# Potential of Sphere, Given Potential of Surface

1. Oct 1, 2016

### RyanTAsher

1. The problem statement, all variables and given/known data

The sphere of radius R has the potential at the surface equal to $$V_0 = \alpha sin^2(\theta) + \beta$$ where $\alpha, \beta$ are some constants. Find the potential inside, and outside the sphere.

2. Relevant equations

$$V(r,\theta) = \sum_{l=0}^{\infty}(A_l r^l + \frac {B_l} {r^{(l+1)}})P_l(cos(\theta))$$

P_l is the legendre polynomials given by

$P_0(x)=1$, $P_1(x)=x$, $P_2(x) = \frac {3x^2-1} {2}$

3. The attempt at a solution

I'm not really too sure how to solve these problems having looked through the examples, so I tried to solve using equal coefficients, this is just my work for the inside of the sphere. Getting rid of $B_l$ so it doesn't blow up inside.

$$\alpha sin^2(\theta) + \beta = \sum_{l=0}^{\infty}(A_l r^l)P_l(cos(\theta)$$

I write out enough legndre polynomials to match the highest term of the given potential, and rewrite $sin^2(\theta)$ in terms of $cos^2(\theta)$.

$$\alpha + \beta - \alpha cos^2(\theta) = A_0 + A_1rcos(\theta) + A_2r^2(\frac {3cos^2(\theta)-1} {2})$$

The A_1 term doesn't effect the potential, given the fact that it is not of the power of any of the given potential, so I get rid of it, then expand the A_2 term.

$$\alpha + \beta - \alpha cos^2(\theta) = A_0 + \frac 3 2 A_2r^cos(\theta)- \frac 1 2 A_2 r^2$$

Now setting $\alpha = \frac 3 2 A_2r^2cos(\theta)$, and $\alpha + \beta = A_0 - \frac 1 2 A_2 r^2$, and get $$A_2 = \frac {2\alpha} {3r^2cos(\theta)}$$ $$A_0 = \alpha + \beta + \frac {\alpha} {3cos^2(\theta)}$$

Adding these into our original Legendre Polynomial expression we get: $$V_{inside} = \alpha(\frac {3cos^2(\theta)+1} {3cos^2(\theta)}) + \beta + \frac {2\alpha} {3r^2cos^2(\theta)}r^2(\frac {3cos^2(\theta) -1} {2})$$

Simplifying: $$V_{inside}= \alpha(\frac {3cos^2(\theta)+1 +3cos^2(\theta) - 1} {3cos^2(\theta)}) + \beta$$ $$V_{inside} = 2\alpha + \beta$$

I just want to know if I am doing this correctly, as I can't find any similar problems in the book, and this answer seemed very simplified, so I'm not sure if correct process, or not.

2. Oct 1, 2016

### TSny

Does this hold for all $r$ or just for a specific value of $r$?

The right hand side is not quite written correctly.

$\alpha$ is a constant, but your right hand side is a function of $r$ and $\theta$.

3. Oct 1, 2016

### RyanTAsher

I'm not too sure on matching the function we are given just of $V(\theta)$ to $V(r,\theta)$, I just set them equal because I couldn't think of any other way to do it.

and I apologize for the typo, the right side should have been $$A_0 + \frac 3 2 A_2r^2cos^2(\theta) - \frac 1 2 A_2 r^2$$ unless you meant it was just incorrect in general.

4. Oct 1, 2016

### TSny

Note that $V_0 = \alpha \sin^2 \left(\theta \right) + \beta$ is the potential "at the surface", not for any arbitrary value of $r$.

OK. Is the left hand side valid for all $r$?

Your expression $\alpha = \frac{3}{2} r^2 A_2 \cos \left(\theta \right)$ from the first post is not correct. The right hand side would vary as you change $r$ or $\theta$. But the left hand side is a constant.

Last edited: Oct 1, 2016
5. Oct 1, 2016

You have a good start, but you need to evaluate at $r=R$. Then you need to set coefficients of like powers of $cos(\theta)$ equal on both sides of the equation. This will allow you to solve for $A_0, A_1, A_2$ $\$ in terms of $\alpha$ and $\beta$.

6. Oct 2, 2016

### RyanTAsher

Okay, so I did some long string of simplification for this over again. I realized I accidentally kept the $cos^2(\theta)$ in my solution earlier.

$\alpha = \frac 3 2 A_2 R^2$
$A_2 = \frac {2\alpha} {3R^2}$
$\alpha + \beta = A_0 - \frac 1 2 A_2R^2 = A_0 - \frac {\alpha} {3}$
$A_0 = \frac 4 3 \alpha + \beta$
-------------------------------
$V(\theta) = \frac 4 3 \alpha +\beta + \frac {2\alpha} {3R^2} R^2 \frac 3 2 cos^2(\theta) - \frac {2\alpha} {3R^2} R^2$
$= \frac 4 3 \alpha + \beta + \alpha cos^2(\theta) - \frac {\alpha} {3} = \alpha + \alpha cos^2(\theta) + \beta$
-------------------------------
$V(\theta) = \alpha(1+cos^2(\theta))+\beta$

So I got it into a theta dependent form only, with r = R at the surface. Not sure if I did everything correctly, but it seems to have simplified decently. As TSny was saying, this seems correct now because alpha, and beta are constants like they should be.

7. Oct 2, 2016

### TSny

I believe you have a sign error in your result for $\alpha$. This will modify your results for $A_2$ and $A_0$.
Your answer for V inside the sphere should be a function of both $r$ and $\theta$. You can check your final answer by seeing if it reduces to the correct form for r = R.

Last edited: Oct 2, 2016
8. Oct 2, 2016

### RyanTAsher

Okay, so I've redone my work, and hopefully have gotten it correct by now.

$- \alpha = \frac 3 2 A_2 R^2$
$A_2 = - \frac {2\alpha} {3R^2}$
$\alpha + \beta = A_0 + \frac 1 2 (\frac {2\alpha} {3R^2}) R^2$
$\alpha + \beta = A_0 + \frac {\alpha} {3}$
$A_0 = \frac 2 3 \alpha + \beta$
------------------------------------------------------------------
$V(r,\theta) = \frac 2 3 \alpha + \beta - \frac 3 2 (\frac {2\alpha} {3R^2})r^2 cos^2(\theta) + \frac 1 2 (\frac {2\alpha} {3R^2})r^2$
$V(r, \theta) = \frac 2 3 \alpha + \beta - \alpha \frac {r^2} {R^2} cos^2(\theta) + \frac {\alpha} {3} \frac {r^2} {R^2}$
$V(r, \theta) = \alpha(\frac 2 3 - \frac {r^2} {R^2} cos^2(\theta) + \frac 1 3 \frac {r^2} {R^2}) + \beta$
------------------------------------------------------------------
CHECK: (Potential at surface, R=r)
$V(\theta) = \alpha(\frac 2 3 - cos^2(\theta) +\frac 1 3) + \beta$
$V(\theta) = \alpha(1 - cos^2(\theta)) +\beta = \alpha sin^2(\theta) + \beta$ $\checkmark$

9. Oct 2, 2016

### TSny

That looks very good.

10. Oct 2, 2016

### RyanTAsher

YES! Thank you both for your help, I was struggling so much, and now I actually think I have a better grasp conceptually of these laplace equation problems.

11. Oct 2, 2016

I think you still need to solve the potential for the case of $V_{out}$ but you seem to understand the method of solution now. (So far you only solved for $V_{in}$. The solution for $V_{out}$ uses similar methods but will have different coefficients and in this case, $V_{out}$ must go to zero as r gets large.) editing... And one additional item=the homework didn't ask for it, but it is worth mentioning: Once you have the expression for the potential, you can compute the electric field everywhere by using $E=-\nabla V$ in spherical coordinates. One of the major reasons for solving for the potential in a problem like this is to find the electric field.