# Similarity between Potential Energy Function of Polarized Sphere and Cylinder

According to this paper, the potential of any point on the surface at a distance '$$h$$' from the center of some simple conducting geometric shape is given by

For a Sphere: $$V(x)=Mh^2\frac{xcos(\alpha)+hsin(\alpha)}{(x^2+h^2)^\frac{3}{2}}$$ For a Horizontal Cylinder: $$V(x)=Mh\frac{xcos(\alpha)+hsin(\alpha)}{x^2+h^2}$$ where, $$M$$ = Electric Dipole Moment of the Polarized body, $$\alpha$$ = Angle of Polarizing axis from the Horizontal, and $$h$$ = Distance of the center of the object from the Horizontal

Now, how to derive this? There exist multiple papers of the same kind which use the formula, but none derived it.

$$My$$ $$Attempt$$ :

1. I tried using the potential energy function of an Electric dipole, $$p$$ $$(U=\frac{kpcos(\theta)}{r^2})$$. Here, $$cos(\theta)$$ can be written as $$\frac{xcos(\alpha)+hsin(\alpha)}{(x^2+h^2)^\frac{1}{2}}$$ from Geometry. But then I cannot figure out how the rest of the terms, in the given formula, can possibly match this Dipole Potential Energy Function.

2. I even tried calculating the $$V(x)$$ for a Sphere by assuming charge distribution on it. Assuming that a "constant" Electric field outside the sphere was responsible in polarizing the sphere, Charge arrangement on sphere can be found. Since Electric Field inside a polarized sphere is zero, charge density of an infinitesimal ring on sphere at an angle $$\theta$$ from the polarizing axis must be of the form $$\frac{dQ}{Rd\theta}=\lambda cos(\theta)$$. But then it got quite complicated.

How to derive these formulas? Are there any other methods?

• IIRC, the electric field inside a uniformly polarized sphere is not zero; instead, it's uniform and points along the polarization axis. Commented Sep 20, 2021 at 18:02
• The Electric Field due to the charge arrangement on the Sphere is uniform and points along the polarization axis. But the net Electric Field inside the conducting Sphere (Constant Electric Field from the outside that polarized the sphere in the first place + Electric Field due to charge arrangement on sphere) must be zero. I didn't mention the constant Electric Field outside the sphere, and that may be the reason of this confusion.
– RSR
Commented Sep 22, 2021 at 3:48
• My confusion arose from the fact that you didn't say that the object was conducting in the original question, just that it was polarized. I would suggest that you edit your question to include this assumptions (and any others you're making.) Commented Sep 22, 2021 at 11:51

One way to approach this is to write out the general solution for Laplace's equation using separation of variables. We assume that the charge distribution on the surface of the sphere $$\sigma$$ is known (or equivalently the polarization $$\vec{P}$$ is known, since $$\sigma = \hat{r} \cdot \vec{P}$$ on the surface), and we solve only for the effects of this charge. If an external field is present, we can add it in after the fact via superposition.
In the case of cylindrical symmetry, the general solution is $$V(r, \theta) = A_0 + B_0 \ln r + \sum_{m = 1}^\infty (A_m r^m + B_m r^{-m}) (C_m \cos (m\theta) + D_m \sin (m\theta)).$$ This holds at all points where there is no net charge, which in this case is all points except on the boundary of the cylinder. We can thereform split this up into two separate solutions, $$V_\text{in}$$ and $$V_\text{out}$$, which hold inside and outside the cylinder respectively. One can then determine values for all of these coefficients using the following observations:
• The situation is symmetric under $$\theta \to - \theta$$.
• Inside the cylinder, $$V$$ remains finite.
• Outside the cylinder, $$V$$ goes to zero as $$r \to \infty$$.
• From the boundary conditions that hold across a surface $$\sigma$$, we have $$\frac{\partial V_\text{out}}{\partial r} - \frac{\partial V_\text{in}}{\partial r} = - \frac{\sigma}{\epsilon_0} = - \frac{\hat{r} \cdot \vec{P}}{\epsilon_0}$$ on the surface of the cylinder.
A similar argument can be made in the context of spherical coordinates. Under the assumption of rotational symmetry about some axis can be written as $$V(r, \theta) = \sum_{m = 0}^\infty (A_m r^m + B_m r^{-m-1}) P_m(\cos \theta),$$ where $$\theta$$ is the angle from the axis of symmetry and $$P_m(x)$$ is the mth Legendre polynomial. The similarities between these two forms (and the fact that $$P_1(\cos \theta) = \cos \theta$$) is what gives rise to the similarities between the two formulas you found.