Potential of a uniformyl charged sphere

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Homework Help Overview

The problem involves finding the electric potential due to a uniformly charged sphere at a point inside the sphere, specifically at a distance r from the center where r is less than the radius R of the sphere.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the potential by integrating the electric field inside the sphere and expresses concern about a discrepancy with the provided answer. Some participants question the simplification of the resulting expression.

Discussion Status

The discussion is ongoing, with participants exploring the integration process and the simplification of expressions. There is no explicit consensus on the correct approach or final answer, but guidance on simplification has been offered.

Contextual Notes

Participants are working under the constraints of the problem statement and are questioning the assumptions made during the integration process. There is an indication that the original poster may have misunderstood the simplification of their expression.

ehabmozart
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Homework Statement



The question asks to find the potential due to a uniformly charged sphere at r<R where R is the radius.


Homework Equations



Va-Vb= ∫ E dr from a → b

The Attempt at a Solution



My attempt was like this

Va is to be a the point of interest r.. Vb to be R... We know Vb to be KQ/R... Now E inside the sphere is KQr/R^3 .. After integration with respect to r we get (R^2-r^2)*Kq/2R^3 ... Thus the net potential at Va is Vb + the integral which is KQ/R + (R^2-r^2)*Kq/2R^3 ... However, this isn't the answer given.. Where was my mistake??
 
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How does it differ from the answer given?
Note that it is possible to simplify your expression.
 
mfb said:
How does it differ from the answer given?
Note that it is possible to simplify your expression.

I am not able to simplify.. Can you help me?
 
Just use (a-b)/c = a/c - b/c for (R^2-r^2)*Kq/(2R^3).
 

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