Simple electric potential and Laplace equation

In summary: Thank you for your help!In summary, the conversation discusses finding the potential generated by four point charges located at the four corners of a square in a 2-dimensional space. The method involves adding the contributions of each individual charge, with the location of each charge being represented as ##\mathbf r_i##. The Laplace equation is used to find the potential at different points, and it is noted that the extrema must occur at the boundary due to the properties of harmonic functions. A plot is used to show that the origin is actually a saddle point rather than a maximum, and the possibility of writing a Laplace equation in 2 dimensions for this problem is discussed.
  • #1
dRic2
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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:
243082


243083

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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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  • #2
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 !

[edit] o:) I missed a ##{1/over 2}## -- twice -- so I ended up with non-zero. Puzzled...

btw: nice picture :smile:
And I dont's see the ##\pm \sqrt 2##, just ##\pm 1## :rolleyes:
 
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  • #3
BvU said:
And I dont's see the ±√2±2\pm \sqrt 2, just ±1±1\pm 1 :rolleyes:

Sorry, I corrected the graphs. It looks flatter but numerically I'm still able to find a maximum value at the origin.

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

[edit] o:) 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 ?
 
  • #4
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
 
  • #5
dRic2 said:
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.

saddle2.jpg
 
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  • #6
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 ?
 
  • #7
Kudos @Dale !

o:) to forget ##{\partial^2 V\over\partial z^2} < 0 ##
 
  • #8
dRic2 said:
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.
 
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  • #9
Yes, I meant Laplace equation as a consequence of 1st maxwell equation (gauss equation) written for the potential instead.
 

FAQ: Simple electric potential and Laplace equation

1. What is electric potential?

Electric potential is a measure of the amount of electric potential energy per unit charge at a given point in space. It is often described as the amount of work needed to move a unit positive charge from an infinite distance away to a specific point in an electric field.

2. How is electric potential related to electric fields?

The electric potential at a point is directly proportional to the strength of the electric field at that point. In other words, the higher the electric potential, the stronger the electric field and vice versa. This relationship is described by the equation V = Ed, where V is electric potential, E is electric field, and d is distance.

3. What is the Laplace equation?

The Laplace equation is a partial differential equation that describes the relationship between the electric potential and the charge distribution in a region of space. It is often used to solve problems involving simple electric potential, such as those in electrostatics.

4. How is the Laplace equation solved?

The Laplace equation can be solved using various mathematical techniques, such as separation of variables, Fourier series, and numerical methods. The specific method used depends on the boundary conditions and symmetry of the problem being solved.

5. What are some real-life applications of simple electric potential and the Laplace equation?

Simple electric potential and the Laplace equation have many practical applications, including in the design of electronic circuits, the calculation of electric fields in medical imaging techniques, and the analysis of electric potential in conductors and insulators. They are also used in the study of fluid flow and heat transfer in engineering and physics.

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