物理代写|Physics6561 Electrodynamics

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Physics6561课程简介

As an undergraduate you learn the language of physics, the laws of physics, and even some physics trivia. And thanks to the skill of your instructors, the subject hums like a well oiled machine. If there’s anything that’s daunting, occasionally, it’s some of the math that comes with the territory. But that’s incidental since you know that, fundamentally, physics is simple and elegant.

But by the time you’re a graduate student you realize that the world doesn’t always seem to play by the rules of the physics laid out in textbooks. There are subtleties, apparent contradictions, paradoxes. You find out that eventually these things got resolved, but when they first appeared on the radar they were called effects. Coming to terms with the effects of physics is a good way to think of the next phase of your education.

Prerequisites 

No doubt some effects, such as Aharonov-Bohm, came up in undergraduate physics. The best effects, such as this one, increase your understanding in a deep way, like the vector potential is a more fundamental description of the electromagnetic field. While not all effects are this famous, every one of them provided an important lesson in physics at some point in time. In fact, the name of the effect often becomes a proxy for a general concept, as for example Hanbury-Brown and Twiss effect might be used to describe an experiment involving electrons when in fact the original effect arose in stellar interferometry.

Physics6561 Electrodynamics HELP（EXAM HELP， ONLINE TUTOR）

In this approach, let’s find the potential approximately, to leading order and subleading order in terms of $\psi$. Also, let’s solve $\psi$ equation perturbatively in terms of $\frac{a}{r}$.
$$
\begin{gathered}
\Phi=\frac{Q}{2} \int_\psi^{\infty} \frac{d \psi^{\prime}}{\psi^{3 / 2}}\left[1-\frac{a^2}{\psi^{\prime}}+\mathcal{O}\left(\frac{a^4}{\psi^{\prime 2}}\right)\right]= \
\Rightarrow \quad \Phi=Q\left(\frac{1}{\sqrt{\psi}}-\frac{1}{3} \frac{a^2}{\psi^{3 / 2}} \cdots\right) \
\frac{x^2}{a^2+\psi}+\frac{y^2}{b^2+\psi}+\frac{z^2}{c^2+\psi}=1
\end{gathered}
$$
For disk, $a=b$, and $c=0$. In terms of spherical coordinates,

$$
\begin{aligned}
& z=r \cos \theta \
& y=r \sin \theta \sin \phi \
& x=r \sin \theta \cos \phi
\end{aligned}
$$
$$
\begin{aligned}
& \frac{x^2+y^2}{a^2+\psi}+\frac{z^2}{\psi}=1 \
& \frac{r^2 \sin ^2 \theta}{a^2+\psi}+\frac{r^2 \cos ^2 \theta}{\psi}=1 \
& \frac{r^2 \sin ^2 \theta}{\psi}\left(1-\frac{a^2}{\psi}+\cdots\right)+\frac{r^2 \cos ^2 \theta}{\psi}=1 \
& \text { Ignoring } \frac{a^2}{\psi} \Rightarrow \psi=r^2 \
& \text { next to leading term } \quad \psi=r^2(1+\delta \psi) \quad \Rightarrow \delta \psi=-\frac{a^2}{r^2} \sin ^2 \theta \
& \psi=r^2\left(1-\frac{a^2}{r^2} \sin ^2 \theta\right)
\end{aligned}
$$
Combining (1.7) and (1.3), one recovers the form of the potential in the problem:
$$
\begin{gathered}
\Phi=Q\left(\frac{1}{\sqrt{r^2\left(1-\frac{a^2}{r^2} \sin ^2 \theta\right)}}-\frac{1}{3} \frac{a^2}{r^3}\right)=Q\left(\frac{1}{r}\left(1+\frac{a^2}{2 r^2} \sin ^2 \theta\right)-\frac{1}{3} \frac{a^2}{r^3}\right) \
=\frac{Q}{r}-\frac{Q a^2}{3 r^3}\left(\frac{3 \cos ^2 \theta-1}{2}\right) \
A_2=\frac{-Q a^2}{3}
\end{gathered}
$$

Gauss law tells us surface charge density for a conductor is given by
$$
\sigma=-\frac{1}{4 \pi} \frac{\partial \Phi}{\partial n}
$$
where $\mathbf{n}$ is the normal direction at that point.
So if you think about disk as an ellipsoid when we sent $c \rightarrow 0$, then disk has two side and for each side, the surface charge density can be found using above equation.
$$
\begin{aligned}
& \sigma(r)^{+}=-\left.\frac{1}{4 \pi} \frac{\partial \Phi}{\partial z}\right|{z=\varepsilon}=\frac{1}{4 \pi} E_z(z=\varepsilon) \ & \sigma(r)^{-}=\left.\frac{1}{4 \pi} \frac{\partial \Phi}{\partial z}\right|{z=-\varepsilon}=-\frac{1}{4 \pi} E_z(z=-\varepsilon)
\end{aligned}
$$
Where $\varepsilon$ is potivie and we’re taking it $\varepsilon \rightarrow 0$.
$$
\begin{aligned}
E_z(x, y, z)= & \frac{\sqrt{2} Q z}{\sqrt{4 a^2 z^2+\left(x^2+y^2+z^2-a^2\right)^2}} \
& \times \frac{1}{\sqrt{-a^2+z^2+x^2+y^2+\sqrt{4 a^2 z^2+\left(x^2+y^2+z^2-a^2\right)^2}}}
\end{aligned}
$$
Now if we expand this expression close to the surface $z=0$,
$$
\begin{aligned}
& E_z(x, y, \varepsilon) \approx \frac{\sqrt{2} Q \varepsilon}{\left(a^2-x^2-y^2\right)} \frac{1}{\sqrt{\frac{2 a^2 \varepsilon^2}{a^2-x^2-y^2}}} \
& \Rightarrow E_z(x, y, \varepsilon) \approx \frac{Q}{\sqrt{a^2-x^2-y^2}} \frac{\varepsilon}{|\varepsilon|} \
& \lim {\varepsilon \rightarrow 0} E(x, y, \varepsilon)=\frac{Q}{\sqrt{a^2-x^2-y^2}} \ & \lim {\varepsilon \rightarrow 0} E(x, y,-\varepsilon)=-\frac{Q}{\sqrt{a^2-x^2-y^2}} \
& \Rightarrow \quad \sigma^{+}(x, y)=\frac{Q}{4 \pi \sqrt{a^2-x^2-y^2}} \
& \Rightarrow \quad \sigma^{-}(x, y)=\frac{Q}{4 \pi \sqrt{a^2-x^2-y^2}}
\end{aligned}
$$
Note that for obtaining (2.5) from (2.4), we used $x^2+y^2 \leq a^2$ mutiple times for simplifying square roots.

Using this, we can verify the total charge on disk indeed is given by $Q$, which is integral of sum of the surface charge densities $\sigma=\sigma^{+}+\sigma^{-}$over the disk:
$$
\int_{x^2+y^2 \leq a^2} \sigma(x, y) d x d y=\int_0^a \rho d \rho \int_0^{2 \pi} d \phi \frac{Q}{2 \pi \sqrt{a^2-\rho^2}}=Q
$$

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