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Potential of Infinite Charge Distribution by First Principle

  1. Aug 28, 2011 #1
    We are presenting you a very perplexing but interesting problem
    which you may have probably encountered in electrostatics.
    We were trying to calculate the potential of an infinite line
    charge distribution at a general point by first principle method i.e.
    the usual integration of the potential of the differential charges on
    the line charge extending from -∞ to +∞ .We got an indeterminate
    form[ln(∞ )-ln(-∞)].
    We realized that the reference for the potential can not be set to
    infinity as we unknowingly did for the differential charge appearing
    in the integration which is wrong for infinite charge
    distributions.However we can not decide how to set the zero of
    potential at some other point. Please help us...
    Regards
    aim1732 and Mayukh Nath.
     
  2. jcsd
  3. Aug 28, 2011 #2

    vanhees71

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    This (and most other problems with a lot of symmetry) are most easily solved not by using the Green's function but by solving the Poisson equation from scratch.

    Here we have

    [tex]\Delta \Phi=-\rho=-\lambda \delta(r).[/tex]

    Here [itex](\rho,\varphi,z)[/itex] denote cylindrical coordinates and [itex]\lambda[/itex] the charge per unit length located at the [itex]z[/itex] axis.

    For symmetry reasons [itex]\Phi[/itex] depends on [itex]r[/itex] only, and the Laplacian thus translates into

    [tex]\Delta \Phi(r)=\frac{1}{r} \frac{\partial}{\partial r} \left (r \frac{\partial \Phi}{\partial r} \right )=0 \quad \text{for} \quad r \neq 0.[/tex]

    Now one can succesively integrate up this equation, leading to

    [tex]r \frac{\partial \Phi}{\partial r}=C_1 \, \Rightarrow \, \Phi=C_1 \ln \left (\frac{r}{r_0} \right ).[/tex]

    Here [itex]C_1[/itex] and [itex]r_0[/itex] are integration constants. The first has to be chosen to get the correct singularity, while [itex]r_0[/itex] is physically irrelevant since it's only an additional constant, and the physically relevant quantity is the field strength, i.e.,

    [tex]\vec{E}=-\vec{\nabla} \Phi=-\frac{C_1}{r} \vec{e}_r.[/tex]

    Now to get [itex]C_1[/itex] we apply Gauss's Law to a cylinder [itex]Z[/itex] of height [itex]L[/itex] and radius [itex]R[/itex] around the [itex]z[/itex] axis, leading to

    [tex]\int_{\partial Z} \mathrm{dd} \vec{A} \cdot \vec{E}=-C_1 \int_{0}^{L} \mathrm{dd} z \int_0^{2 \pi} \mathrm{dd}\varphi=-2 \pi L C_1 \stackrel{!}{=}\lambda L.[/tex]

    Then you finally get the solution

    [tex]\Phi=-\frac{\lambda}{2 \pi} \ln \left (\frac{r}{r_0} \right )[/tex]

    and

    [tex]\vec{E}=\frac{\lambda}{2 \pi r} \vec{e}_r.[/tex]
     
  4. Aug 28, 2011 #3

    clem

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    There are several ways to get phi.
    The simplest is to get E by Gauss's law, and then integrate -E.dr from a radius a_0 (where you set phi(a_0)=0) to r.
    You can also use the Coulomb integral by integrating
    [tex]\phi(r)=\int_{-\infty}^{+\infty}[\frac{1}{\sqrt{x^2+r^2}}-\frac{1}{\sqrt{x^2+a_0^2}}].[/tex] so [tex]\phi(a_0)=0.[/tex]
     
    Last edited: Aug 28, 2011
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