Question

2. Consider the system of two identical plates at x = ±a each with charge density σ that we studied in class and is in the lecture notes. Starting with the Poisson Boltzmann equation dr2 kTe we obtained 1 (dV 2 (de) -4π1Ba(0) exp(-V)=c with c a constant. This can be written 2 dr kTe or kTe a) Take x2 = 0 and zi = a because you know the electric field --(kT/e)dV/dz at these two places. Show from this that 2 dr kTe 2dz o2 a(a) = ρ(0) + 2ckT Note that the charge density of counter ions at the surface, ρ(a), never falls below σ2/ekT (e.g. for two surfaces far apart, p(0) → zero.) b) Consider σ 0.2C/m2 which is typical for a fully ionized surface and which corresponds to one elementary charge per 0.8nm2 or 1.25x1018 charges per square meter. Calculate in this case the limiting value of the charge density, σ2/2ckT at the surface. If these counterions are considered to occupy a layer of thickness δ = 0.2nm, what charge per unit area, σ20/2ekT, does this correspond to? You should find that it is almost the same as the charge density σ itself. This is an interesting result, for it shows that regardless of the charge profile away from the surface, most of the counterions that effectively balance the surface charge are located in the first few angstroms from the surface. (For lower surface charge densities, however, the diffuse layer extends well beyond the surface.)

Answer 1

Since, (dv/dx)^2 = (e tshe/KT)^2 thatis (2) Sub: (2) in (1) implies 1/2 (e/KT)^2 (E^2 /_1 - E^2 /_2) = q^2 /KT Epsilon (rho (x_1) - rho (x_2)) x_2 = 0, x_1 = a implies e^2 /2 (KT)^2 (E^2 /_a - E^2 /_0) = q^2 /KT Epsilon (rho (a) - rho (0)) E = sigma/epsilon_0 implies rho (a) = rho (0) + KT Epsilon e^2 /2q^2 (KT)^2 (E^2 /_a - E^2 /_0) = rho (0) + Epsilon/2 KT (sigma/E)^2 = rho (0) + sigma^2 /2 Epsilon KT (b) The value of sigma^2 /2 Epsilon KT = (0.2)^2 /2 times 1 times 1.25 times 10^+ 18 v_0 similarly, for delta sigma^2 /2 Epsilon KT ~ 0