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Numerical modelling of electrostatic potential energy of a system

  1. Sep 6, 2010 #1
    Hi there,

    Lately, I've been trying to gain a deeper understanding of the relationship between the Coulomb force F, electric field E, electric potential V, and electrostatic potential energy U.

    In regards to electrostatic potential energy, as I understand it, if you had, say (made up a few values):

    stationary point charge Q:

    [tex]Q = 1.9258\cdot10^{-5} C[/tex]

    you bring in another charge q:

    [tex]q = 1.9258\cdot10^{-5} C[/tex]

    to a distance of 66.6cm and hold it, the electrostatic potential energy of this 2 charge system is:

    [tex]U = \frac{Qq}{4\pi\varepsilon_0r}[/tex]
    [tex]U = \frac{(1.9258\cdot10^{-5})^2}{4\pi\varepsilon_0(0.666)} = 5 J[/tex]

    If I then released q, it would gain 5 J of kinetic energy and move with a velocity calculable from [tex]\frac{1}{2}mv^2[/tex] (relativistic corrections made as needed).

    Is this overall idea correct?

    Then, in learning about capacitance and electric field energy, I come across the equation for the total energy of an electric field:

    [tex]U = \int_{V}{\frac{1}{2}\varepsilon_0|E|^2dV}[/tex]

    I was wondering if I could numerically compute the 5 J total system energy using this equation somehow. My initial thought was stepping over a uniform volume (a cube) around the system, slicing it into tiny chunks which serve as my differential volume elements, and calculating the vector sum of the individual E field contributions from the 2 charges:

    [tex]E_{net} = E_Q + E_q[/tex] and then a "differential energy" I guess: [tex]dU = \frac{1}{2}\varepsilon_0|E|^2dV[/tex] where dV = (dimension of a slice)3. The dU's are calculated inside a triple nested for-loop, one for each coordinate, and totaled. My hope was that total would be approximately 5.0 J, but I soon realized something was conceptually quite wrong with this method.

    For one, if I simulate and compute the field energy of a system with just the original stationary charge Q, the square of the field amplitude is positive everywhere and my total energy can reach enormous values. According to textbooks I'm reading, an isolated charge does not have electric field energy since it took no work to bring that charge to its location. This is an intuitive idea since the Coulomb force acts between two charges, but I can't seem to wrap my head around an intuitive way to discretize the volume around a system in order to numerically compute the results obtained analytically.

    Perhaps I have some fundamental misunderstanding of the field energy equation, or maybe I'm just too tired :confused:

    Anyone have any pointers for this one?
  2. jcsd
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