# Field of a cylinder with surface charge density $\sigma= \sigma_0 \cos(\theta)$

I have this problem, where I have a cylinder with the surface charge density in the title. The angle $\theta$ refers to the angle on the horizontal axis. We had seen a similar exercise with spheres, and the way to do that is to consider two uniformly charged spheres of charge density $\rho$ and $-\rho$ whose centers are shifted of $d$, and where $\rho d=\sigma_0$. The point is, inside the sphere the field is simple, you just sum the fields of those spheres. Outside of it, however, it seems complicated, but I think that seeing the two spheres as two points with the same charge at distance $d$, I could treat them as a dipole.

Now, however, while it should be simple to evaluate the field inside the cylinder (it should be $\frac{\sigma_0}{2 \epsilon_0}$), I have no idea on how to find the field outside of it (I am assuming a relatively tall cilinder, and I am working in the middle to avoid border effects)

• Are you familiar with the general solution of Laplace's equation in cylindrical coordinates? Nov 12, 2017 at 12:24
• @eranreches I'm afraid I'm not... Nov 12, 2017 at 14:43
• You may want to consult section 3.3 in Griffiths' Introduction to Electrodynamics. Nov 12, 2017 at 15:06

I think the handiest trick to dealing with these situations is by noting that, if you have a solution $\varphi$ to the Poisson equation $\nabla^2 \varphi = -\rho/\varepsilon_0$, then you can get new ones by applying the cartesian derivatives $\partial/\partial x$ and so on. (This is because $\partial/\partial x$ commutes with the laplacian; the trick fails in curvilinear coordinates where that's no longer the case.)
Many standard solutions can be seen as arising in this way: for example, the point-dipole field is obtained by taking the point-charge field, $$\varphi(\mathbf r) = \frac{q}{4\pi\varepsilon_0}\frac{1}{r}, \qquad\text{obeying}\qquad \nabla^2\varphi = q\delta(\mathbf r)/\varepsilon_0,$$ and differentiating it with respect to $x$, giving you $$\partial_x\varphi(\mathbf r) = -\frac{q}{4\pi\varepsilon_0}\frac{x}{r^3}, \qquad\text{obeying}\qquad \nabla^2\partial_x\varphi = q\partial_x\delta(\mathbf r)/\varepsilon_0,$$ where $\partial_x\delta(\mathbf r) = \delta'(x)\delta(y)\delta(z)$ is a point dipole's representation as a distribution-valued charge density. (The higher multipoles can also be obtained in this way, though if you just use cartesian derivatives then you'll need to be careful with how you handle their linear combinations.)
Similarly, the two-displaced-spheres calculation has a natural home in this formalism, where you take as the starting point the field of a uniform ball of charge: $$\varphi(\mathbf r) = \begin{cases}\frac{Q}{4\pi\varepsilon_0}\frac{1}{r} & r>a \\ \frac{Q}{4\pi\varepsilon_0} \frac{r^2}{a^3} & r<a\end{cases}, \qquad\text{obeying}\qquad \nabla^2\varphi = \rho_0\theta(a-r)/\varepsilon_0,$$ where you now need to differentiate the Heaviside step function to give you $$\frac{\partial}{\partial x} \theta(a-r) = \frac{\partial r}{\partial x} \theta'(a-r) = \frac{x}{r} \delta(a-r),$$ and the internal derivative $\partial_x r^2 = 2x$ gives you a linear potential and therefore a constant electric field. (I'm possibly fudging some constants. Work this stuff through in detail to check it's right.)
Moreover, it should be clear why this derivative trick works for this situation: if you want to model a uniform ball of charge centered at $(\Delta x,0,0)$, you can say that it's the result of a uniform ball of charge at the origin plus the $\sigma\propto\cos(\theta)$ double-shell thing, or you can do a Taylor expansion in $\Delta x$ to get $$\varphi_{\Delta x}(\mathbf r) = \varphi_{0}(\mathbf r) + \Delta x \left[ \frac{\partial}{\partial \Delta x}\varphi_{\Delta x}(\mathbf r) \right]_{\Delta x=0} + \mathcal O((\Delta x)^2),$$ where the derivative along $\Delta x$ is equivalent to a derivative along $x$ (in the same way that active and passive frame transformations are equivalent) and the neglected terms in $\mathcal O((\Delta x)^2)$ are also thrown away in the hand-wavy example.