Calculate a charge distribution given an electric potential.

  • #1

Homework Statement


Find the distribution of charge giving rise to an electric field whose potential is $$\Phi (x,y) = 2~tan^{-1}(\frac{1+x}{y}) + 2~tan^{-1}(\frac{1-x}{y})$$where x and y are Cartesian coordinates. Such a distribution is called a two-dimensional one since it does not depend on the third coordinate z.[/B]


Homework Equations


Poisson's equation: $$\nabla^2 \Phi = -4 \pi \rho$$[/B]


The Attempt at a Solution


The first thing I noted while attempting to solve this problem is that there is a singularity when x = ±1 and y = 0. Either way, I wanted to compute the laplacian of the electric potential to see what it would result in and it reduced to zero. I thought I had made an algebraic mistake and put it in Mathematica, again reducing to zero. Then I actually took my original thought and tried to work out a solution to avoid the singularities using the Dirac delta function, the only problem is that I don't know how I could operate the Dirac delta function. Is is possible to work out the following equation? $$ \Phi (x,y) \delta(x) \delta(y) = \Phi (0,0) \delta(x) \delta(y) \\ \nabla^2 \Phi (x,y) \delta(x) \delta(y) = \nabla^2 \Phi(0,0) \delta (x) \delta(y) $$ Although I don't think my steps will lead to something useful.
 

Answers and Replies

  • #2
The first thing I noted while attempting to solve this problem is that there is a singularity when x = ±1 and y = 0.
The potential does not have a singularity at these points, it is finite valued.
Either way, I wanted to compute the laplacian of the electric potential to see what it would result in and it reduced to zero.
Then the charge distribution must be zero. The reason the potential is not zero or some other constant is likely due to some imposed boundary conditions.
 
  • #3
I suppose you could also find the the charge distributions on the bounding surfaces by relating the first derivative of the potential to the surface charge density on a bounding plane at ##y=0## and ##x=\pm1##. The surface charge density can then be related to the volume charge density using a delta function.
 
  • #4
Orodruin
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Then the charge distribution must be zero. The reason the potential is not zero or some other constant is likely due to some imposed boundary conditions.

This is not correct. The potential may not approach infinity, but it does have a discontinuity over the entire segment ##-1 < x < 1## at ##y = 0##. The appropriate way of going about this is to note that this discontinuity generally will lead to ##\partial_y V = A(x) \delta(y) + f(x,y)##, where ##f(x,y)## is the continuous function you would get by naively performing the differentiation. Not to give too much away, I will leave it at that for the OP.
 
  • #5
The appropriate way of going about this is to note that this discontinuity generally will lead to ##\partial_y V = A(x) \delta(y) + f(x,y),## where ##f(x,y)## is the continuous function you would get by naively performing the differentiation. Not to give too much away, I will leave it at that for the OP.
I understand I would need the Dirac delta function, what I don't quite understand is your solution method. What would ##A(x)## represent? Also, as I have mentioned before, what you name ## f(x,y) ## is actually zero once I calculate the laplacian, so I will be left with just the expresion ## \partial_y V = A(x) \delta(y) ,## which, if I understand correctly, is the first partial derivative with respect to ##y##.
 
  • #6
Orodruin
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##A(x)## would generally be a function of ##x## to be determined by looking at the size of the discontinuity of the potential. ##f## would not be zero, it represents the y-component of the field (this is the first derivative wrt y, not ##\nabla^2V##). You will need to differentiate one more time to compute ##\partial_y^2 V##.
 

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