct expression for the potential difference is derived as -(ρ0A²)/(3ε0). Participants highlight the importance of distinguishing between the electric field of a uniformly charged shell and that of a solid sphere, emphasizing the application of Gauss' law to determine the electric field inside the shell.

PREREQUISITES
  • Understanding of electric potential and electric fields
  • Familiarity with Gauss' law
  • Knowledge of spherical coordinates in electrostatics
  • Basic calculus for integration
NEXT STEPS
  • Review Gauss' law applications in electrostatics
  • Study the differences between electric fields of charged shells and solid spheres
  • Practice problems involving electric potential calculations
  • Explore the concept of charge density and its effects on electric fields
USEFUL FOR

Students studying electrostatics, physics educators, and anyone seeking to deepen their understanding of electric potential in spherical charge distributions.

epicrux
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Homework Statement


A spherical shell with inner radius A and outer radius 3A which has a uniform charge density, i.e charge per unit volume, p0. Find difference in electric potential between the center of the shell and a point a distance 2A from the center.

Homework Equations


The answer given is. -(p0A2)/3ε0

The Attempt at a Solution


I am completely lost as to what I'm supposed to do here. I would think you'd find the potential at 2A and subtract it from that at the center. But how exactly do you do that?
I integrated the elctric field due to a charged sphere along 2A to A. So that makes:
∫E dr= ∫(ρr)/3ε0)dr=V
That makes:

[(ρr2)/6ε)] from 2A to A which does not equal the answer.
I think I'm stumped here. Any help would be greatly appreciated!
 
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Your expression for the electric field does not look correct. E for a uniformly charged shell is different than E for a uniformly charged solid sphere.
 
I was not aware of the difference. I'll look it up.
 
epicrux said:
I'll look it up.
Or, apply Gauss' law to find E inside the shell.
 

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