Mean value theorem in elelctrostatics

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The mean value theorem in electrostatics indicates that the electrostatic potential at any point in charge-free space equals the average potential over a surrounding sphere. A user expresses confusion over their derivation, finding that the potential appears independent of the surface shape and center location. They utilized Green's function and Neumann boundary conditions to arrive at their conclusion, stating V(x)=s, where s can be any surface. Other participants question the validity of this result and encourage sharing calculations for further clarification. The discussion highlights the complexities of applying the mean value theorem in electrostatics.
hyperspace
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The mean value theorem in electrostatics states that for charge free space the value of the electrostatic potential at any point is equal to the average of the potential over the surface of any sphere centered at that point.
In its derivation I'm getting a kind of strange result that is not satisfactory. What I am getting is that the potential at required point is independent of the type of surface taken (spherical or not) and that it may not be the center as well.
It would be better if you use Green's function to do this.

TIA
 
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So what are you asking exactly? (Also if you want help with your derivation, it'd be nice to see it ;-)
 
Ok, so the result I'm getting is V(x)=<V>s, where s represents the surface, of any kind actually. So there's no reference to spherical shape also it doesn't follow that x has to be the center of the sphere if s is at all a sphere. I got this using the Green's function and Neumann boundary conditions
 
hyperspace said:
Ok, so the result I'm getting is V(x)=<V>s, where s represents the surface, of any kind actually. So there's no reference to spherical shape also it doesn't follow that x has to be the center of the sphere if s is at all a sphere. I got this using the Green's function and Neumann boundary conditions

That result doesn't look right...if you post your calculations for it, we can tell you wjere you are going wrong.
 

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