Cylinder charged to a potential

[gijkhama]

What does it actually mean that a cylinder was charged to a potential of V?
Let's consider the following situation as seen on image below:
[URL]http://files.droplr.com/files/5666792/zkMA.Screen%20shot%202010-11-11%20at%2023%3A07%3A18.png[/URL]
Upper half-cylinder is charged to potential V and the one below to -V.

 

[Mindscrape]

It means that the surface has some charge on it that creates an equipotential.

 

[gijkhama]

But does it mean that the charge is equaly distributed along a cylinder?

 

[Mindscrape]

No, you can't assume that just because the potential is equally distributed that the charge will be, and vice versa. You would have to calculate dv/dn.

 

[gijkhama]

Would you be so kind and show me just how to start with it?

 

[Mindscrape]

Generally, for potentials with no source charges you want to use a laplacian

$$\nabla^2 V =0$$

 

[gijkhama]

Solution to Laplace equation in 2D (in polar cords) is:

$$V(r, \varphi ) = \sum_{l= 0}^{+\infty} \sum_{m = -l}^l C_{lm} \left( \alpha_r r^l + \frac{\beta}{r^{l+1}} \right) P_l^m (\cos \varphi )$$

Because for $$\varphi \in [0, \pi]$$ we want $$V(R,\varphi) \equiv const$$ than one associated legendre polynomial can stay, so it simplifies to the case when $$(l,m) = (0,0)$$ and hence: $$V(R, \varphi) = \frac{C}{R} \equiv \mathcal{V} \Rightarrow C = \mathcal{V} R$$.

And that's all. I can't see how this can lead to showing what is the charge distribution over the cyllinder.

 

[Mindscrape]

Well, you used the solution for spherical coordinates with azimuthal symmetry, not polar coordinates. Furthermore, you didn't even consider the cases of potential inside the shape (is it a cylinder or a sphere?) compared to outside. Also, like I said before, the charge distribution will be

$$\frac{\partial V}{\partial n}=\sigma$$

if this really is a sphere it will be dV/dr.

I don't really even know what question you're asking. You started off by asking how a cylinder with separate potentials in the top and bottom would work, and now all you want is the charge distribution.

 

[gijkhama]

Now I'd like to see how is charge distributed over the cylinder (it's not a sphere even it's darken inside).
I must say that I don't know what this $$n$$ means in th equation you gave :|.

 

[Mindscrape]

n is for normal. So for the charge distribution you want dV/dr for a cylinder or a sphere. Still, as I said above, you haven't worked out the right potential.

$$\frac{\partial V_{out}}{\partial r}-\frac{\partial V_{in}}{\partial r}=\sigma$$

I am having a bit of trouble fully comprehending you, we must have some language differences.

 

[gijkhama]

Ok, so the general solution of Laplace equation is:

$$V(r, \phi) =\sum_{l = 0}^{+\infty} \sum_{m = -l}^{l} C_{lm} \left( \alpha r^l + \frac{\beta}{r^{l+1}} \right) ( a_m \cos m \phi + b_m \sin m \phi )$$

Now, should I use the boundary condition $$V(R,\phi) = \pm \mathcal{V}$$ and compute some unknown coefficients (it leads me to the case when m=l=0 :/)?

Honestly, I don't have any idea how to derive forms of $$V_{out}$$ and $$V_{in}$$ from this general case.

 

[Mindscrape]

Here is what separation of variables in cylindrical coordinates gives for the general solution

$$V(s,\phi)= a_o + b_o ln s + \sum_{k=1}^\infty [s^k(a_k cos(k \phi)+ b_k sin(k \phi) + s^{-k}(c_k cos(k \phi)+ d_k sin(k \phi)]$$

From here you will need to consider how the solution will look on the inside of the cylinder and outside the cylinder, and you will need to apply boundary conditions. Though from your drawing the inside of the cylinder appears to be solid (it actually looks like a sphere to me), so it should be at equipotential, and you just have to consider Vout.

 

[gijkhama]

Thank you for your patience. I think that I'm finally getting the idea.

Actually the inside of the cylinder (s<R) is empty. It's just filled with the color for a purpose of another task.

Back to problem, in s=0 there should be no singularity. Therefore I can write:

$$V_{in}(s,\phi) = \sum_{k = 0}^{+\infty} s^k (a_k \cos k \phi + b_k \sin k \phi)$$

If I now scale both coefficients as it follows: $$\alpha_k = a_k R^k / V_0 , \beta_k = b /( V_0 R^k)$$, than:

$$V_{in}(R, \phi) = V_0 \sum_{k = 0}^{+\infty} (\alpha_k \cos k \phi + \beta_k \sin k \phi) = \begin{cases} V_0 & \phi \in (0,\pi)\\ -V_0 & \phi \in (\pi,2\pi)\end{cases}$$

So I see here a Fourier series and I can compute necessary coefficients.
$$\alpha_k \equiv 0$$ due to the fact that the function is odd and:
$$\beta_k = \frac{2 (1- (-1)^k)}{k \pi}$$ therefore:

$$V_{in} (s,\phi) = V_0 \sum_{k = 0}^{+\infty} \left( \frac{s}{R} \right)^k \frac{2 (1- (-1)^k)}{k \pi} \sin k \phi$$

That's of course the case when s<R, for s>R the series contains only of negatives powers of s but the reasoning is the same, I believe. So:

$$V_{out} (s, \phi) = V_0 \sum_{k = 0}^{+\infty} \left( \frac{R}{s} \right)^k \frac{2 (1- (-1)^k)}{k \pi} \sin k \phi$$

Now I try to see how the charge is distributed. I compute two derivatives:

$$\frac{\partial V_{in}}{\partial r} = V_0 \sum_{k = 1}^{+\infty} \left( \frac{s}{R} \right)^{k-1} \frac{2 (1- (-1)^k)}{ \pi R} \sin k \phi = V_0 \sum_{k = 0}^{+\infty} \left( \frac{s}{R} \right)^{k} \frac{2 (1- (-1)^{k+1})}{ \pi} \sin [(k+1) \phi]$$
$$\frac{\partial V_{out}}{\partial r} = - V_0 R \sum_{k = 0}^{+\infty} \frac{R^k}{s^{k+2}} \frac{2 (1- (-1)^{k+1})}{ \pi } \sin [(k+1) \phi]$$

Finally:

$$\sigma = -V_0 R \sum_{k = 0}^{+\infty} \frac{R^k}{s^{k+2}} \frac{2 (1- (-1)^{k+1})}{ \pi } \sin [(k+1) \phi] - V_0 \sum_{k = 0}^{+\infty} \left( \frac{s}{R} \right)^{k} \frac{2 (1- (-1)^{k+1})}{ \pi R} \sin [(k+1) \phi]$$
$$\sigma = -V_0 \sum_{k = 0}^{+\infty} \left( \frac{R^{k+1}}{s^{k+2}} + \frac{s^k}{R^{k+1}} \right) \frac{2 (1- (-1)^{k+1})}{ \pi } \sin [(k+1) \phi]$$

But charge is only distributed on the cylinder for which s=R, so in fact:

$$\sigma(R,\phi) = - \frac{2V_0}{\pi R} \sum_{k = 0}^{+\infty} \frac{2 (1- (-1)^{k+1})}{ \pi } \sin [(k+1) \phi]$$

I'm a bit confused because this series doesn't seem to converge for any value of angle.

 

[Mindscrape]

Hmm, everything you did looks right. I'll try to take a more in depth look at it this weekend.