Why integrate at these points? (Electric potential of a sphere)

[tappling]

Homework Statement

Find the potential inside and outside a uniformly charged solid sphere of radius R and total charge q.

Homework Equations

V(r) = -∫E dl

The Attempt at a Solution

I just have a question about finding the potential inside the sphere. Why integrate from infinity to the surface of the sphere (infinity to R) and add the integral of inside the sphere (R to r, whatever radius is inside the sphere). I'm just having trouble visualizing this integration, and why the integral is structured the way it is. Thanks in advance!

 

[TSny]

This is tied to the question: Where are you choosing V = 0?

Remember, when you integrate -∫E[itex]\cdot[/itex]dl between two points, you get the potential difference between those two points. So, if you want the potential at a point P, you can integrate from the point of zero potential to the point P.

 

[tappling]

Thanks for the reply!

So, if I'm interpreting this correctly, since I choose V=0 to be at infinity (the potential goes to 0 really, really far away from the charged sphere), I can integrate from infinity to point P for the potential difference. If point P is outside the sphere, I'm finding the potential difference outside the sphere. To find the potential inside the sphere, I would need to find the potential outside to radius R AND from radius R to some arbitrary r inside the sphere, because the potential outside the sphere still affects the inside potential?

 

[TSny]

To find the potential at a point inside the sphere, you need to integrate E from infinity to the point inside the sphere. Since the mathematical expression for E outside is different from the expression for E inside the sphere, you have to break up the integration into two parts.

 

[Nickie]

Integrating at these points allows us to consider the potential at any arbitrary point within the sphere. By integrating from infinity to the surface of the sphere, we are taking into account the potential due to the entire charge distribution of the sphere. This is necessary because the potential at any point within the sphere is influenced by the charge distribution both inside and outside of that point. By also integrating from the surface of the sphere to a specific radius within the sphere, we are able to isolate and calculate the potential due to the charge distribution within the sphere at that specific point. This integration method allows us to accurately determine the potential at any point within the sphere.