# Image method

1. Jan 5, 2016

### aaaa202

I'm solving a problem numerically where I have some charge density above an infinite, grounded conducting plane and want to determine the electrical potential at a given point. My intuition says that this is not simply given by the potential of the charge density above the plane, since this will also attract charges of opposite sign to the grounded plane.
To solve the problem I employ the method of Images such that the same charge density with opposite sign is found below the conducting plane. I then wish to calculate the electric potential. To do so I use Poissons equation:
2φ = -ρ/ε
From this I can find the charge density by inverting the laplacian, which I have in matrix form (remember all this is done numerically, my density is a vector of values).
The problem with doing so is that I get the exact same potential, as I would if I had just ignored the mirror charge distribution and worked with only the original charge distribution (which I guess is not so surprising). But I am pretty sure this is wrong. If you for example do the problem with a point charge above the grouned plane it is easy to see, that the mirrored charge will affect the potential.
What am I doing wrong?

2. Jan 6, 2016

### andrewkirk

I'm not familiar with the method of images but, drawing some pics on the back of an envelope and mulling over it, it seems to me that we need to calculate the vertical and horizontal components of the electric field separately.
I think the horizontal component would be the same as calculated by ignoring the conducting plate and assuming a charge of opposite sign at the mirror image position.
But in calculating the vertical component I think the assumed image charge should have the same sign. Why? Because the field needs to be zero everywhere on the plane, and only a same sign charge will achieve that in the vertical direction.

3. Jan 6, 2016

### BvU

DId you take the boundary condition (at the grounded plate $\vec E = E\perp$) into consideration ?

4. Jan 6, 2016

### aaaa202

hmm no, but how should I do that? I thought that was automatically taken care of by the mirror charge density.

5. Jan 6, 2016

### BvU

This is highly suspicious. What do you get when you use just one charge and one (opposite polarity) mirror charge ? You should get a dipole field, but according to your statement the mirror charge is ignored ?

6. Jan 6, 2016

### BvU

Well, apparently you don't get V = 0 on the grounded plate.

7. Jan 6, 2016

### aaaa202

I do. I get the following results. I mirror the charge density on the grounded plane at x=0. I then calculate the electrostatic potential by inverting the second order derivative in Poissons equation. The problem is that I get the exact same potential as I would have, if I had simply ignored the mirror charge density. In the case I work on the domain of only the positive position values above and invert the second order derivative given by the usual matrix expression for this in the finite element method. What am I doing wrong?

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8. Jan 6, 2016

### BvU

What are we looking at ? Position in which direction ?
A drawing, perhaps?
I sure don't expecet pictures like that !

9. Jan 6, 2016

### aaaa202

Well my problem is one dimensional so call the position whatever coordinate you like, say x. The direction is normal to the grounded plane at x=0.

10. Jan 6, 2016

### BvU

One dimensional ? You are solving $${d^2\phi\over dx^2}=-\rho(x)/\epsilon_0$$ ? What rho and what boundary conditions ?

11. Jan 6, 2016

### aaaa202

My rho is the part of the density of the picture defined on positive positive values. It is a vector of values, i.e. the problem is numeric.

12. Jan 6, 2016

### vanhees71

I don't understand what you mean by "onedimensional problem". In 3D it's straight forward.

Let's take the $xy$ plane as your conducting infinite plate with some charge distribution at $z>0$ and none at $z<0$. Then your potential is $V(\vec{x})=0$ for $z<0$, and we have to solve the problem for the half-space $z>0$ with the boundary condition $V(\vec{x})|_{z=0}=0$.

First we construct the Green's function of this problem, i.e., the potential for a unit charge at $\vec{x}'$, fullfilling
$$G(\vec{x},\vec{x}')=-\delta(\vec{x}-\vec{x}'), \quad z'>0, \quad G(\vec{x},\vec{x}')|_{z \geq 0}=0.$$
Then your problem is solved by
$$V(\vec{x})=\int_{z'>0} \mathrm{d}^3 \vec{x}' G(\vec{x},\vec{x}') \rho(\vec{x}'),$$
because then
$$\Delta V(\vec{x})=-\rho(\vec{x}), \quad V(\vec{x})|_{z \leq 0}=0.$$
To find $G$, indeed the image-charge method is very convenient in this case, because obviously you get the correct boundary condition by just putting a unit charge at the mirror point $(x',y',-z')$ in the lower plane. Then for any $\vec{x}$ in the upper plane the potential reads
$$G(\vec{x},\vec{x}') = \frac{1}{4 \pi \sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}-\frac{1}{4 \pi \sqrt{(x-x')^2+(y-y')^2+(z+z')^2}}.$$
Now you only need to integrate your given charge distribution (perhaps numerically).

13. Jan 6, 2016

### BvU

Well, you found the solution. So what is the problem ?
Apparently let you $\rho$ be approximatey constant and V = 0 at 0 and at $\pm$ 20 nm (why is that ?)