Homework Help: Determining Electric Potential with Charge Density

1. Jun 3, 2017

an_single_egg

1. The problem statement, all variables and given/known data
A nonconducting sphere of radius r0 carries a total charge Q. The charge density ρE increases as the square of the distance from the center of the sphere, and ρE=0 at the center.

a) Determine the electric potential as a function of the distance r from the center of the sphere for r > r0. Take V=0 for r=∞.

b) Determine the electric potential as a function of the distance r from the center of the sphere for r < r0.

2. Relevant equations

Vb-Va = -∫E⋅dl

3. The attempt at a solution

a) Because it is a sphere with total charge Q,
E=kQ/r2
From ∞ to r,
-∫kQ/r2 dr
=-kQ * [-1/r](∞→r)
= kQ/r

Part a I think I understand OK.

b) I don't know where to begin here really. Because r<r0, I can't use E=kQ/r^2.
The question (I think) is saying that Q isn't distributed evenly.
So would ρE = dQ/dV?
How do I find E with a changing charge AND changing radius?

Let me know if any of my formatting is weird, this is only my second time trying to post something.

2. Jun 3, 2017

diredragon

It's not really that complicated as $ρ_e$ is dependent on radius and your changing the radius so you have in total one thing that changes. For your part a you took your $Q$ for granted and that's why you were confused with the second part where they ask of you to calculate the partial sphere.
Let me give you a hint. Your total charge: $Q_t = \int_0^{r_0} ρ_e \, dV$ and your $dV$ is simply a incremental of the sphere's volume which is the surface of the sphere times a little increment in your radius. Algebraically $dV = 4 \pi r^2 dr$. You can check that this is true if you integrate $dV$ from 0 to some $r$ to get the volume of the sphere which i leave to you to find or calculate by the way described.

In the Image above you can see a simpler version of your sphere which is a circle where we look for an increment in the surface. You go one dimension down and multiply the length of the circle and the small increment in radius and get the small surface increment. In your problem you do the same exept that everything is a dimension up and you multiply the surface with the increment to get the increment in the volume.
How this correlates to your problem? You have to find the total charge in your partial sphere so that you can put that $Q_p$ (function that depends on the radius) in your original integral equation for the potential ( not the one you derived at at a) ) and integrate from 0 to some radius which is constant set to $r_s$. I'm sure you can do this as you know the integral function for $Q_t$ as simply $Q_t = \int_0^{r_0} ρ_e \, dV$. What's different? You're nt integrating the full sphere. Just some part. Try it :)

3. Jun 3, 2017

an_single_egg

Ok, so:

Q = (0→r)∫ρEdV
=4/3ρEπr3

Then,
E⋅∫dA = Qenc0
E(4πr2)=(4ρEπr3)/(3ε0)
E=(ρEr3)/(3r2ε0)
E=(ρEr)/3ε0

And then,

Vba = -(r→r0)∫(ρEr)dr/3ε0
Vba = -ρE/6ε0 * (r02-r2)

Is that right?

4. Jun 3, 2017

diredragon

The left side of the Gauss's law is fine. E is constant at distance $r$ from the center but the right side is all wrong. You have to determine the ammount of Q you take in by closing it with a Gaussian surface of radius r. Now, look at your expression for Q on the right involving the charge density. You basically integrated from 0 to r and said that the charge density is constant along the integration. Doesn't the problem specifically say that it depends on the 1/r^2.
What that means is that the density is some function dependent on the square of the radius. So generally something like k/r^2 where k is a constant determined by the initial conditions. The initial condition states that charge Q is present when r = r0 so your k is (Qr0^2) You have to use this expression for density when you integrate from 0 to r because the density is not constant as you assumed it was in the reply. Does this make things clearer?
Also check out https://www.physicsforums.com/help/latexhelp/ for some LaTeX information. Some pretty useful information is to be found there :)

5. Jun 3, 2017

SammyS

Staff Emeritus
The problem actually states the following.
Thus, the charge density is proportional to r2 .

OP should use ρ = k⋅r2, where k is chosen so that $\displaystyle \ \int_\text{Sphere} {k\cdot r^2}\, dV =Q\,.$

6. Jun 3, 2017

diredragon

Oh yeah, i thought it said inverse proportional like the electric field, my bad xD.

7. Jun 3, 2017

an_single_egg

Sorry, I guess I didn't fully understand what that portion of the problem meant. I also realized I'm supposed to be solving in terms of Q, r, and r_0 and constants, so I'll be trying to solve it correctly now, oops!

Does this still apply?

If I use that integration, then $$Q = \int_0^r kr^2 dV$$ $$Q = \frac 4 5 k \pi r^5$$
Right? Is the rest of my process okay once I plug in for Q (and then I guess I'd need to solve for it one I got an answer, so I wouldn't be left with a k)?

(Thanks for the LaTeX guide :) )

8. Jun 3, 2017

SammyS

Staff Emeritus
That's almost correct.

If r < r0 , then that works for Q enclosed .

If r = r0 , then that works for Q, the total charge on the sphere. Use this to find the value of k in terms of Q and r0 .

9. Jun 3, 2017

an_single_egg

Just solving for k,

$$k = \frac {5Q} {4\pi r^5}$$

Or did you mean something in my integral wasn't right? I'm a little confused how to get k in terms of $Q$ and $r_0$ , if the integral is from $0$ to $r$. Do I need to be taking it from $r$ to $r_0$ ?

I'm really sorry, I'm just trying to make sure I've got everything right, this has been my problem of the day, so to speak.

10. Jun 3, 2017

SammyS

Staff Emeritus
That should be $\displaystyle \ k = \frac {5Q} {4\pi r_0^5}\,.$

11. Jun 3, 2017

an_single_egg

My book is telling me that the answer for part b is $$V = \frac {Q} {16 \pi \epsilon_0 r_0} (5 - \frac {r^4} {r_0^4})$$

If I plug in $\frac {5Q} {4 \pi r_0^5}$ for k, I'm not left with anything to give me that answer... I am definitely missing something but don't know what. :(

12. Jun 3, 2017

SammyS

Staff Emeritus
What are you plugging in and what are you plugging it into ?

13. Jun 3, 2017

SammyS

Staff Emeritus
By the way, it's also true that $\displaystyle \ k = \frac {5Q_\text{Enclosed }} {4\pi r^5}\,.$

14. Jun 4, 2017

an_single_egg

I'm trying to plug into Gauss's Law, $$E \int dA = \frac {Q_{enc}} {\epsilon_0}$$
$$E \cdot (4\pi r^2) = \frac {4k \pi r^5} {5\epsilon_0}$$

And then solving for $E,$ I tried to find $V_{ba}$ by integrating E from $r \rightarrow r_0$, but that's not correct.

I think this is confusing me also. If $k$ can be equal to something in terms of both $r$ and $r_0$, how do I put that into an equation?

15. Jun 4, 2017

SammyS

Staff Emeritus
But previously, you found that $\displaystyle \ k = \frac {5Q} {4\pi r_0^5}\,,\$ which only has constant factors.

Plug that into $\displaystyle \ E \cdot (4\pi r^2) = \frac {4k \pi r^5} {5\epsilon_0} \$ and solve for $\ E\,.$

16. Jun 4, 2017

an_single_egg

Ok, I think I've got it:

$E = \frac {Qr^3} {4\pi \epsilon_0 r_0^5}$
$V_{ba} = \int_r^{r_0} \frac {Qr^3} {4\pi \epsilon_0 r_0^5}$
$= \frac {Q} {16\pi \epsilon_0 r_0^5} (r_0^4 - r^4)$
$= \frac {Q} {16 \pi \epsilon_0 r_0} (1 - \frac {r^4} {r_0^4})$

Now... I know that 1 should be a 5... but I'm close :)

Edit: I am plugging in $k$ in terms of $r_0$ because $k$ needs to be a constant, correct?

17. Jun 4, 2017

SammyS

Staff Emeritus
You're plugging in for k, because you need the answer in terms of the given quantities, Q and r0 .

The result, Vba that you have for potential looks like the potential at r, relative to the potential at r0 .

What is the potential at r0 relative to the potential very far from the sphere (at infinite distance).

Last edited: Jun 4, 2017
18. Jun 4, 2017

an_single_egg

Why am I thinking about infinite distances for part b? If $r_0$ is a finite radius and r<$r_0$, isn't everything contained within the sphere?

The potential for any $r$ > $r_0$ is $\frac {KQ} {r^2}$.

19. Jun 4, 2017

SammyS

Staff Emeritus
Where Is the potential considered to be zero? It's considered to be zero at r = ∞. You inherit this from part (a) and can use part a to determine the potential at r = r0 .