10th of September, 2008

- The electric potential $\varphi $
must satisfy Poisson’s equation.
$${\nabla}^{2}\varphi =-\frac{\rho}{{\epsilon}_{0}}$$
The spatial charge density can be expressed in terms of the concentration $n$ and charge $q$ of positive and negative ions:

$$\rho ={n}_{+}{q}_{+}+{n}_{-}{q}_{-}$$For simplicity, consider the case when ${q}_{+}=-{q}_{-}=q$, and the concentration of both kinds of ions is ${n}_{0}$ far from the colloidal particle.

$${n}_{+}={n}_{0}{e}^{-q\varphi \u2215kT}\phantom{\rule{2em}{0ex}}{n}_{-}={n}_{0}{e}^{q\varphi \u2215kT}$$For $kT\gg q\varphi $ the approximation ${e}^{x}\approx 1+x$ can be used. Thus

$$\rho \approx -\frac{2{n}_{0}{q}^{2}\varphi}{kT}.$$Substituting this into Poisson’s equation and using the spherical symmetry of the problem we get

$$\frac{{\partial}^{2}\varphi}{\partial {r}^{2}}+\frac{2}{r}\frac{\partial \varphi}{\partial \rho}-\underset{C}{\underbrace{\frac{2{n}_{0}{q}^{2}}{kT}}}\varphi =0.$$This differential equation can be written as

$$\frac{{\partial}^{2}\left(r\varphi \right)}{\partial {r}^{2}}-C\left(r\varphi \right)=0,$$whose general solution is $r\varphi ={C}_{1}{e}^{-\sqrt{C}r}+{C}_{2}{e}^{\sqrt{C}r}$. For the potential to be zero at infinity, it is required that ${C}_{2}=0$. When $r\to 0$, the potential must be the same as in the absence of any ions: ${C}_{1}=Q\u22154\pi {\epsilon}_{0}$, where $Q$ is the charge of the colloidal particle.

$$\varphi \left(r\right)=\frac{1}{4\pi {\u03f5}_{0}}\frac{Q}{r}{e}^{-\sqrt{2{n}_{0}{q}^{2}\u2215kT}r}$$ - The problem can be solved using the method of images. The electric
potential inside a conductor is constant, and the field vector near a
conducting surface is always perpendicular to the surface. The charges
inside the conducting sheet redistribute themselves in such a way that
the field created by them balances that component of the dipole’s field
which is parallel to the sheet.

Figure 1: Using the method of images: the electric field in the right half-space will be the same as a field that would be created by the mirrored dipole ${\overrightarrow{p}}^{\prime}$.

Let us imagine that the sheet is removed, and another electric dipole ${\overrightarrow{p}}^{\prime}$ is added in such a way that it is the mirror image of $\overrightarrow{p}$ with respect to the plane of the sheet, but its direction is reversed (see figure 1). It is easy to see that the electric potential created by $\overrightarrow{p}$ and ${\overrightarrow{p}}^{\prime}$ together is constant in the mirror plane. We know that the solution of Laplace’s equation is unique if the boundary conditions are fixed. Thus the field created in the right-hand-side half-space by the charges on the conducting sheet must be the same as the field created by a mirror dipole ${\overrightarrow{p}}^{\prime}$ in the absence of the sheet. Now all we need to do is find the torque acting on $\overrightarrow{p}$ in such a field.

The electric potential created by a dipole ${\overrightarrow{p}}^{\prime}$ is

$$V\left(\overrightarrow{r}\right)=\frac{1}{4\pi {\epsilon}_{0}}\frac{{\overrightarrow{p}}^{\prime}\cdot \overrightarrow{r}}{{r}^{3}}.$$The electric field vector is

$$\overrightarrow{E}\left(\overrightarrow{r}\right)=-\nabla V=\frac{1}{4\pi {\epsilon}_{0}}\frac{3({\overrightarrow{p}}^{\prime}\cdot \overrightarrow{r})\overrightarrow{r}-{r}^{2}{\overrightarrow{p}}^{\prime}}{{r}^{5}},$$and the potential energy of a dipole $\overrightarrow{p}$ in this field is

$$W=-\overrightarrow{E}\cdot \overrightarrow{p}=\frac{1}{4\pi {\epsilon}_{0}}\frac{{r}^{2}(\overrightarrow{p}\cdot {\overrightarrow{p}}^{\prime})-3(\overrightarrow{r}\cdot \overrightarrow{p})(\overrightarrow{r}\cdot {\overrightarrow{p}}^{\prime})}{{r}^{5}}.$$Let $\theta $ and ${\theta}^{\prime}$ denote the angles between $\overrightarrow{r}$, and the vectors $\overrightarrow{p}$ and $\overrightarrow{p}$’, respectively. If both $\overrightarrow{p}$ and ${\overrightarrow{p}}^{\prime}$ have the same magnitude, then the potential energy can be written as

$$W=\frac{{p}^{2}}{4\pi {\epsilon}_{0}}\frac{cos(\theta +{\theta}^{\prime})-3cos\theta \phantom{\rule{3.26212pt}{0ex}}cos{\theta}^{\prime}}{{r}^{3}}.$$The magnitude of the torque acting on the dipole $\overrightarrow{p}$ is just

$$T=-\frac{\partial W}{\partial \theta}=-\frac{{p}^{2}}{4\pi {\epsilon}_{0}}\frac{-sin(\theta +{\theta}^{\prime})+3sin\theta \phantom{\rule{3.26212pt}{0ex}}cos{\theta}^{\prime}}{{r}^{3}}.$$Let us apply this result to the dipole and its mirror image. We have $r=2s$ and $\theta ={\theta}^{\prime}=\phi $, so

$$\begin{array}{cc}\begin{array}{rl}{T}_{\mathrm{\text{dipole}}}& =-\frac{{p}^{2}}{4\pi {\epsilon}_{0}}\frac{sin2\phi}{16{s}^{3}}\\ \end{array}& \end{array}$$ If the dipole is free to rotate, it will adopt one of the positions where the torque is 0, i.e. $\phi =0$, $\phi =\pi \u22152$, or $\phi =\pi $. Of these three, $\phi =\pi \u22152$ is an unstable equilibrium, so the dipole will be perpendicular to the plane.

- We have essentially a one dimensional problem, because
$A\gg d$. We can write
Poisson’s equation for $V$:
$$\frac{{d}^{2}V\left(x\right)}{d{x}^{2}}=-\frac{\rho \left(x\right)}{{\epsilon}_{0}}$$
In a steady state the current density $I\u2215A=\rho \left(x\right)v\left(x\right)$ between the plates is constant. Here $v\left(x\right)$ denotes the velocity of the electrons when they are at distance $x$ from the cathode. The electrons gain their kinetic energy from the electric field, so ${E}_{kinetic}=m{v}^{2}\u22152=eV\left(x\right)$, where $e$ is the elementary charge. Using these two relations we find that $\rho \left(x\right)=(I\u2215A)\sqrt{m}\u2215\sqrt{2eV\left(x\right)}$. Substituting this result into Poisson’s equation,

$$\frac{{d}^{2}V\left(x\right)}{d{x}^{2}}\sqrt{V\left(x\right)}=-\frac{I\sqrt{m}}{A{\epsilon}_{0}\sqrt{2e}}=C$$Note that $C>0$ because the physical current is flowing from the anode to the cathode: $j<0$. We seek the solution of this differential equation in the form $V\left(x\right)=\beta {x}^{\alpha}$. Substituting this back into the equation we find that $\alpha =4\u22153$ and $\beta ={(9C\u22154)}^{2\u22153}$. This is not a general solution as it does not contain any arbitrary constants but it satisfies the boundary conditions of the problem: $V\left(0\right)=0$ and ${\left.\frac{dV}{dx}\right|}_{x=0}=0$ (the electric field is $0$ at the cathode), so we need seek no further. The final results are

$$V\left(x\right)={\left(\frac{81}{32}\frac{{I}^{2}m}{{A}^{2}{\epsilon}_{0}^{2}e}\right)}^{1\u22153}{x}^{4\u22153}$$$$v\left(x\right)={\left(\frac{9}{2}\frac{Ie}{A{\epsilon}_{0}m}\right)}^{1\u22153}{x}^{2\u22153}$$$$\rho \left(x\right)={\left(\frac{2}{9}\frac{{I}^{2}{\epsilon}_{0}m}{{A}^{2}e}\right)}^{1\u22153}{x}^{-2\u22153}$$Putting $x=d$ to find ${V}_{0}$, we get

$$I=\frac{4}{9}\sqrt{\frac{2e}{m}}\frac{A{\epsilon}_{0}}{{d}^{2}}{V}_{0}^{3\u22152}$$