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Effect of electric field on a wire segment

  1. Jul 14, 2015 #1
    Hello. I was trying to understand the mathematics underlying Faraday cages. After hours of research on the web, It turns out that there are very few mathematical analysis related to Faraday cages, even in the electrostatic case, and those that I read do not satisfy me. So, I tried to solve the problem by myself in the electrostatic case (the dynamic case of waves, which probably involves reflection phenomenons and complicated physics is beyond my skills). While trying to solve this problem, I have come to the following much simpler problem, that I cannot solve as well.

    PROBLEM : In a system of axes x,y,z, assume that there is a rectilinear segment of copper wire located at [0 1]. Assume an electric field parallel to the x direction is applied in the space (we do not assume it is uniform, but this can be assumed in first analysis). For example, one can put a charge at x = -1 to create the electric field. Actually, for what we have to say, only the electric field on the x axis will play a role. Finally, we assume that the system has reached electrostatic equilibrium : no more charges move inside the wire.

    QUESTIONS :
    1) what is the distribution of the charges in the wire ?
    2) is it really true, as asserted by many, that the electrical potential inside the wire is constant (not so simple as it looks !)

    WHAT I HAVE READ AND WHAT I HAVE TRIED :
    Well, it is often said that in conductors, the electrical potential is constant (electrostatic case), for otherwise free charges would move from the point of higher potential to the lower one. Even if this seems a good argument, it is subject to caution : the free electrons in the wire are ultimately limited in number, and there may be no more free charges that can move at some point. So, if the field is very strong, there may be a non constant potential inside the wire : the free electrons move somewhere in the wire and do their best to annihilate this potential, but they cannot do it completely.
    Let us model the charges inside the wire as a perfect gas of positively and negatively charged particles, forced to lie inside the segment [0 1]. Let ##\rho_+(x)## and ##\rho_-(x) (>0) ## be the positive and negative charge density at x, and ##\rho(x)=\rho_+(x) - \rho_-(x)##; so the charge inside a segment ##dl## around a point x is ##\rho(x)dl##. Actually, since the (positive) nucleus of the atoms inside the wire cannot move, this model is somewhat incorrect, but I neglect this issue (it can be rendered by the equation ##\rho_+(x) < K##, where ##K## is the maximal positive charge density at a point in the wire).
    I denote by ##V_0(x,y,z)## the electrical potential associated to the applied field at a point ##(x,y,z)##, by ##V(x,y,z)## the total electrical potential at any point, and by ##E(x,y,z)## the electric field.
    We have several equations :
    *##\int_0^1 \rho_+(x) dx = \int_0^1 \rho_-(x)dx = C##, for some constant ##C## (the numbers of positive and negative charges are finite and equal);
    * in particular, ##\int_0^1 \rho(x) dx = 0##;
    * ##\nabla^2 V(x,0,0) = {\rho(x)\over \varepsilon}##, with ##x\in [0,1]## and ##\varepsilon## the conductivity inside the wire (Maxwell equation);
    * With ##r=(x,y,z)##, ##V(r) = V_0(r) + \int_0^1 {\rho(w)\over 4\pi \varepsilon |r-(w,0,0)|} dw## (the total potential is the sum of the applied field potential with the contributions of the potentials created by the infinitesimal charges ##\rho dw## inside the wire). Notice that this integral is singular at w = x, and has to be understood in the Cauchy principal value sense.

    That's all, and this leads to an integral equation with respect to ##\rho(x)##, together with the constraints expressed in the first equations. I have no Idea of how to solve for ##\rho##, even assuming some simple form for the electric field ##E##. As noted above, under reasonable applied field strength, there should hold ##V=0## along the wire, but this may not be true if the field is very strong, and in this case, I don't know how to solve for the potential ##V## too. Any idea or critical observations will be welcome.
     
    Last edited: Jul 14, 2015
  2. jcsd
  3. Jul 14, 2015 #2
    This is just an idea. Maybe, when the external field is applied, the charges in the wire will separate into + and - and will crowd towards the ends (electrons drawn towards the + direction). There is no shortage of free charges in a metal. The wire has capacitance between its ends, so as the charges move to the ends, the voltage between the ends will rise, as V=Q/C. When it rises so that the voltage gradient along the wire is equal and opposite to the external applied E-field, there will be no further movement and the field outside the wire will be zero. (If all the available electrons were to move, it is still possible for them react to the field by crowding more densely towards the ends).
     
  4. Jul 15, 2015 #3
    I'm not sure I understand what you mean by "there is no shortage of free charges in a metal". Anyway, there is something that is likely to be wrong in your method: you are using the capacitance of the wire to assert that V = Q/C; but Q is limited by the number of free charges inside the wire, otherwise you could "load" a wire with a charge as large as desired, submitting it to a sufficiently strong voltage. This is obviously impossible. Notice that in a real capacitor, there is a breakdown voltage above which the capacitor "breaks" and let the current flow; so V=Q/C is correct only inside certain limits.
     
  5. Jul 15, 2015 #4
    As far as I can calculate, the charge of the free electrons in a c.c. of copper is 1.4 x 10^5 Coulombs, about the same as a car battery, and this is why I thought we could say there was no shortage of free charges. Even if there were few free charges, maybe the material could still respond to an applied field because the charges would bunch up closer together in response to the field? As the charges try to repel each other, it may be similar to compressing a gas at each end of the wire. Of course, I do agree that the air will break down above a certain field strength.
     
  6. Jul 15, 2015 #5
    1) I am interested to know how you obtained the value of 1.4 x 10^5 Coulombs.
    2) Of course, the material could still respond (and does respond) to an applied field (and this is what I have written from the beginning), but what I say is that even if the totality of the charges is supposed to be able to cluster around the extremity of the wire, this may not be sufficient to annihilate the potential inside the wire.
     
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