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Let's suppose I have two finite lines of charge, each of them at some voltage $V$. The first one would be at $x=0$ and the second one at $x=a$, and each of them has the same length $l$.

If I want to calculate the electric potential map I would have to solve the Laplace equation with the boundary conditions $V$ at $x=0,a$. Naively I could say that I can calculate the electric potential map of the line at $x=0$, then do the same with the other one and add them together. This would be wrong, because after calculating the field for the line at $x=0$, $V(x=a)>0$, and when I add the field of the second line, $V(x=a)>V$, which would violate one of the boundary conditions of the problem.

However, if I am given both lines and their respective voltages and asked for the electric field potential at any point, I can use the superposition principle and add the contribution of each one separately.

Why can I use the superposition principle in the second scenario but not when calculating the electric potential map?

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  • $\begingroup$ Wires or plates? And if wires, what does it mean for a wire to be at x=0? $\endgroup$ – nasu May 19 at 22:42
  • $\begingroup$ @nasu Sorry I meant wires, I changed it. Imagine a wire at x=0 parallel to the y axis $\endgroup$ – Psyphy May 19 at 23:36
  • $\begingroup$ Are the charges fixed on the line? I mean, the "line" it's a metalic wire or a linear charge density wich does not change due to the external fields? $\endgroup$ – nasu May 20 at 4:42
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The solution for a line of charge, is a boundary problem with two boundaries; the line segment (at voltage V), and the presumed voltage 0 (at infinity).

The solution will give, among other things, a charge distribution on the line.

The second line at a given voltage, though, will have a similar solution, and similar charge distribution. It will NOT superimpose with the first, however, because the charge distributions aren't fixed; they interact. Basically, if you made the second solution to a problem with THREE boundaries (the original line, the boundary at infinity, and the new line) it could superimpose, as long as the charges in the first line were 'frozen', i.e. not allowed to shift.

The boundaries for one-conducting-line-of-charge and for two-conducting-lines-of-charge are not the same boundaries. Constant voltage in those lines is itself a boundary condition, not a charge distribution. Multiple charge patterns ARE suitable for a superposition, but multiple different boundary conditions are not.

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  • $\begingroup$ Let me know if I understand it correctly. Let's say that I solve the system for the line at $x=0$, and get some map $V_1(\vec{r})$ for a charge density $\lambda$. When adding a second line at some voltage $C$, I am imposing the condition that $V(a)=C$. In order to satisfy this condition, the first line will have to change its charge distribution to some $\lambda'$, therefore the solution of the system can't be obtained as some superposition like $V_1+V_2$ $\endgroup$ – Psyphy May 19 at 23:51
  • $\begingroup$ Yes, the charge distribution (quantity and positions) would change,and violate the 'voltage constant along this segment' boundary condition when the second line-segment-with-charges is added. Added charge induces charge redistribution in a conductor, and a constant-voltage condition means, basically, that the first line is such a conductor. Fields can add, so superposition of charges-causing-fields works, but the constant-voltage line requirement is a boundary condition rather than a charge distribution. $\endgroup$ – Whit3rd May 20 at 6:39
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The problem with this approach is that in the situation with two charged wires, the potential at each of the wires is not just caused by the charge on that wire, but is also influenced by the charge on the other wire.

If you have a situation with two wires at the same potential $V$ and then "magically" remove one of the wires, the potential at the remaining wire will generally no longer be $V$, but will change. How much it will change depends on the exact configuration.

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if I am given both plates and their respective voltages and asked for the electric field potential at any point, I can use the superposition principle and add the contribution of each one separately.

That's not what the superposition principle is meant to be, you can't add solutions of the Laplace equation with different (!) boundary condition and expect the resulting sum to obey both boundary conditions. This wouldn't work even if there was single boundary condition; the sum of two obeying functions can't obey it, unless the potential on the boundary is zero.

Here is what the superposition principle says: generally, there is a source/excitation (in this case, charge density) and there is the effect (in this case, the Coulomb potential function at all points of space). The principle says this: if charge density function $\rho_1(\mathbf x)$ causes potential function $\phi_1(\mathbf x)$, and charge density function $\rho_2(\mathbf x)$ causes potential function $\phi_2(\mathbf x)$, then charge density function $\rho(\mathbf x) = \rho_1(\mathbf x) + \rho_2(\mathbf x)$ will cause potential function $\phi(\mathbf x) = \phi_1(\mathbf x) + \phi_2(\mathbf x)$.

To your question, I don't see a difference between the two scenarios, they seem to be the same, and you can't find the solution for two wires easily from solution valid for one wire. We do not know the charge distribution.

The easiest way to solve this kind of problem is numerically.

The simplest method is to iteratively calculate potential at each point of the discretized space region containing the two wires: for each point (except those corresponding to wires where we already know the potential), the next estimate of potential will be average of values in the neighboring points. After many iterations, the values will approach the solution and obey the boundary conditions.

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