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How to calculate the electric potential inside of a sphere

  • Thread starter Vosegus
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  • #1
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Homework Statement


The density of the charge inside of the sphere is given-
ro=a*r+b/r
The electric potential on the outer layer of the sphere is phi=0
The radius of the sphere is 58.4m
r-is the distance from the center of the sphere
What is the electric potential when r=15.4m?

Homework Equations




The Attempt at a Solution


Basically, what I tried to do the to do the field equation and integrate it between 58.4 and 15.4.
That didnt work so im clueless to what I need to do.
4*pi*r^2*E=(4*pi*R^3)ro/3epsilon
 

Answers and Replies

  • #2
54
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Like potential, the Electric field also varies with the distance from center. Did you get your electric field right? Your integration parameters seem to be fine, it may be possible that your Electric field is not right.
 
Last edited:
  • #3
rude man
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Homework Statement


The density of the charge inside of the sphere is given-
ro=a*r+b/r
The electric potential on the outer layer of the sphere is phi=0
The radius of the sphere is 58.4m
r-is the distance from the center of the sphere
What is the electric potential when r=15.4m?

Homework Equations




The Attempt at a Solution


Basically, what I tried to do the to do the field equation and integrate it between 58.4 and 15.4.
That didn't work so i'm clueless to what I need to do.
4*pi*r^2*E=(4*pi*R^3)ro/3epsilon
What is R? This equation makes no sense.
You have to do two integrations:
the first is to get the charge within a given r to get E(r), along the lines of your equation but fixed up;
then you integrate E(r) from the outer surface to r = 15.4m.
Watch your signs.
 
  • #4
rude man
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Just FYI this problem can also be solved by solving the Poisson equation. In this case it becomes the second-order Euler-Cauchy equation. Boundary conditions are the potential and the E field at the surface. Just thought I'd throw it out in case you go on to advanced methods (actually not that advanced, let's call them "alternative approaches" though the math is more demanding than what you need to solve this problem your way.
 

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