10th of September, 2008

  1. The electric potential ϕ must satisfy Poisson’s equation.
    2ϕ = -ρ ε0

    The spatial charge density can be expressed in terms of the concentration n and charge q of positive and negative ions:

    ρ = n+q+ + n-q-

    For simplicity, consider the case when q+ = -q- = q, and the concentration of both kinds of ions is n0 far from the colloidal particle.

    n+ = n0e-qϕkT n - = n0eqϕkT

    For kT qϕ the approximation ex 1 + x can be used. Thus

    ρ -2n0q2ϕ kT .

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

    2ϕ r2 + 2 r ϕ ρ -2n0q2 kT Cϕ = 0.

    This differential equation can be written as

    2(rϕ) r2 - C(rϕ) = 0,

    whose general solution is rϕ = C1e-Cr + C 2eCr. For the potential to be zero at infinity, it is required that C2 = 0. When r 0, the potential must be the same as in the absence of any ions: C1 = Q4πε0, where Q is the charge of the colloidal particle.

    ϕ(r) = 1 4πϵ0 Q r e-2n0 q2 kTr

  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.


    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 p.

    Let us imagine that the sheet is removed, and another electric dipole p is added in such a way that it is the mirror image of 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 p and p 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 p in the absence of the sheet. Now all we need to do is find the torque acting on p in such a field.

    The electric potential created by a dipole p is

    V (r) = 1 4πε0 pr r3 .

    The electric field vector is

    E(r) = -V = 1 4πε0 3(pr)r - r2p r5 ,

    and the potential energy of a dipole p in this field is

    W = -E p = 1 4πε0 r2(p p) - 3(r p)(r p) r5 .

    Let θ and θ denote the angles between r, and the vectors p and p’, respectively. If both p and p have the same magnitude, then the potential energy can be written as

    W = p2 4πε0 cos(θ + θ) - 3 cos θ cos θ r3 .

    The magnitude of the torque acting on the dipole p is just

    T = -W θ = - p2 4πε0 - sin(θ + θ) + 3 sin θ cos θ r3 .

    Let us apply this result to the dipole and its mirror image. We have r = 2s and θ = θ = φ, so

    T dipole = - p2 4πε0 sin 2φ 16s3

    If the dipole is free to rotate, it will adopt one of the positions where the torque is 0, i.e. φ = 0, φ = π2, or φ = π. Of these three, φ = π2 is an unstable equilibrium, so the dipole will be perpendicular to the plane.

  3. We have essentially a one dimensional problem, because A d. We can write Poisson’s equation for V :
    d2V (x) dx2 = -ρ(x) ε0

    In a steady state the current density IA = ρ(x)v(x) between the plates is constant. Here v(x) 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 Ekinetic = mv22 = eV (x), where e is the elementary charge. Using these two relations we find that ρ(x) = (IA)m2eV (x). Substituting this result into Poisson’s equation,

    d2V (x) dx2 V (x) = - Im Aε02e = 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 (x) = βxα. Substituting this back into the equation we find that α = 43 and β = (9C4)23. This is not a general solution as it does not contain any arbitrary constants but it satisfies the boundary conditions of the problem: V (0) = 0 and dV dx x=0 = 0 (the electric field is 0 at the cathode), so we need seek no further. The final results are

    V (x) = 81 32 I2m A2ε02e13x43

    v(x) = 9 2 Ie Aε0m13x23

    ρ(x) = 2 9 I2ε 0m A2e 13x-23

    Putting x = d to find V 0, we get

    I = 4 92e m Aε0 d2 V 032

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