{ "cells": [ { "cell_type": "markdown", "id": "4dd47827", "metadata": {}, "source": [ "# Problem set 7\n", "\n", "1. This will lead to\n", "\n", " $$\\nabla^2 \\psi = - \\frac{\\rho}{\\epsilon_\\mathrm{w}}$$\n", "\n", "2. The electric potential energy is\n", "\n", " $$\\begin{aligned}\n", " V_+(z) &= e \\psi(z)\\\\\n", " V_-(z) &= -e \\psi(z)\n", " \\end{aligned}$$\n", "\n", "3. $$\\begin{aligned}\n", " \\rho_+(z) &= \\rho_{+0} e^{-\\beta V_+(z)} = \\rho_{+0} e^{-e\\psi(z)} \\\\\n", " \\rho_-(z) &= \\rho_{-0} e^{-\\beta V_-(z)} = \\rho_{-0} e^{e\\psi(z)}\n", " \\end{aligned}$$\n", "\n", " with $\\rho_+(z)$ in units of number per volume, $V_+(z)$ the\n", " potential energy of a positive ion at position $z$, and $\\rho_{+0}$\n", " the density far away from the plate, where $V_+(z)$ is set to zero.\n", " These expressions are approximately correct, if the interactions\n", " between the ions are weak enough. Of course, ions interact with each\n", " other, and that makes it very difficult to solve the question\n", " exactly. This type of approximation is called a “mean field”\n", " approximation, and is a topic for an advanced statistical mechanics\n", " course.\n", "\n", "4. $$\\rho(z) = e \\rho_+(z) + (-e)\\rho_-(z) = e(\\rho_+(z) - \\rho_-(z))$$\n", "\n", "5. Far away from the plate, the charge density has to be zero,\n", " otherwise the system would not be charge neutral. A globally charged\n", " system has a very high potential energy, that is difficult to\n", " maintain (think about lightning). Locally, a system can be charged,\n", " e.g. near the plate.\n", "\n", "6. Substituting the equations for $\\rho_+$ and $\\rho_-$:\n", "\n", " $$\\begin{aligned}\n", " \\psi''(z) &= -\\frac{\\rho(z)}{\\epsilon_\\mathrm{w}} \\\\\n", " &= - \\frac{e}{\\epsilon_\\mathrm{w}}(\\rho_+(z) - \\rho_-(z)) \\\\\n", " &= -\\frac{e}{\\epsilon_\\mathrm{w}}(\\rho_0 e^{-e\\psi(z)} - \\rho_0 e^{e\\psi(z)}) \\\\\n", " &= \\frac{2e\\rho_0}{\\epsilon_\\mathrm{w}}\\sinh{\\beta e \\psi(z)} \n", " \\end{aligned}$$\n", "\n", "7. This result is obtained after multiplying both sides of equation\n", " {eq}`eq:PB_psi` with $\\beta e$:\n", "\n", " $$\\beta e\\psi''(z) = \\frac{2\\beta e^2\\rho_0}{\\epsilon_\\mathrm{w}}\\sinh{\\beta e \\psi(z)}$$\n", "\n", " and making the substitution. The parameter $\\kappa$ is therefore:\n", "\n", " $$\\kappa^2 = \\frac{2\\beta e^2\\rho_0}{\\epsilon_\\mathrm{w}}$$\n", "\n", "8. $$\\kappa = \\sqrt{\\frac{2 e^2\\rho_0}{k_\\mathrm{B}T\\epsilon_\\mathrm{w}}}$$\n", "\n", " ($k_\\mathrm{B}T = \\beta^{-1}$)\n", "\n", "9. If the equation reduces to\n", "\n", " $$\\phi''(z) = \\kappa^2 \\phi(z)$$\n", "\n", " the general solution is\n", "\n", " $$\\phi(z) = A e^{-\\kappa z} + B e^{\\kappa z}$$\n", "\n", "10. If the potential is set to zero (or any finite value) in the limit\n", " to infinity, the constant $B$ has to be zero, so\n", "\n", " $$\\phi(z) = A e^{-\\kappa z}$$\n", "\n", " With the second condition one can relate $A$ to the surface charge.\n", "\n", " $$E_0 = \\frac{\\sigma}{\\epsilon_\\mathrm{w}} = - \\psi'(0) = - \\frac{\\phi'(0)}{\\beta e} = \\frac{\\kappa A}{\\beta e}$$\n", "\n", " so\n", "\n", " $$A = \\frac{\\beta e \\sigma}{\\kappa \\epsilon_\\mathrm{w}} = \\sqrt{\\frac{\\sigma^2}{2\\rho_0k_\\mathrm{B}T\\epsilon_\\mathrm{w}}}$$\n", "\n", " which is a dimensionless number.\n", "\n", "11. This is an exponentially decaying function. See\n", " {numref}`fig:PB_densities` for the densities.\n", "\n", "12. Using that assumption\n", "\n", " $$\n", " \\begin{aligned}\n", " \\rho_-(z) &\\approx \\rho_0 (1 - A e^{-\\kappa z}) \\\\\n", " \\rho_+(z) &\\approx \\rho_0 (1 + A e^{-\\kappa z})\n", " \\end{aligned}\n", " $$ \n", "\n", "13. There are a few conversions: $k_\\mathrm{B}T = 4 \\cdot 10^{-21}$ J at\n", " room temperature, 0.1 M is equal to $6.022\\cdot10^{25}$ particles\n", " per m$^3$ (convert from liter to get the S.I. unit), the dielectric\n", " constant of water is (depending on circumstances)\n", " $\\epsilon_\\mathrm{w} = 80 \\epsilon_0 = 7\\cdot10^{-10}\\ \\frac{\\mathrm{C}^2}{\\mathrm{N}\\mathrm{m}^2}$.\n", " The screening length at 0.1M at room temperature and ambient\n", " pressure is therefore $\\lambda_\\mathrm{D} \\approx 1$ nm.\n", "\n", " The screening length\n", " $\\lambda_\\mathrm{D} \\propto \\frac{1}{\\sqrt{\\rho_0}}$, so if the\n", " density decreases by a factor 100, the screening length increases by\n", " a factor 10, so $\\lambda_\\mathrm{D} \\approx 10$ nm at 1 mM." ] }, { "cell_type": "code", "execution_count": 1, "id": "8be53c7b", "metadata": { "tags": [ "hide-input", "remove-output" ] }, "outputs": [ { "ename": "ModuleNotFoundError", "evalue": "No module named 'numpy'", "output_type": "error", "traceback": [ "\u001b[0;31m---------------------------------------------------------------------------\u001b[0m", "\u001b[0;31mModuleNotFoundError\u001b[0m Traceback (most recent call last)", "\u001b[0;32m\u001b[0m in \u001b[0;36m\u001b[0;34m\u001b[0m\n\u001b[1;32m 1\u001b[0m \u001b[0mget_ipython\u001b[0m\u001b[0;34m(\u001b[0m\u001b[0;34m)\u001b[0m\u001b[0;34m.\u001b[0m\u001b[0mrun_line_magic\u001b[0m\u001b[0;34m(\u001b[0m\u001b[0;34m'config'\u001b[0m\u001b[0;34m,\u001b[0m \u001b[0;34m\"InlineBackend.figure_formats = ['svg']\"\u001b[0m\u001b[0;34m)\u001b[0m\u001b[0;34m\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n\u001b[0;32m----> 2\u001b[0;31m \u001b[0;32mimport\u001b[0m \u001b[0mnumpy\u001b[0m \u001b[0;32mas\u001b[0m \u001b[0mnp\u001b[0m\u001b[0;34m\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n\u001b[0m\u001b[1;32m 3\u001b[0m \u001b[0;32mimport\u001b[0m \u001b[0mmatplotlib\u001b[0m\u001b[0;34m.\u001b[0m\u001b[0mpyplot\u001b[0m \u001b[0;32mas\u001b[0m \u001b[0mplt\u001b[0m\u001b[0;34m\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n\u001b[1;32m 4\u001b[0m \u001b[0;32mfrom\u001b[0m \u001b[0mmyst_nb\u001b[0m \u001b[0;32mimport\u001b[0m \u001b[0mglue\u001b[0m\u001b[0;34m\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n\u001b[1;32m 5\u001b[0m \u001b[0;34m\u001b[0m\u001b[0m\n", "\u001b[0;31mModuleNotFoundError\u001b[0m: No module named 'numpy'" ] } ], "source": [ "%config InlineBackend.figure_formats = ['svg']\n", "import numpy as np\n", "import matplotlib.pyplot as plt\n", "from myst_nb import glue\n", "\n", "# Densities\n", "A = 0.5 # choose a convenient value for A\n", "z = np.linspace(0,5,100)\n", "negative = 1 - A*np.exp(-z)\n", "positive = 1 + A*np.exp(-z)\n", "\n", "## Make the plot\n", "fig, ax = plt.subplots(figsize=(6,4))\n", "\n", "ax.plot(z, negative, color='#e96868', label='Negative ions')\n", "ax.plot(z, positive, color='#6a8ba4', label='Positive ions')\n", "\n", "ax.legend()\n", "\n", "# Labels\n", "ax.set_xlabel(r'$z/\\lambda_D$'), ax.set_ylabel(r'$\\rho/\\rho_0$')\n", "\n", "# Limits\n", "ax.set_xlim([0, z[-1]])\n", "ax.set_ylim([0.4, 1.6])\n", "\n", "# Save graph to load in figure later (special Jupyter Book feature)\n", "glue(\"PB_densities\", fig, display=False)" ] }, { "cell_type": "markdown", "id": "f2b1a1e8", "metadata": {}, "source": [ "```{glue:figure} PB_densities\n", ":name: \"fig:PB_densities\"\n", "\n", "The density of positive and negative ions near a charged wall.\n", "```" ] } ], "metadata": { "jupytext": { "text_representation": { "extension": ".md", "format_name": "myst", "format_version": 0.13, "jupytext_version": "1.10.3" } }, "kernelspec": { "display_name": "Python 3 (ipykernel)", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.6.8" }, "source_map": [ 12, 120, 151 ] }, "nbformat": 4, "nbformat_minor": 5 }