E&M - calculating the energy stored in a sphere w/ nonuniform charge density

[mooeypoo]

I am so confused, specially (AGAIN!!) about my vectors in this equations..

 

The question:

 

A solid sphere has a charge density

[math]\rho (r, \theta , \phi ) = kr \text{ for } 0 <= r < R[/math]

Find the energy stored in this electrostatic this electrostatic configuration by integrating

[math]\frac{1}{2} \int d^3 r \rho V[/math]

(V is potential).

 

The way I understood it, the energy stored in this configuration is the one involving the potential inside + the one involving potential outside of this sphere. They're both, obviously, different.

 

So, I tried to see if I can first define the two Vs (in and out) and then integrate the W equation. For the potential inside the sphere, I created a "gaussian surface" with r smaller than R, here's what I did:

 

[math]

V® = \int_{0}^{r} 4\pi \left( \frac{1}{3\pi \epsilon} \right) \frac{kr'}{r'} r'^2 dr' = \frac{1}{\epsilon} \int_{0}^{r} kr'^2 dr = \frac{kr^3}{3\epsilon}

[/math]

 

That's the potential inside the sphere. Now, outside the sphere, I'm getting stuck, primarily because the rho outside the sphere is zero (??). but I figured I might have a chance calculating it from the other way, by figuring out the Electric Field E:

 

[math]

\int E \dot da = \frac{Qenc}{\epsilon}

[/math]

 

[math]

Qenc = \int_{0}^{R} \rho(r') d\tau ' = 4\pi k \int_{0}^{R} r' r'^2 dr = 4\pi k\frac{r'^4}{4} |_{0}^{R} = \pi k R^4

[/math]

 

[math]

da = 4\pi r'^2

[/math]

 

The electric field E is in the same direction as r, so the dot product is settled, and we have:

 

[math]

4\pi r'^2 E = \frac{2\pi kR^4}{\epsilon}

[/math]

[math]

E = \frac{kR^4}{2\epsilon r'^2}

[/math]

 

And now I find the potential:

[math]

V(r')=- \int E dr = - \int^{\infty}_{R} \frac{kR^4}{2\epsilon r'^2} dr' = -\frac{kR^4}{2\epsilon} \int^{\infty}_{R} \frac{1}{r'^2} = \frac{kR^4}{2\epsilon r'}

[/math]

 

So, I guess i have both my potentials, but now I am having trouble figuring out how I deal with \rho in my W integral..

 

[math]W = \frac{1}{2} \int d^3 r \rho \text{(inside)} V \text{(inside)} + \frac{1}{2} \int d^3 r \rho \text{(outside)} V \text{(outside)}[/math]

 

Help!

 

(1) Did I do it correctly up until now? Is that the way to "attack" the question or should I use another method? (I know of "poisson's equation" but.. err can't figure out if or how to use it here?)

(2) I don't quite have a \rho outside... what do I do???

 

 

Help me make sense of these type of questions, pleaaase!

 

~moo

[timo]

I am not sure if you really understood what the question asks: If you have two charges at some distance then there is some energy gained by seperating the to an asymptotically infinite distance (a negative value if their charges are opposite). The same holds true for three charges, four charges, ... . That is this pretty much the definition of potential energy. The same is supposed to hold true for infinitesimal charges [math]\rho \, dV[/math]. Calculating the potential energy for a set of charges [math]\rho \, dV[/math] is what the question asks for. For two charges, the potential energy is something like [math]C \frac{q_1 q_2}{\| \vec x_1 - \vec x_2 \|}[/math], with C being some suitable constant. For n charges the expression becomes [math] \sum_{i<j} C \frac{q_i q_j}{\| \vec x_i - \vec x_j \|} = \frac{1}{2} \sum_{i \neq j} C \frac{q_i q_j}{\| \vec x_i - \vec x_j \|} [/math]. In the limit of infinitesimal charges this becomes [math] \frac{1}{2} \int_V dV_1 \rho(\vec x_1) \underbrace{\int_V dV_2 C \frac{\rho(\vec x_2)}{\|\vec x_1 - \vec x_2 \|}}_{=: V(\vec x_1)} = \frac{1}{2} \int_V dV_1 \rho(\vec x_1) V(\vec x_1)[/math]; the expression you are asked to calculate.

So your first step, finding an analytic expression for V, is fine (I did not check the calculation; from my experience most students are better at rearranging equations and integrating stuff than me, anyways). However, since the charge density is zero outside the sphere, I do not understand why you bother with this volume at all - it will contribute a plain zero. I think your problem might lie in understanding what you are supposed to calculate and why so - hence my little handwaving explanation above.

[mooeypoo]

I am not sure if you really understood what the question asks: If you have two charges at some distance then there is some energy gained by seperating the to an asymptotically infinite distance (a negative value if their charges are opposite). The same holds true for three charges, four charges, ... . That is this pretty much the definition of potential energy. The same is supposed to hold true for infinitesimal charges [math]\rho \, dV[/math]. Calculating the potential energy for a set of charges [math]\rho \, dV[/math] is what the question asks for. For two charges, the potential energy is something like [math]C \frac{q_1 q_2}{\| \vec x_1 - \vec x_2 \|}[/math], with C being some suitable constant. For n charges the expression becomes [math] \sum_{i<j} C \frac{q_i q_j}{\| \vec x_i - \vec x_j \|} = \frac{1}{2} \sum_{i \neq j} C \frac{q_i q_j}{\| \vec x_i - \vec x_j \|} [/math]. In the limit of infinitesimal charges this becomes [math] \frac{1}{2} \int_V dV_1 \rho(\vec x_1) \underbrace{\int_V dV_2 C \frac{\rho(\vec x_2)}{\|\vec x_1 - \vec x_2 \|}}_{=: V(\vec x_1)} = \frac{1}{2} \int_V dV_1 \rho(\vec x_1) V(\vec x_1)[/math]; the expression you are asked to calculate.

So your first step, finding an analytic expression for V, is fine (I did not check the calculation; from my experience most students are better at rearranging equations and integrating stuff than me, anyways). However, since the charge density is zero outside the sphere, I do not understand why you bother with this volume at all - it will contribute a plain zero. I think your problem might lie in understanding what you are supposed to calculate and why so - hence my little handwaving explanation above.

 

Yeah, I just thought of this again.

 

My professor did this question in class using the other method (integral with E^2) , and calculated both mediums (inside and out).

 

But... I think I overthought this. In the case of using E, I *do* have a medium outside and inside... in this case I don't.. so.. there shouldn't be a need to calculate the outside.

 

I think...

 

Does that suond reasonable?

 

I'll try to solve it and post the steps and answers, see if I got this right (at least in method).

 

Thanks

 

Okay, so here are my computations, thinking that there is only one medium to calculate from:

 

[math]

V® = \frac{kr^3}{3\epsilon}

[/math]

 

[math]

W = \frac{1}{2} \int \rho ® V® d\tau = \frac{1}{2} \int (kr) \frac{kr^3}{3\epsilon} r^2 dr sin\theta d\theta d\phi = \frac{4\pi k^2}{2\epsilon} \int r^6 dr = \frac{4\pi k^2 r^6}{6\epsilon}

[/math]

 

Does that look reasonable?

[timo]

- I do not understand your calculation for the potential V®: The (3 pi epsilon)^-1 probably is just a typo, so no problem with that. But I do not understand where the kr'/r' part comes from. What is the r' in the denominator?

- You are missing integration ranges in your final calculation. This causes your result to be expressed in terms of an arbitrary variable r rather than the given radius of the sphere R.

[mooeypoo]

- I do not understand your calculation for the potential V®: The (3 pi epsilon)^-1 probably is just a typo, so no problem with that. But I do not understand where the kr'/r' part comes from. What is the r' in the denominator?

 

Yes, sorry, the (3\pi\epsilon)-1 is supposed to be (4\pi\epsilon)^-1 --- sorry, that was a typo, indeed.

 

The kr'/r' comes from the calculation of E with a gaussian surface r>R

[math]

4\pi r'^2 E = \frac{2\pi kR^4}{\epsilon}

[/math]

 

but I didn't use it in my final answer (as I didn't need, I think, an "Outside" v®.

 

My final calculation was

[math]

W = \frac{1}{2} \int \rho ® V® d\tau = \frac{1}{2} \int (kr) \frac{kr^3}{3\epsilon} r^2 dr sin\theta d\theta d\phi = \frac{4\pi k^2}{2\epsilon} \int r^6 dr = \frac{4\pi k^2 r^6}{6\epsilon}

[/math]

 

Using [math]

V® = \frac{kr^3}{3\epsilon}

[/math]

 

 

- You are missing integration ranges in your final calculation. This causes your result to be expressed in terms of an arbitrary variable r rather than the given radius of the sphere R.

 

Right.. so just to make sure -- my limits are "inside" the sphere only, from R to 0, right? That would make the final answer:

 

[math]

W = \frac{4\pi k^2 R^6}{6\epsilon}

[/math]

 

((OOPS! wrong.. ))

 

 

[math]

W = \frac{1}{2} \int \rho ® V® d\tau = \frac{1}{2} \int (kr) \frac{kr^3}{3\epsilon} r^2 dr sin\theta d\theta d\phi = \frac{4\pi k^2}{6\epsilon} \int r^5 dr = \frac{\pi k^2 r^6}{9\epsilon} |_{0}^{R} = \frac{\pi k^2 R^6}{9\epsilon}

[/math]

Edited October 19, 2008 by mooeypoo
math error