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Suppose there are two charges (4uC each) fixed in the horizontal axis. One is in x=0 and the other in x=8m.

I've obtained the electric field:

$E=-k\cdot4\mu C \cdot [\frac{1}{x^2}+\frac{1}{(x-8m)^2}], x<0$

$E=k\cdot4\mu C \cdot [\frac{1}{x^2}-\frac{1}{(x-8m)^2}], 0<x<8m$

$E=k\cdot4\mu C \cdot [\frac{1}{x^2}+\frac{1}{(x-8m)^2}], x>8$

Now I had to find the points in space where the field is 0. So I've solved each part and obtain x=4m (the middle point between the two charges). But looking on the given answers it says that $x=\pm\infty$ is also a solution.

I know that the limit is 0, but I'm not sure how to arrive to that solution and if it has a physical meaning.

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closed as off-topic by John Rennie, Jon Custer, stafusa, Kyle Kanos, ZeroTheHero Aug 8 at 12:55

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  • $\begingroup$ The physical meaning would be that if you are as you go away from the source the electric field tends to 0. Also, you could think that any function f(r) ~ 1/r^n, will go to 0 as r tends to infinity, so a solution to f(r)=0 will always be +- infinity $\endgroup$ – IvanMartinez Aug 7 at 4:02
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I would personally have marked $x=\pm\infty$ wrong (though possibly made a note and not taken any points off), unless I had specifically asked you to include them along with the real-valued solutions.

There is physical meaning there, but it is in the limiting sense, as you say. Ultimately this is a discussion to have with your instructor, but I think you have the right idea.


Incidentally, the electric field due to a point charge in a 1D universe is not inversely proportional to the square of the distance - it would actually be constant, which would drastically affect your answers. This is outside the scope of your course, so I'm just mentioning it as an aside; a better way to phrase the question would be to consider a standard 3D universe like the one we live in, and to simply restrict your attention to the $x$-axis (though there would be no solutions off of the $x$-axis anyway).

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  • $\begingroup$ Thanks for the answer. I will discuss it with my instructor. As you suggested I rephrased the question to avoid future confusions. $\endgroup$ – Giuliano306 Aug 7 at 4:56
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Maxwell's equations for electrostatics give you the curl and divergence of the electric field in terms of the sources. However, like any other differential equation for a scalar function, these two differential equations for the vector field do not determine the field uniquely. What is required to completely specify the solution is boundary conditions.

To derive the electric field for the point charge, the boundary condition imposed is that electric field goes to zero at infinities(or equivalently, the potential becomes constant at infinities)-which makes sense if you think about it. Note that you cannot impose this boundary condition for a charge configuration that is itself infinite.

If you want to make sense of why the field of a local charge should die off, think about Gauss's law. Roughly, the flux(which can be interpreted to be the number of field lines) is constant as you take larger and larger surfaces around the local charge. Thus, the electric field, which is measured by the number of lines per unit area, goes on decreasing.

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