Subfield: Electromagnetism

5. Polarization and Oscillation 

In this problem, we will understand the polarization of metallic bodies and the method of images that simplifies the math in certain geometrical configurations.

Throughout the problem, suppose that metals are excellent conductors and they polarize significantly faster than the classical relaxation of the system.
Context question:
(a) Explain in words why there can't be a non-zero electric field in a metallic body, and why this leads to constant electric potential throughout the body.
Context answer:
开放性回答


Context question:
(b) Laplace's equation is a second order differential equation

$$
\nabla^{2} \phi=\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}}+\frac{\partial^{2} \phi}{\partial z^{2}}=0
\tag{8}
$$

Solutions to this equation are called harmonic functions. One of the most important properties satisfied by these functions is the maximum principle. It states that a harmonic function attains extremes on the boundary.

Using this, prove the uniqueness theorem: Solution to Laplace's equation in a volume $V$ is uniquely determined if its solution on the boundary is specified. That is, if $\nabla^{2} \phi_{1}=0$, $\nabla^{2} \phi_{2}=0$ and $\phi_{1}=\phi_{2}$ on the boundary of $V$, then $\phi_{1}=\phi_{2}$ in $V$.

Hint: Consider $\phi=\phi_{1}-\phi_{2}$.
Context answer:
\boxed{证明题}


Question: (c) The uniqueness theorem allows us to use "image" charges in certain settings to describe the system. Consider one such example: There is a point-like charge $q$ at a distance $L$ from a metallic sphere of radius $R$ attached to the ground. As you argued in part (a), sphere will be polarized to make sure the electric potential is constant throughout its body. Since it is attached to the ground, the constant potential will be zero. Place an image charge inside the sphere to counter the non-uniform potential of the outer charge $q$ on the surface. Where should this charge be placed, and what is its value?

Solution:
Take the origin of our coordinate system at the center of the sphere and align the axes so that the coordinates of our charge is $(x, y)=(L, 0)$. Now suppose we place an image charge $q_{0}$ at $(x, 0)$ where $0<r<R$. Then the potential at the point $(R \cos \theta, R \sin \theta)$ on the surface becomes

$$
V(\theta)=\frac{q_{0}}{x_{\theta}}+\frac{q}{L_{\theta}}
$$

where

$$
\begin{aligned}
x_{\theta} & =\sqrt{x^{2}+R^{2}-2 x R \cos \theta} \\
L_{\theta} & =\sqrt{L^{2}+R^{2}-2 L R \cos \theta}
\end{aligned}
$$

We want this image charge to satisfy $V(\theta)=0 \forall \theta \in[0,2 \pi]$. Then we have,

$$
\begin{aligned}
\frac{q}{q_{0}} & =-\frac{\sqrt{L^{2}+R^{2}-2 L R \cos \theta}}{\sqrt{x^{2}+R^{2}-2 x R \cos \theta}} \\
& =-\sqrt{\frac{L}{x} \frac{\sqrt{\frac{L^{2}+R^{2}}{L}-2 R \cos \theta}}{\sqrt{\frac{x^{2}+R^{2}}{x}-2 R \cos \theta}}}
\end{aligned}
$$

Consequently,

$$
\begin{aligned}
& \frac{L^{2}+R^{2}}{L}=\frac{x^{2}+R^{2}}{x} \text { and } \\
& \frac{q}{q_{0}}=-\sqrt{\frac{L}{x}}
\end{aligned}
$$

Solution to the first equation with $x<R$ condition gives $x=\frac{R^{2}}{L}$, and from the second equation, $q_{0}$ can be found as $q_{0}=-q \frac{R}{L}$. By azimuthal symmetry, we conclude that if the



potential is zero on surface crossing the $x-y$ plane with the image charge $q_{0}=-q \frac{R}{L}$ at $x=\frac{R^{2}}{L}$, then it is zero everywhere throughout the surface.

Final answer: $x=\frac{R^{2}}{L}$ , $q_{0}=-q \frac{R}{L}$