# I Simple electric potential and Laplace equation

#### dRic2

Gold Member
Imagine to be in 2 dimensions and you have to find the potential generated by 4 point-charges of equal charge located at the four corners of a square.

To do that I think we simply add all the contributions of each single charge:
$$V_i(x, y) = - \frac k {| \mathbf r - \mathbf r_i|}$$
$$V(x, y) = \sum_i^4 V_i(x,y)$$
where $\mathbf r_i$ is the location of each charge. In particular if I choose the origin of the cartesian coordinates at the center of the square I get (the side of the square was set equal to 2):
$\mathbf r_1 = (-1, -1)$
$\mathbf r_2 = (-1, +1)$
$\mathbf r_3 = (+1, +1)$
$\mathbf r_4 = (+1, -1)$

Now. If I plotted it correctly I get something like this:

Which clearly has a maximum.

Now consider a circle centered at the origin but smaller than the square so that it contains no charges. Here I can write the Laplace equation:
$$\Delta V = 0$$
$$+ \text{boundary conditions}$$

A particular property of solutions of the Laplace equation is that they can have no local minimum or maximum: al extrema must occur at the boundary. This follows from the property of harmonic functions.

What am I doing wrong here ?

Thanks fro the help.

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#### BvU

Homework Helper
wikipedia said:
other than the exceptional case where f is constant
work out $V(0,0)$ and conclude that ${\partial^2 V\over\partial x^2} = 0$ (idem $y$) there !

 I missed a ${1/over 2}$ -- twice -- so I ended up with non-zero. Puzzled...

btw: nice picture
And I dont's see the $\pm \sqrt 2$, just $\pm 1$

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#### dRic2

Gold Member
And I dont's see the ±√2±2\pm \sqrt 2, just ±1±1\pm 1
Sorry, I corrected the graphs. It looks flatter but numerically I'm still able to find a maximum value at the origin.

work out $V(0,0)$ and conclude that ${\partial^2 V\over\partial x^2} = 0$ (idem $y$) there !

 I missed a ${1/over 2}$ -- twice -- so I ended up with non-zero. Puzzled...
Sorry, I'm not following. Can you explain a little more please ?

#### BvU

Homework Helper
I added the four $V_i$ at $(\varepsilon,0)$ and figured the $\varepsilon$ and $\varepsilon^2$ terms would vanish -- but they don't ...

Let the four charges be located at $(1, 0), (0,1), (-1,0), (0,-1),$ then $$V_i(\varepsilon,0)={1\over 1-\mathstrut\varepsilon}+{1\over \sqrt{1+\varepsilon^2}}+{1\over 1+\varepsilon}+ {1\over \sqrt{1+\varepsilon^2}}$$

use Taylor and I'm left with $4+\varepsilon^2\qquad$ -- logical because of symmetry, but I expected 2nd and 3rd order terms to vanish ... they don't and that leaves me puzzled

#### Dale

Mentor
A particular property of solutions of the Laplace equation is that they can have no local minimum or maximum: al extrema must occur at the boundary. This follows from the property of harmonic functions.

What am I doing wrong here ?
You are looking in only 2D for a function which is in 3D. If you plot the solution in the z direction you will see that the origin is not in fact a maximum, it is a saddle point. It is a maximum along lines parallel to x or y, but a minimum along lines parallel to z, thus it is a saddle point. The maxima and minima do indeed occur on the boundaries.

Here is a plot in x and z, where I plotted from -0.5 to +0.5 on each axis instead of -1 to +1 so that the shape is more visible. The same trend holds further out but it is harder to see.

#### dRic2

Gold Member
I see. Thank you. But this leads me to think that I can't write a Laplace equation in 2 dimension for the the above problem, right ?

#### BvU

Homework Helper
Kudos @Dale !

to forget ${\partial^2 V\over\partial z^2} < 0$

#### Dale

Mentor
I see. Thank you. But this leads me to think that I can't write a Laplace equation in 2 dimension for the the above problem, right ?
Well, you certainly can write a Laplace equation in 2 dimensions, but I don't know how you would write Maxwell's equations in 2 spatial dimensions. Also, if you did write Maxwell's equations in 2 dimensions I don't know what the potential would look like and I don't know if it would still satisfy Laplace's equation (I suspect not). So the 2-D issue isn't actually with Laplace's equation but Maxwell's.

#### dRic2

Gold Member
Yes, I meant Laplace equation as a consequence of 1st maxwell equation (gauss equation) written for the potential instead.

"Simple electric potential and Laplace equation"

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