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Problem on Potential of Sphere

  • Thread starter echau
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  • #1
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Hi,

Could somebody please help me on a problem regarding Potential?

Calculate the potential inside and outside a sphere of radius R and charge Q,in which the charge is distributed uniformly throughout. [Hine: The additive constant for the potential inside the charged sphere must be chose so that the two potentials inside and outside, agree at r=R]

So far, I know that V(outside)=Q/(4*PI*e0*r) when r>R.

How would I find the V(outside) when r<R?
The answer given in the back of the book is V=[Q/(8*PI*e0*R)][3-(r/R)^2]
I have no idea how to obtain it.

Also how would I go about finding the V(inside) for when r>R, and r<R?

Any help would be greatly appreciated!
 

Answers and Replies

  • #2
Physics Monkey
Science Advisor
Homework Helper
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You seem to be a bit confused. The potential "outside" refers to r>R while the potential "inside" refers to r<R. Your outside potential looks good. In order to find the inside potential perhaps you should begin by calculating the electric field inside (r<R) ...
 
  • #3
mukundpa
Homework Helper
524
3
The potential due to a thin spherical shell inside the shell is same as that on the surface.

Consider a point P at distance r (< R) from the center of the sphere. The sphere can be considered as a solid sphere of radius r, and a hollow sphere of inner radius r and outer radius R. Potential at point P is due to both.
1 Due to inner sphere and
2 due to outer part.

For first calculate charge and solve as on sphere of radius r.
For second consider a thin shell of radius x and thickness dx (R > x > r) calculate charge dq on the shell, potential at P due to this charge is dq/4*pi*e0*x
Integrate this for x = r to x = R
add the two potentials.

Hope this much is sufficient.
 
  • #4
18
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okay so the Electric Field inside is = (Q/4*PI*e0)(r/R^3). to find Potential, what would i do?

After i find this potential, i add it to the potential outside to get the total?

Thanks for all the input guys!
 
  • #5
18
0
ah! nevermind. i got it!

first i had to find the electric field inside and integrate it, since dV=-E ds

I just integrate it from the interval [R,r] and that gives me the answer!

Thanks for all the help!
 

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