## The equivalent electrostatics problem等效的靜電學問題

In our previous episode (see: http://biophys3min.blogspot.tw/2016/08/67-gradient-sensing-ii.html) we have derived the particle flow across the surface of a perfectly absorbing sphere without a background gradient. Today, however, we will talk about the expected particle flow across the surface of a perfectly absorbing sphere with a background gradient.

Recall our friends, the Fick's law:

Under steady state, the Fick's second law becomes a Laplace equation.
That shall remind you in electrostatics, the following 2 things:

So our original diffusion problem now becomes a standard electrostatics problem:

What is the induced surface charge for an uncharged conducting sphere placed in a uniform electric field $\textbf{E}=E_0\widehat{z}$ ?

## The solution of Laplace equation in spherical coordinate 球座標中Laplace equation的解

To answer this question, we need the solution of Laplace equation in spherical coordinate. (The following would be a little bit mathematically intensive. However, since it is useful in many fields from electrostatics, diffusion to quantum mechanics, I guess it's still worthwhile to pay some attention to it.) That is the solution of

This looks crazy. However, we could break down the monster into pieces by standard technique of separation of variables. By setting$V(r,\theta,\phi)=R(r)\Theta(\theta)\Phi(\phi)$. After some manipulation our monster becomes:

The RHS of equation is a constant while the LHS of equation has 3 terms with different variable-dependency. Therefore we could readily assume:

And our Φ is obviously

The m must be an integer because we expect the function to satisfy$\Phi(\phi+2\pi)=\Phi(\phi)$.

The rest part of equation could be manipulated in basically the same way, by:

The l(l+1) might look strange to you. Nevertheless, it will prove its own benefit soon. Let's see the R part first:

The 3 terms has the same power of r (the Cauchy-Euler equation), therefore the solution could be assumed to be R = r^s. Substitute it back would be:

The Θ is definitely the "boss" of the game. It now looks like:
Θ顯然在這裏是最難解的大魔王，他的方程式現在長得像這樣：
It would be too messy to demonstrate the full process of solving. If you are interested in that, try to assume w = cosθ and assume $\Theta=(1-w^2)^{|m|/2}F(w)$. And it will turns out that the solution is an associated Legendre polynomial which could be also defined by the Rodrigues formula:

w = cosθ然後再假設  $\Theta=(1-w^2)^{|m|/2}F(w)$，用級數解展開然後算一算，最後的答案會是
associated Legendre polynomial。他有好幾種表達方式，其中比較好用的是用Rodrigues formula表示：
For a system with azimuthal symmetry, m = 0 and the solution could then be expressed as Legendre polynomials.

The solution expressed by sum of (associated) Legendre polynomials is cool. However, We hope to have some "basis" that is orthonormal (that is both orthogonal and normalized) rather than merely orthogonal. So it is also common to express our solution using spherical harmonics instead:

## Now back to our electrostatics problem...現在回到電磁學問題

Because our problem possesses azimuthal symmetry, the solution could therefore expressed as:

It still does not look easy. Let's calm down and think about it. Because there is a background electric field and we know the electric field is basically the gradient of electric potential. So in somewhere far far away, the potential would definitely forget he has encountered a crazy sphere floating somewhere. That means:
hmm他看起來還是有點複雜欸，我們稍微冷靜思考一下。因為我們有一個均勻電場，其實就是均勻的電位梯度。想想看在距離導體球無窮遠的彼方，電場怎麼可能還會記得說他曾經遇到這麼一個小小的導體球呢？也就是說當z→無窮大時，電位可以表示成：
Since a conducting ball is an equipotential and the system possess symmetry, we could assume the potential of entire x-y plane and the sphere to be 0. So the constant C must be 0, too. And because r cosθ = z, so only the term with l = 1 would survive. Check the table of Legendre polynomial and we will know that $P_1(\cos\theta)=\cos\theta$, so the solution we are craving for now becomes

It now looks much more friendly, right? Now by knowing that V = 0 when r = R, and V→-Ez, it is easy to show that

$A_1=-E_0$
So the electric potential and the induced surface charges come out

$V&space;=&space;-E_0(r-\frac{R^3}{r^2})\cos\theta$
$\sigma&space;=&space;-\epsilon_0&space;\frac{\partial&space;V(r=R)}{\partial&space;r}&space;=&space;3\epsilon_0&space;E_0\cos\theta$

By some comparison, the solution of our original diffusion problem will be:

And this is left for you to derive.