Potential due to a uniformly charged sphere

[gracy]

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?

 

[Doc Al]

I don't know why are we subtracting these two?

You are actually adding two pieces: (1) change in potential from ∞ to R, (2) change in potential from R to r.

 

[gracy]

You are actually adding two pieces:

Just because
$E$=$-\frac{∂V}{∂r}$

Negative sign!

 

[Doc Al]

Negative sign!

Yep!

 

[gracy]

You are actually adding two pieces

And why adding them gives potential at r<R?

 

[Doc Al]

And why adding them gives potential at r<R?

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.

 

[gracy]

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?

 

[Doc Al]

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?

Right.

 

[gracy]

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?

 

[Doc Al]

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?

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