Finding electric potential between two concentric spheres.

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Homework Help Overview

The problem involves calculating the electric potential difference between the center of a spherical shell and a point outside the shell, specifically at a distance of 2A from the center. The shell has a uniform charge density and defined inner and outer radii.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to find the electric potential by integrating the electric field, but expresses confusion about the correct approach and the application of the electric field formula. Some participants suggest that the electric field expression used may be incorrect and recommend looking up the differences between a uniformly charged shell and a solid sphere.

Discussion Status

The discussion is ongoing, with participants exploring the correct application of Gauss' law and clarifying the differences in electric field calculations for different charge distributions. There is no explicit consensus yet, but guidance has been offered regarding the need to verify the electric field expression.

Contextual Notes

Participants are navigating the complexities of electric fields in relation to spherical charge distributions and the implications of using Gauss' law. The original poster expresses uncertainty about the integration process and the correct setup for the problem.

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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