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}


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


Context question:
(d) Argue from the uniqueness theorem that the electric field created by this image charge outside the sphere will be the same as the field created by the complicated polarization of the sphere.
Context answer:
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Context question:
(e) Find the force of attraction between the charge and the sphere.
Context answer:
\boxed{$F=\frac{1}{4 \pi \epsilon_{0}} \frac{q^{2} \frac{R}{L}}{\left(L-\frac{R^{2}}{L}\right)^{2}}$}


Question: (f) Now suppose that we attach the point-like charge to a wall with a rod of length a. Any perturbation from the equilibrium will cause a perturbation of the polarization of the sphere. Prove that this equilibrium is stable and find the frequency of oscillation around it. The charge and rod have masses $m$ and $M$, respectively and assume that $L>R+a$.

<img_4529>

Solution:
Force is attractive. Therefore, any deviation from the equilibrium $(\theta=0)$ induces an attractive torque towards $\theta=0$.

Force at the equilibrium would be the same as in part (e) with $L$ replaced by $L-a$. Since we are looking for small oscillations, we need to calculate the torque in the first order approximation of $\theta$. Denote $\beta$ as the angle between the line segment from the center of the sphere to the particle and from the particle to the wall. Then the equations of motion become

$$
\begin{aligned}
\tau=-F(\theta) a \sin (\beta) & \approx-F(0) a \frac{L}{L-a} \theta \\
& =I \frac{d^{2} \theta}{d t^{2}} \\
& =\left(m+\frac{M}{3}\right) a^{2} \frac{d^{2} \theta}{d t^{2}}
\end{aligned}
$$

Thus it follows that

$$
\omega=\frac{q}{(L-a)^{2}-R^{2}} \sqrt{\frac{R L}{4 \pi \epsilon_{0} a\left(m+\frac{M}{3}\right)}}
$$

In the limit $a<<L$, we obtain the formula

$$
\omega=\frac{q}{L^{2}-R^{2}} \sqrt{\frac{R L}{4 \pi \epsilon_{0} a\left(m+\frac{M}{3}\right)}}
$$