I have a conducting plate on $x$-$y$ plane. So I have a boundary condition at $z=0$ $\Phi=0$ but, for $z>0$ I have a point charge at z=a which is expected to create a potential. $$ \nabla^2\Phi=\frac{\rho}{\varepsilon_0} $$ I need a Green function which can be assigned as : $$ G(r,r')=\frac{1}{|\sqrt{(x-x')^2+(y-y')^2+(z-a)^2}|} $$ But this Green function does not satisfy Laplace $\nabla^2G=0$ at $z = 0$. Right? So I need an additional function to also satisfy Laplace such as : $$ G(r,r')=\frac{1}{|r-r'|}+F(r,r') $$ here is my question starts : What is proper function for F and how am I supposed to assign it ? How should I use the Laplace and Poisson properly?

I have chosen this unique and common problem to understand the Green function solutions but mainly my aim is to understand the use of Laplace and Poisson with boundary conditions.

  • $\begingroup$ Are there any charges in your problem? $\endgroup$
    – jinawee
    Jan 11, 2015 at 20:50
  • $\begingroup$ aah yes my bad, I forgot to add point charge which is at z=a and the image opposite. $\endgroup$
    – aQuestion
    Jan 11, 2015 at 20:52
  • $\begingroup$ And the Green function seems wrong, shouldn't the denominator have a square root of squares? $\endgroup$
    – jinawee
    Jan 11, 2015 at 20:54
  • $\begingroup$ Hi Major_Tom. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$
    – Qmechanic
    Jan 11, 2015 at 21:55
  • $\begingroup$ I don't understand the question. You state that you know that $\vec \nabla^2 G(r,r') = 4\pi\delta(r-r')$, and then you say "But this equation should also satisfy Laplace", but an equation cannot "satisfy" another equation. I don't know what you are trying to do. $\endgroup$
    – ACuriousMind
    Jan 11, 2015 at 21:59

1 Answer 1


I have to say that I don't fully understand your question, and there still seem to be inconsistent factors of $4\pi$. But I want to get at the heart of the Green function for your boundary conditions, and then you can always ask another question about how to handle it for other boundary conditions.

$$G(r,r')=\frac{1}{\sqrt{(x-x')^2+(y-y')^2+(z-a)^2}}$$ is not your Green function, and neither is

$$G(r,r')=\frac{1}{\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}$$, but I think


is the function you are looking for. It tells you the potential at $r$ due to a charge at $r'$, and when $z=0$ it gives $G=0$. Now for a charge distribution $\rho$ you can integrate to get $\int G(r,r')\rho(r')dx'dy'dz'$

The point of a Green function is that if you can find the solution at $r$ due to a single unit charge at an arbitrary point $r'$ that meets your boundary conditions, and call that function $G(r,r')$ then the work you did to get $G$ now allows you to solve for any charge distribution $\rho$ by doing an integral to get $V(r)=\int G(r,r')\rho(r')dx'dy'dz'$.

One Green function $G(r,r')$ that allows you to get $V(r)$ for many different $\rho$ is the whole point.

  • $\begingroup$ I remember that I have learned if you are working in 3 dim. Green should give you something with factor $4\pi$ and in 2 dim. factor of $2\pi$ with log function. For 1-dim it is basically $\frac{1}{|r-r'|}$ as your answer above. My question is how can we come up with these solutions, indeed. And how can you prove your solution gives you poisson for z > 0 and zero at z = 0 $\endgroup$
    – aQuestion
    Jan 11, 2015 at 22:34
  • $\begingroup$ and briefly doesn't your Green function solution hold $\phi=\phi_1+\phi_2$ which $\phi$ should be simply sum up of Coulomb potential for point and the image charge? And should be factor of $\frac{1}{4\pi}$ ? $\endgroup$
    – aQuestion
    Jan 11, 2015 at 22:46
  • $\begingroup$ If you want $\nabla^2 \Phi = \rho/\epsilon_0$ then you want a factor of $\epsilon_0$ too, putting in a constant factor isn't a big deal (you need it, but it is easy to do). Finding a Green function for different boundary conditions is a very complex task, what kinds of regions and boundary conditions are you looking at? $\endgroup$
    – Timaeus
    Jan 11, 2015 at 23:26
  • $\begingroup$ correct me if I am wrong. I take the second derivatives of your solution and try to get Laplace. But my calculations does not hold it. I make mistake I think. The order of denominator should get bigger and bigger.. $\endgroup$
    – aQuestion
    Jan 12, 2015 at 0:03
  • $\begingroup$ Sorry laplace should hold at z' = 0 and it does $\endgroup$
    – aQuestion
    Jan 12, 2015 at 0:13

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