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two infinite parallel conducting planes grounded and are separated by a distance d. place a point charge "q" between the two planes, using the "green teoerma reciprocity" how I show that the total charge induced is "-q" by the product of the fraction of the total perpendicular distance between the plane and the load?

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If you place a charge q between two conducting plates, the total charge induced is -q on both plates, but the ratio of the charge induced on either plate is as the ratio of the distance of the point to the plate, so that if the distance between the plates is L, and the point is at a distance pL from the left plate, the charge on the left plate is -pq, and on the right plate is -(1-p)q.

One way to see this is that the problem of solving Laplace's equation has a probability interpretation. If you start a random walk at the position of the charge, the induced charge on the left plate is equal to q times the probability that the random walk will hit the left plate before the right plate. This probability is the classical problem of a Brownian motion in 1d confined between two absorbing points, and this gives the answer.

The solution of the 1d random walk problem allows you to understand that this problem is really one dimensional. If you smear the charge q into a parallel plane of charge, each infinitesimal charge on the plane induces the same charge on the plates, by symmetry. The solution for a plane of charge between two conductors is very simple, and it reproduces the given answer, in a way completely parallel to the probability argument, but without introducing probability concepts.

"Using" Green's reciprocity theorem

Green's reciprocity theorem is integration by parts twice.

$$ \int \phi_1 \nabla^2 \phi_2 = \int \phi_2 \nabla^2 \phi_1 $$

It has the interpretation that the potential energy from the field of 2 acting on body 1 is equal to the potential energy from the field of 1 on 2, and it is clearly true, because the potential is from pairwise interaction, and this potential is all the pairs in the separate bodies.

By itself, this theorem proves nothing, because, being just integration by parts, it cannot be used to solve any differential equation. But if you use the additional fact that the potential between two uniformly charged plates is linear (this is the central fact used to get the result), you learn that if you add a charge density $\sigma_1$ to one plate and $\sigma_2$ to the other, then, up to units, the potential energy of the point charge is

$$ (\sigma_1) qx + \sigma2 q(L-x) $$

By Green's reciprocity, this is the energy of the uniform charge densities in response to the field of the charge. The derivative with respect to each charge density then tells you how many electric field lines end on each plate, proportionally. The work is all in the solution of the linear 1d problem, and the Green theorem is adding nothing particularly remarkable.

This is another "guess what I was thinking" problem all too common in education. This is possible, but it requires a knowledge not of physics, but of physicist psychology.

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I understand the problem from this point of view, but I do not know is how to solve it using the "green reciprocity theorem." –  jormansandoval Oct 16 '11 at 13:30
    
@jormansandoval: If you define the "green reciprocity theorem", it would help –  Ron Maimon Oct 16 '11 at 19:34
    
Let $\rho$ y $ \sigma $, y $\rho_a$ y $\sigma_a$ charge distributions, and whether and $\phi $ y $\phi_a$ potential created by $\rho, \sigma, \rho_a , \sigma_a$ respectively. Then we must have: $\int \phi.\rho_a dV + \int \phi . \sigma_a = \int \phi_a.\rho dV + \int \phi_a . \sigma. dS $ –  jormansandoval Oct 16 '11 at 19:55
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Spanish "y" == English "and", it is hard to read what you write. This is an exercise of "guess what I am thinking". Green's reciprocity theorem is nothing more than integration by parts twice, and it cannot establish a result of this sort by itself, nontrivially, since this result requires solving a 1d differential equation. Still, it is possible to guess what the person was thinking. I will add it to the answer. –  Ron Maimon Oct 17 '11 at 1:48
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