I know the charge density at the plates of a capacitor, from which the applied voltage and potential profile is to be calculated. Using the finite difference method in MATLAB script I solved the Poisson's equation that gives the correct result for all kind of continuous charge profiles. I checked constant (step function), linearly increasing/decreasing charge profile, in all cases numerical results agree with analytically calculated potential profile. But when I put the delta-function charge to study the original problem (in figure), it gives spooky results. (With $\rho=13.2$$\mu$C/$cm^2$, $\epsilon_r$=30, $x$=20nm, I would expect the voltage to be 9.92V,while I am getting 40nV!)

In the finite different approach:

$d^2V/dx^2$=-$\rho(x)$/$\epsilon$ in matrix form is $Au$=$h^2$$f$, where $f$=-[$\rho(i)$/$\epsilon$]' and $i$=1,2,......n are the discrete nodes. In my problem, I have $-\rho(1)$=$\rho(n)$=$13.2$$\mu$C/$cm^2$, and $\rho(i)=0$ elsewhere. Therefore I take $f$=$-\rho(i)$/$\epsilon$[-1 0 0 0 .......... 0 0 0 1]'. Then potential is obtained by $u$=$A^{-1}$$f$

My question is, what is the physical reason for such anomaly?

How can I solve this issue and make this code work? enter image description here enter image description here



clear all;

t=20;  %Thickness value in nm


epsil=epsil_r*8.854*10^-14; %In CGS unit

n=500;              %Number of grid points

x1=0; x2=t*10^-7; %Thickness in cm


rho=13.2*10^-6;     %Strength of delta-function charge, C/cm^2


A(n,n)=-1;        %Neumann boundary at right node

f=-rho/(epsil)*[-1;zeros(n-2,1);1]; %Right hand side of Poisson.

u=inv(A)*[h^2*f];              %Potential
  • 1
    $\begingroup$ Would Computational Science be a better home for this question? $\endgroup$ – Qmechanic Apr 30 '17 at 16:40
  • $\begingroup$ I think people will be hesitant to discuss the physical behaviour of a simulation whose code they could not verify ... ;) $\endgroup$ – Sanya Apr 30 '17 at 18:59
  • $\begingroup$ "What is the physical reason for such anomaly?" Either your computer is broken, or (more likely) there is a bug in your code. If you want the code debugged without showing it to anybody you need a mind-reading forum, not SE. $\endgroup$ – alephzero Apr 30 '17 at 23:48
  • $\begingroup$ Dear altruist, I added the code in the question. I was wondering if there is anything fundamental for which delta-function charge can not be used, or it is all about my code. I am expecting your guideline. $\endgroup$ – Alam May 2 '17 at 9:32
  • $\begingroup$ I don't understand the code in this question. There's a bunch of definitions, but I don't see what actually acts on those objects. How you plot things is not important, what's important is how you solve things. $\endgroup$ – Emilio Pisanty May 2 '17 at 10:10

Your units don't add up. You're solving for the equation $$ \frac{\mathrm d^2 u}{\mathrm dx^2} = f $$ where $f = \rho/\epsilon$. If you take your solution at face value, then $u$ will have the dimensionality \begin{align} [u] & = [h^2 f] = [h^2\rho/\epsilon_0] = [Q/\epsilon_0]. \end{align} Since $[\epsilon_0]=[Q/(VL)]$, this means that $[u]=[VL]$, a voltage times length, which is obviously not what you intended. Since the units don't match up, it is meaningless to talk about inconsistencies in the numbers - if you did this in some other unit system you would get a different answer.

The problem is in your handling of the surface charge associated with the delta function. The right-hand side of the Poisson equation requires a legitimate volumetric charge density, or something dimensionally equivalent to that. If $\sigma$ is a surface charge density, with $[\sigma]=[Q/L^2]$, then $$\rho(x,y,z) = \sigma\delta(x-x_0)$$ is a correct volumetric charge density because $[\delta(x-x_0)] = [1/L]$, but your discretization procedure has dropped this fact out.

To fix this, I would suggest dropping the pretense that you have a true singular charge density and instead model it as a finite slab of thickness $d$ and volumetric charge density $\sigma/d$, where you can set $d$ to your discretization length if you want. Ultimately, the value of $d$ should be unimportant to your finite result - but it's crucial that you get your units right.

  • $\begingroup$ Thank you very much, its giving perfect result now. I just needed to convert the surface charge density into "per volume" unit. @Emilio Pisanty $\endgroup$ – Alam May 2 '17 at 16:09

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