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Potential due to a uniformly charged sphere

  1. Nov 9, 2015 #1
    http://www.phys.uri.edu/gerhard/PHY204/tsl93.pdf
    To find electrical potential at r<R
    I want to know why should we subtract
    ##V##=-##\int_∞^R\frac{kQ}{r^2}\,dr##-##\int_R^r(0)\,dr##=##\frac{kQ}{R}##
    I don't know why are we subtracting these two?
     
  2. jcsd
  3. Nov 9, 2015 #2

    Doc Al

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    Staff: Mentor

    You are actually adding two pieces: (1) change in potential from ∞ to R, (2) change in potential from R to r.
     
  4. Nov 9, 2015 #3
    Just because
    ##E##=##-\frac{∂V}{∂r}##

    Negative sign!
     
  5. Nov 9, 2015 #4

    Doc Al

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    Yep!
     
  6. Nov 9, 2015 #5
    And why adding them gives potential at r<R?
     
  7. Nov 9, 2015 #6

    Doc Al

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    The work done against the electric field (which is the potential) in going from "A" to "C" equals the work done from A to B plus the work done from B to C.
     
  8. Nov 9, 2015 #7
    potential at infinity is zero,as we move in the direction of electric field the potential increases .Moving from infinity to R gives potential at R (integration with upper and lower limits i.e final and initial positions being infinity and R),right?
     
  9. Nov 9, 2015 #8

    Doc Al

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    Right.
     
  10. Nov 9, 2015 #9
    Then Moving from R to r (initial and final points being R and r respectively)gives
    1)potential at r
    OR
    2)Change in potential while going from R to r
    which one?
     
  11. Nov 9, 2015 #10

    Doc Al

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    Number 2: The change in potential. (The potential itself is defined as work from infinity to r.)
     
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