Potential of Sphere, Given Potential of Surface

[Destroxia]

Homework Statement

[/B]
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.

Homework 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

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

The Attempt at a Solution

[/B]
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.

 

[TSny]

this is just my work for the inside of the sphere.
$$ \alpha + \beta - \alpha cos^2(\theta) = A_0 + A_1rcos(\theta) + A_2r^2(\frac {3cos^2(\theta)-1} {2})$$

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

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

The right hand side is not quite written correctly.

Now setting $\alpha = \frac 3 2 A_2r^2cos(\theta)$

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

 

[Destroxia]

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 $\theta$.

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.

 

[TSny]

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.

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

the right side should have been $$ A_0 + \frac 3 2 A_2r^2cos^2(\theta) - \frac 1 2 A_2 r^2 $$

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.

 

[Charles Link]

Homework Statement

[/B]
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.

Homework 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

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

The Attempt at a Solution

[/B]
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 $cos^2(\theta)$.

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

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 $\beta$.

 

[Destroxia]

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 $\beta$.

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.

 

[TSny]

$\alpha = \frac 3 2 A_2 R^2$

I believe you have a sign error in your result for $\alpha$. This will modify your results for $A_2$ and $A_0$.

-------------------------------
$V(\theta) = \alpha(1+cos^2(\theta))+\beta$.

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.

 

[Destroxia]

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 $\beta$.

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.

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)
$V(\theta) = \alpha(\frac 2 3 - cos^2(\theta) +\frac 1 3) + \beta$
$V(\theta) = \alpha(1 - cos^2(\theta)) +\beta = \alpha sin^2(\theta) + \beta$ $\checkmark$

 

[TSny]

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

That looks very good.

 

[Destroxia]

That looks very good.

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.

 

[Charles Link]

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.

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.