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Finding potential at the center of metal sphere 
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#55
Mar1412, 02:51 AM

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I think, you convinced me about the validity of the solution with the image charge method. ehild 


#56
Mar1412, 02:58 AM

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The potential function we use in calculations must be distinguished from the real potential. Infinite line, infinite plane do not exist, but the potential can be approximated near them with a function that do not tend to zero at infinity. ehild 


#57
Mar1412, 03:07 AM

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Like I said  one must be careful with infinities. For all finite cases, kq/r holds, I believe. 


#58
Mar1512, 06:59 PM

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As I think about this problem, I think the image approach will work even in situations where
the inner spherical cavity of radius [itex]R_1[/itex] is not centered at the center of the whole sphere. As long as we have a spherical cavity within a metallic sphere, the above image approach will work. So after all the discussion on this problem, we can generalise the situation and we can think of applying this to any spherical cavity within any metallic sphere. 


#59
Mar1712, 07:05 AM

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rude man or ehild, can you confirm my opinion in post # 58



#60
Mar1712, 08:25 AM

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I have confirmed that the image charge method can be applied for a cavity in post #55.
Find the potential at an arbitrary point when the charge q is very near to the centre of the cavity and see the limit a >0. You can get the potential easily in case of a central charge. Check if the potentials are equal. ehild 


#61
Mar1712, 08:32 AM

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ehild, in post # 58, I am talking about the inner spherical cavity which is not necessarily concentric with the outer spherical shell. You were replying to different question in post # 55. I just generalised that in post # 58.



#62
Mar1712, 09:45 AM

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I see. How do you get the potential at the outer surface then?
ehild 


#63
Mar1712, 10:17 AM

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Ok
Irrespective of the shape of the cavity inside the metallic sphere, if we place the charge ,q, inside the cavity, then equal charge q is distributed uniformly on the outer spherical surface. The cavity and the outside are different worlds altogether as long as we are talking about electrostatic case. So potential on the outer surface would just be [itex]\frac{1}{4\pi\epsilon_o}\frac{q}{R_2}[/itex] 


#64
Mar1712, 11:17 AM

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As for your earlier query: as long as you can wrap a spherical Gaussian surface around the cavity and fully contained in metal, your statement must be true, otherwise the E field could not be zero everywhere about the surface. So it makes no difference what the shape of the metal is  it can be quite irregularly shaped  the charges in the cavity have to be distrtibuted so as to null the net field at every point on the Gaussian shell just beyond R1, just as in the concentric case. So the image would not change from the concentric c ase. 


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