1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Finding the equation a potential must satisfy in order to minimize E

  1. Mar 12, 2014 #1
    1. The problem statement, all variables and given/known data

    Find an expression involving the function [itex] ϕ(x_1, x_2, x_3) [/itex] that has a minimum average value of the
    square of its gradient within a certain volume V of space.

    If ϕ is the electric potential, [itex] \vec{E} = -\nabla ϕ [/itex] is the electric field, and [itex] ρ = \frac{1}{2} ϵ_0|\vec{E}\cdot\vec{E}| [/itex] is the energy density of
    the electric field, this result tells us what equation the electric potential must satisfy to minimize
    the total energy.

    2. Relevant equations

    (1)[itex] \frac{∂f}{∂ϕ}+\sum_{i=1}^3\frac{∂}{∂x_i}\frac{∂f}{∂ϕ'_i} = 0 [/itex] where [itex] ϕ'_i [/itex] are the partial derivatives of [itex] ϕ [/itex]

    (2)min[itex](\frac{1}{V}\iiint ϕ dV) = \nabla ϕ \cdot \nabla ϕ [/itex]

    3. The attempt at a solution

    First, since minimizing the average value of [itex] ϕ [/itex] yields a function [itex] \nabla ϕ \cdot \nabla ϕ [/itex] I will use this as my functional since it must satisfy (1). Doing so yields:

    [itex] \frac{∂f}{∂ϕ}= 0 [/itex] and [itex] \nabla \cdot [\frac{∂(\nabla ϕ \cdot \nabla ϕ)}{∂(\nabla ϕ)}] = 0 [/itex]

    The second part of (1) yields:

    [itex] 2\nabla \cdot \nabla ϕ =0 [/itex] or alternatively, [itex] \nabla^2 ϕ = 0 [/itex] (Laplace's Equation)

    Now this makes sense to me ( or at least tells me that I'm on the right track) since I know that the electrostatic potential does satisfy this equation; however, I know that Laplace's equation is a simpler case of Poisson's equation and when I plug my result in using [itex] \vec{E} = -\nabla ϕ [/itex] I get [itex] \nabla \cdot \vec{E} = 0 [/itex] which is only true if the charge density inside of the object is zero. So here's my question should I have gotten the more general Poisson equation, or am I overthinking this whole thing?
  2. jcsd
  3. Mar 14, 2014 #2
    I think I have figured it out I actually need to use (2) as a constraint on (1):

    [itex] f+μg = \frac{\phi}{V} + μ\nabla \phi \cdot \nabla \phi [/itex] and (1) becomes [itex] \frac{∂(f+μg)}{∂\phi}+\frac{∂}{∂x_i}\frac{∂(f+μg)}{∂\phi'_i} [/itex]

    The first term of (1) is just [itex] \frac{1}{V} [/itex] and the second term of (1) is the same as before except multiplied by μ.

    Now I have: [itex] \frac{1}{V}+μ\nabla^2\phi = 0 → \nabla^2\phi = \frac{-1}{μV}[/itex]

    Because [itex] μ [/itex] is an arbitrary constant I'll define it to be: [itex] μ \equiv \frac{ε_0}{Vρ} [/itex]

    Then I get the Poisson equation: [itex] \nabla^2\phi = \frac{-ρ}{ε_0} [/itex]

    Now if I make the substitution [itex] \vec{E} = -\nabla \phi [/itex] I get the desired result:

    [itex] \nabla \cdot \vec{E} = \frac{ρ}{ε_0} [/itex]
  4. Mar 16, 2014 #3


    User Avatar
    Homework Helper

    I think your first answer was most likely what they were looking for. you minimised the average value of the square of the vector norm of the gradient of the potential, over a given volume. And it gave you Laplace's equation. Which is good. intuitively, it means that in a region with zero charge distribution, the EM field will hold the least amount of energy possible, while still respecting the boundary conditions.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted