Electric potential: point charge in a hollow charged conductor

[misa]

[solved] electric potential: point charge in a hollow charged conductor

Homework Statement

A hollow spherical conductor, carrying a net charge +Q, has inner radius r₁ and outer radius r₂ = 2r₁. At the center of the sphere is a point charge +Q/2.

d) Determine the potential as a function of r for 0 < r < r₁.

Homework Equations

(π = pi)
For r > r₂, the electric field is (3Q)/(8πε₀r²).
For r₁ < r < r₂, the electric field is 0 (ie, field inside conductor is zero in static situations).
For 0 < r < r₁, the electric field is Q/(8πε₀r²).


The potential as a function of r for r > r₂ (where voltage is taken to be 0 when r is infinite) is (3Q)/(8πε₀r).

The potential as a function of r for r₁ < r < r₂ is (3Q)/(16πε₀r₁).

The Attempt at a Solution

My first instinct was to add the potential (3Q)/(16πε₀r₁) to Q/(8πε₀r), which is the potential from infinity to r if the shell wasn't present. However, the answer is wrong. I also made many other fruitless attempts at this problem, but none of them very logical.

Can someone tell me what I'm doing wrong, and how to find this potential when r is within the cavity of the conductor?

I would greatly appreciate your help with this problem (and thank you in advance)!

 

[gabbagabbahey]

When in doubt, go back to the definition of the potential at a point \vec{r}:

V(\vec{r})=-\int_{\infty}^{\vec{r}}\vec{E}\cdot\vec{dl}

The electric field has different values in different regions, so you break the integral into pieces:

V(\vec{r})=-\int_{\infty}^{\vec{r}}\vec{E}\cdot\vec{dl}=-\int_{\infty}^{r_2}\vec{E}_{\text{outside}}\cdot\vec{dl}+-\int_{r_2}^{r_1}\vec{E}_{\text{between}}\cdot\vec{dl}+-\int_{r_1}^{\vec{r}}\vec{E}_{\text{inside}}\cdot\vec{dl}

 

[misa]

So are you saying that I should add all three potentials like this?
ie, (3Q)/(16πε₀r₁) + [Q/(8πε₀)](1/r - 1/r₁)

-I realize I derived the potential from r to r₁ incorrectly.

-Also, I didn't include the potential in the region from r₂ to infinity when I first solved the problem because I thought that (3Q)/(16πε₀r₁) basically came from the total potential from infinity to r₁ (ie, potential is constant in that region aka the integral representing the change in potential is zero, so potential anywhere in r₂ to r₁ is just the potential as we approach r₂). Am I right in assuming this or do I need to add (3Q)/(16πε₀r) to the sum of the integrals too?

 

[gabbagabbahey]

Well, what does the middle integral actually add to the potential inside r<r1 ?

 

[misa]

No, it doesn't add to the integral because E is 0 in that region.

Okay, thank you for your help. :)