Problem on Potential of Sphere

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The discussion revolves around calculating the electric potential inside and outside a uniformly charged sphere of radius R and charge Q. The potential outside the sphere is correctly identified as V(outside) = Q/(4*PI*e0*r) for r > R. To find the potential inside the sphere (r < R), the user learns to calculate the electric field first and then integrate it to determine the potential. The correct formula for the potential inside the sphere is given as V = [Q/(8*PI*e0*R)][3-(r/R)^2]. The user successfully resolves their confusion by integrating the electric field to find the potential.
echau
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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!
 
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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) ...
 
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.
 
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!
 
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!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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