Electric Potential V inside nonconducting sphere with cavity

[WK95]

Homework Statement

A nonconducting sphere of radius r_2 contains a concentric spherical cavity of radius r_1. The material between r_1 and r_2 carries a uniform charge density rho_E(C/m^3). Determine the electric potential V, relative to V=0 at r= infinity, as a function of the distance r from the center for r_1 < r < r_2

Homework Equations

The Attempt at a Solution

http://i.imgur.com/bOMD7ej.png

The link is just a screenshot of the answer and steps given in the solution manual. I have similar work but I am confused about why the solution involves integrating from r_2 to r. How is that the potential relative to V=0 at r=infinity?

And what happened to V_R_2? Since it disappears later on, I assume it's because it's equal to 0V but how come?

 

[ehild]

The link is just a screenshot of the answer and steps given in the solution manual. I have similar work but I am confused about why the solution involves integrating from r_2 to r. How is that the potential relative to V=0 at r=infinity?

And what happened to V_R_2? Since it disappears later on, I assume it's because it's equal to 0V but how come?

You know that the negative gradient of the potential is equal to the electric field. Getting the potential function, find the electric field at the different domains. You get the potential at a point r if integrating between r and a point where you know the potential. Do you know the potential at r2, the radius of the outer surface of the shell? What is the electric field outside the shell?

 

[haruspex]

the solution involves integrating from r_2 to r. How is that the potential relative to V=0 at r=infinity?

It isn't. The potential at r is that at r₂ minus the integral from r to r₂. The potential at infinity is the same as it would be at r if the body were uncharged, or if r₁ = r = r₂.

And what happened to V_R_2?

It was evaluated as (something like) ##\frac{4\pi}3\left(r_2^2-\frac{r_1^3}{r_2}\right)\frac{\rho}{4\pi\epsilon_0}## and combined with the result of the integral. Write out the steps and you'll see the result is correct.