10th of September, 2008

1. 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.

   $$n_{+}=n_0{e}^{-q\varphi ∕kT}{n}_{-}=n_0{e}^{q\varphi ∕kT}$$

   For $kT\gg q\varphi$ the approximation $e^x\approx 1+x$ can be used. Thus

   $$\rho \approx -\frac{2n_0{q}^2\varphi }{kT}.$$

   Substituting this into Poisson’s equation and using the spherical symmetry of the problem we get

   $$\frac{{\partial }^2\varphi }{\partial r^2}+\frac{2}{r}\frac{\partial \varphi }{\partial \rho }-\underset{C}{\underbrace{\frac{2n_0{q}^2}{kT}}}\varphi =0.$$

   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: $C_1=Q∕4\pi {\epsilon }_0$, where $Q$ is the charge of the colloidal particle.

   $$\varphi \left(r\right)=\frac{1}{4\pi {ϵ}_0}\frac{Q}{r}{e}^{-\sqrt{2n_0{q}^2∕kT}r}$$

2. 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.

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   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 ${\overrightarrow{p}}^{\prime }$.

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   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

   $$W=\frac{p^2}{4\pi {\epsilon }_0}\frac{cos(\theta +{\theta }^{\prime })-3cos\theta cos{\theta }^{\prime }}{r^3}.$$

   The magnitude of the torque acting on the dipole $\overrightarrow{p}$ is just

   $$T=-\frac{\partial W}{\partial \theta }=-\frac{p^2}{4\pi {\epsilon }_0}\frac{-sin(\theta +{\theta }^{\prime })+3sin\theta cos{\theta }^{\prime }}{r^3}.$$

   Let us apply this result to the dipole and its mirror image. We have $r=2s$ and $\theta ={\theta }^{\prime }=\phi$, so

   $$\begin{array}{cc}\begin{array}{rl}{T}_{\mathrm{\text{ dipole}}}& =-\frac{p^2}{4\pi {\epsilon }_0}\frac{sin2\phi }{16s^3}\\ \end{array}& \end{array}$$

   If the dipole is free to rotate, it will adopt one of the positions where the torque is 0, i.e. $\phi =0$, $\phi =\pi ∕2$, or $\phi =\pi$. Of these three, $\phi =\pi ∕2$ is an unstable equilibrium, so the dipole will be perpendicular to the plane.

3. 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 $I∕A=\rho \left(x\right)v\left(x\right)$ 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 $E_{kinetic}=m{v}^2∕2=eV\left(x\right)$, where $e$ is the elementary charge. Using these two relations we find that $\rho \left(x\right)=(I∕A)\sqrt{m}∕\sqrt{2eV\left(x\right)}$. 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 $\alpha =4∕3$ and $\beta ={(9C∕4)}^{2∕3}$. 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

   $$V\left(x\right)={\left(\frac{81}{32}\frac{I^{2}m}{A^2{\epsilon }_0^{2}e}\right)}^{1∕3}{x}^{4∕3}$$
   $$v\left(x\right)={\left(\frac{9}{2}\frac{Ie}{A{\epsilon }_{0}m}\right)}^{1∕3}{x}^{2∕3}$$
   $$\rho \left(x\right)={\left(\frac{2}{9}\frac{I^2{\epsilon }_{0}m}{A^{2}e}\right)}^{1∕3}{x}^{-2∕3}$$

   Putting $x=d$ to find $V_0$, we get

   $$I=\frac{4}{9}\sqrt{\frac{2e}{m}}\frac{A{\epsilon }_0}{d^2}{V}_0^{3∕2}$$

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