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For simplicity, consider the one dimensional Poisson equation expressed in Cartesian coordinates: $$\frac{d^2V}{dx^2}=-\frac{\rho}{\epsilon_0}.$$ Now, we observe in this case that the charge density is not unique. For a potential of $V=2x+1$ and a potential of 5, we get the same charge distribution. What does is physically mean? Where am I going wrong?

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  • $\begingroup$ Problems in electrostatics, generally assume that the charge distribution is known/given and the potential is computed using Poisson's equation with appropriate boundary condition. For a given $\rho(x)$, $V(x)$ is not uniquely determined unless you specify the boundary condition. Moreover, $V(x)$ is unique only up to a constant. $\endgroup$
    – SRS
    Commented Jan 15, 2017 at 6:17
  • $\begingroup$ but If I know the exact Potential in the first place, shouldnt it produce a unique charge distribution? and in this case how can a charge distribution which is zero everywhere produce an electric field of -2? $\endgroup$
    – Chandrahas
    Commented Jan 15, 2017 at 6:21
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    $\begingroup$ What is your unique potential? If it is $V(x)=2x+1$, then $\rho=0$ and if it is $V(x)=5$ then too $\rho=0$. Both the potentials correspond to solutions of Poisson's equation with $\rho=0$ but with different boundary conditions. So in each case your charge distribution is unique and unambiguous (but same). $\endgroup$
    – SRS
    Commented Jan 15, 2017 at 6:28
  • $\begingroup$ but where is the potential and electric field coming from? $\endgroup$
    – Chandrahas
    Commented Jan 15, 2017 at 6:30
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    $\begingroup$ $\rho=0$ can produce non-zero electric field too. But $\rho=0$ only where you find the solution. There must be other charge distributions which produces this field. You know the information of this charge in terms of the boundary condition. Consider the region between two charged plates: there is no charge in between. You may not know the charge distribution of the plates, but the potential on the plates are known. Then you can have a non-zero electric field in between. Hope it helps! $\endgroup$
    – SRS
    Commented Jan 15, 2017 at 6:43

4 Answers 4

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One problem is the lack of dimensions. (You have one). Let's skip the potential and work with a field, normally:

$$ {\bf \nabla E} =\rho / \epsilon_0 $$

In 3 dimensions, the free space field falls off as $1/r^2 \ (and\ V \propto 1/r)$. You can imagine conserved flux line spreading out in space as the pass through a surface of radius $\propto r^2$.

In 2 dimensions you expect the field to fall off as $1/r$, with $V \propto \ln r$, again you have conserved lines of flux spreading out.

Now you have chosen 1 dimension, so any amount of flux can't spread out. A point charge any where leads to a constant field pointing towards (or away from) it, and it never falls off. That means a point charge anywhere leads to a linear ramp on $V(x)$--this seems counter intuitive, that a distance charge can lead to a linear diverging potential, but it's only because there is only 1 dimension working.

As mentioned in other answers, this term is fixed with boundary conditions. If there is a point charge at $x_0$, then

$$ V(x) \propto |x-x_0| $$

Of course, the constant term is meaningless in any number of dimensions.

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Your 1-dimensiomal Poisson equation is an inhomogeneous linear ordinary differential equation of second order for the potential $V(x)$. For the specification of a unique solution for $V(x)$, you have to specify the inhomogeneity (charge density) $\rho(x)$ and two boundary conditions for the potential. Then you obtain the unique potential solution by adding two independent solutions of the homogeneous equation ($\rho=0$) to one solution of the inhomogeneous equation so that the two boundary conditions are fulfilled. The two potential solutions you mention are both solutions to the homogeneous equation and thus correspond only to the case where the charge density is zero ($\rho=0$).

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The very first thing is the potential is a relative quantity, just like the gravity potential, you can only talk about the physical meaning before choosing some right reference point. So plus a constant in your potential has no affection.

The second thing is the Poisson equation is a differential equation, so what you are looking for is the potential if given the density distribution. Of course, you can treat it as a mathematical equation so you can pick up any mathematical function "V", but there is no correspondence between that and physical electron distribution.

Hope it helps.

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The point to note is that when the RHS=0 you obtain the Lapalace Eq. In that case the potential V will not have any local maxima. That is the average value of V about any point cannot be lower or higher than V at that point. When $\rho$ is not zero there are local maxima or minima at those ponits. The LHS is simply the curvature of V.

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