Electric Potential Inside Uniformly Charged Sphere

  • Thread starter nabeel17
  • Start date
  • #1
nabeel17
57
1
I wanted to know how to find the electric potential inside a uniformly charged sphere of radius R. What i understand is that my textbook uses a reference point as infinity and then expresses the potential as the difference of 2 integrals.

Sooo,

V(r)= -∫E dr and the electric field is k(qr)/R^3 r is where you are from the centre of the sphere.

So the method I have seen is
. . . . r. . . . . . . . . . . . . . . . R
V = - ∫ (1/(4πεo)) qr/R³ dr - ∫ (1/(4πεo)) q/r² dr
. . . .R. . . . . . . . . . . . . . . .∞

but I was wondering if there is a way to express it as one integral from 0-R
like
. . . . R. . . . . . . . . . . . . . . .
V = - ∫ (1/(4πεo)) qr/R³ dr
. . . .o. . . . . . . . . . . . . . . .

but this of course produces a different answer. Where am I wrong in my thinking?
 

Answers and Replies

  • #2
Muphrid
834
2
Remember that the potential at any given point is not significant; only the potential difference between two points is. Does your definition produce the same potential difference between two arbitrary points?
 
  • #3
andrien
1,024
32
since the potential at ∞ is zero it is better to take reference there.if you will choose center of sphere ,then potential is not zero as a reference.
 
  • #4
nabeel17
57
1
So, is it possible to find the potential using the reference as the center (origin) of the sphere and going out to the radius?
 
  • #5
Muphrid
834
2
There is nothing seriously illegal in doing so, but some vector calculus theorems that we generally assume will hold only do so when the field falls off to zero at infinity. That doesn't stop us from considering, say, infinite constant electric fields and what that would do in some very limited cases, though. So for the most part, I see no problem. Again, the potential between two points is meaningful; the potential function itself is not.
 
  • #6
nabeel17
57
1
Oh ok, thank you for clearing that up, makes sense now. I'm just trying to set up the right integral to get the same answer if I use the centre as reference or infinity.
 

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