In a solid sphere , a point is located inside . While calculating potential at that inernal point , we follow two steps . We have a sphere above which the point lies and a part of sphere below which the point lies . We calculate potential separately for both cases and add them up .

But my confusion lies in the procedure of the ways .In first case , we simply express mass by density and put in the potential formula . But in the second case , we follow calculus . We divide the sphere into infinitesimal rings and integrate them .

But my question is , why are separate methods followed ?

  • $\begingroup$ -1 Not clear what you are asking. The method of calculation does not look sensible. Can you provide an image or a link to an example in which it is used? $\endgroup$ – sammy gerbil Feb 8 '18 at 19:06
  • $\begingroup$ @NehalSamee Thank you. I now understand what is meant by "above" and "below". But I do not understand why "separate methods" are being used for the inner sphere and the outer shell. If density is not uniform then calculus must be used for the inner sphere also. As I suggested in my 1st comment, an example would be useful. $\endgroup$ – sammy gerbil Feb 10 '18 at 14:13
  • $\begingroup$ @sammygerbil...The density is uniform everywhere...(actually a proof from textbook ) ... Textbooks provide such methods...ρ is uniform... $\endgroup$ – Nehal Samee Feb 10 '18 at 15:58

Look , you need to keep in mind the shell theorems . In first case , it's done using density ρ...The point P lies on the surface of the inner subsphere.So , it's whole mass contributes to the potential on the point . This mass is 4/3*pir cubeρ...

But integration is used in the second step...Because the point P lies inside the outer subsphere.Force varies with displacement...Integration is used to add up the potential...enter image description here

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