# Solving Poisson's equation ##\nabla^2\psi##

## Homework Statement:

Consider the situation of a point charge q a distance d above an infite
plane that is maintained at a potential \V_0. Solve the Poisson's equation for all points
above the plane. Also compute the surface charge density and the total charge on the
plane.

## Relevant Equations:

Poisson equation
I know the separation of variables is gonna be used but I am confused because we usually solve a Laplace equation for this

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Paul Colby
Gold Member
Have you had or seen the method of images?

• PhDeezNutz
yes I have seen it.

Paul Colby
Gold Member
Cool, so the expression obtained by reflecting the charge in the ground plane (AKA ##V = 0## plane) is the solution to Poission's equation. One is using two known solutions and the linearity of PE to construct a solution meeting the stated boundary conditions which are ##V=0## on the plane and ##V=0## at infinity.

Do you know the general relation of surface charge to the normal electric field?

• etotheipi and PhDeezNutz
yes its the differential of Potential wrt the normal vector gives the surface charge density.

But here I need to solve the Poisson equation because my plane has a constant potential (V)

I am actually confused we can use the method of images but here we need to solve the posions equation straight forwardly?

One thing about Poisson’s equation (and virtually any other partial differential equation to my knowledge) is that the solution inside the region of interest (in your case above a plane) is unique as long as the boundary conditions are maintained. So you can change anything outside the region of interest as long as the boundary conditions are maintained (Uniqueness Theorems).

Let me ask you if a point charge (+q) is a distance (d) above an infinite plane and a point charge (-q) is a distance (d) below it aren’t all points on the plane at the same potential (conventionally designated as zero)?

Hence why the poster above suggested method of images.....it takes advantage of the uniqueness theorem to solve Poisson’s equation.

• etotheipi and Delta2
I am actually confused we can use the method of images but here we need to solve the posions equation straight forwardly?
method of images is a way of solving it directly in my opinion. It takes advantage of the uniqueness theorem and without uniqueness theorems physics would be pointless; different solutions to the same thing would be invalid.

just my two cents.

I am almost certain your instructor wants you to use method of images but if you’re concerned you should email him/her.

• etotheipi
he says perhaps there is another source setting the potential \V_o..
so maybe he's talking about image charges but here he mentions no such thing

• PhDeezNutz
he says perhaps there is another source setting the potential \V_o..
so maybe he's talking about image charges but here he mentions no such thing
That’s exactly what he’s talking about (image charges). He doesn’t mention it because he was hoping for it to “jump out at you”.

• etotheipi
you mean I need to write the potential directly according to the situation using the method of images? and put that in the Poisson Eq. to show it is satisfied. I also put the boundary condition to show they are also satisfied?

you mean I need to write the potential directly according to the situation using the method of images? and put that in the Poisson Eq. to show it is satisfied. I also put the boundary condition to show they are also satisfied?

Yes but it’s easier than you think If you just appeal to the uniqueness theorems.

1) Write down the combined potential of two equal but opposite point charges a distance (d) away from each other. Set up the coordinate axes however you like. I prefer to put (+q) at (0,0,d) and (-q) at (0,0,-d).

Do this part on your own.

2) Mention the uniqueness theorem that says you can change the exterior as long as the interior/region of and boundary conditions are maintained.

3) Then your answer from 1) automatically satisfies Poisson’s equations because you didn’t change anything in the interior/region of interest nor the boundary conditions.

• etotheipi and Delta2
in that combined potential I need to write a /V_o term as well,right?

in that combined potential I need to write a /V_o term as well,right?
I don't think that's actually necessary. You could if you wanted to; just add "+V_0" at the end. But remember a scalar potential is uniquely determined up to a constant.

I could call a first point A and a second point B.

You could call them C and D.

As long as B-A = D - C we both agree on the same answer.

vela
Staff Emeritus
Homework Helper
plane that is maintained at a potential \V_0. Solve the Poisson's equation for all points
he says perhaps there is another source setting the potential \V_o..
in that combined potential I need to write a /V_o term as well,right?
Just out of curiosity, you keep putting a slash before V_0 (or V_o). What do you mean by that?

I try to write in Latex form but I don't know why it doesn't show up like that

Paul Colby
Gold Member
What does the potential being constant on the ##z=0## plane say about the direction of the electric field on the plane?

BTW solving Poisson's equation doesn't imply separation of variables or any other specialized method must be used. There are 11 orthogonal coordinate systems in which separation of variables may work with this general problem, ##\nabla^2 V = 0##. Even so the resulting solution could well involve an infinite series of special functions which converges to a simple form (I know because the method of images will yield a simple closed form directly) but you'll still be stuck at summing some god awful series.

• etotheipi, PhDeezNutz and Delta2

The exact value of the potential (or potential energy) is arbitrary up to a constant, the only quantity that really matters is the change in potential (or potential energy) from one point to another.

If I drop a ball from 10 meters to 5 meters I've dropped it 5 meters. What if you are 5 meters beneath me? From your perspective it dropped from 15 meters to 10 meters. Either way the change in potential energy is the same (mass)(-9.8)(5).

I feel like you're making this much more difficult than it is, i'll try my best to help you without solving it for you

Do we agree that for two point charges the potential is the following?

##V = \frac{1}{4 \pi \epsilon} \frac{q_1}{\sqrt{\left(x - x_1 \right)^2 + \left(y - y _1\right)^2 + \left(z - z_1\right)^2}}
+
\frac{1}{4 \pi \epsilon} \frac{q_2}{\sqrt{\left(x - x_2 \right)^2 + \left(y - y_2\right)^2 + \left(z - z_2\right)^2}}##

Given the situation from your problem what would be the most convenient choices for ##q_1,q_2, x_1, y_1, z_1, x_2, y_2, z_2##? My advice is to let the plane of constant potential be in the ##xy## plane.

How would you then evaluate the potential at the plane and what do you get? How would this then be amended to account for ##V_0##?

Last edited:
• etotheipi, AHSAN MUJTABA and Paul Colby
yes, I agree with your first point.
But if lets say x,y goes to infinity then ##V_o## will also become zero. Just asking
##q_2=-q_1## and ##z_2=+d and -d##

I will say my ##V_0## outside the volume of interest in 0. Right?

yes, I agree with your first point.
But if lets say x,y goes to infinity then ##V_o## will also become zero. Just asking
##q_2=-q_1## and ##z_2=+d and -d##
I agree with your choices of ##q_1,q_2, z_1, z_2##

The problem says the entire xy-plane (no matter how far away we are) is at the same potential that’s the point. In your case it’s zero with the formula I previously mentioned. But you don’t want zero at the xy-plane you want V_0. Wouldn’t you just add V_0 to your previous answer?

okay but will the boundary conditions satisfy? because adding ##V_0## they won't be satisfied

okay but will the boundary conditions satisfy? because adding ##V_0## they won't be satisfied
the only boundary conditions to satisfy is that at the middle plane the potential is V_0. Without the +V_0 term the potential would be zero at the middle plane.