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Find charge density of an electric field

  1. Jan 18, 2014 #1
    1. The problem statement, all variables and given/known data

    Let the electric field in a certain region of space be given by E([itex]\vec{}r[/itex])=C[itex]\vec{}r[/itex]/ε0a3, where a has dimension length and C is a constant. The charge density is given by?

    2. Relevant equations

    ∇.E=ρ/ε

    3. The attempt at a solution

    I've been searching this on the net, from what I understood the divergence is not defined at the origin and is zero everywhere else except the site of a local divergence created by the presence of a charge, which is then calculated by a surface integral of the field. In this question a field equation is given and charge density is asked. The divergence theorem is the only relation between the two (that is, I didn't find any more) but I'm unable to calculate anything from the given field equation that would result in a reasonable answer to the charge density. Could you please point out where my understanding/searching has failed me?
     
  2. jcsd
  3. Jan 18, 2014 #2

    vanhees71

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    I can't read your electric field well. Does it mean
    [tex]\vec{E}(\vec{r})=\frac{C \vec{r}}{\epsilon a^3}?[/tex]
    Then there's no problem at the origin, but it's well-defined everywhere.

    Just calculate the divergence, i.e.,
    [tex]\vec{\nabla} \cdot \vec{E}=\partial_x E_x+\partial_y E_y + \partial_z E_z.[/tex]
    You've already written how this is related with the electric charge density via Gauß's Law.
     
  4. Jan 18, 2014 #3
    Yes, that's the equation, new here, having trouble with writing complex equations.
    I didn't get that. No problem at origin means its undefined there or well-defined even at the origin? If yes, how does Gauss' law remain intact?

    That's the whole problem, no boundaries for the surface integral are defined and the equation is not broken down into the 3 co-ordinate system. How do I extract a 3 dimensional representation of the equation? I'm not asking for a solution, just nudge me in the right direction.

    Thanks
     
  5. Jan 19, 2014 #4
    In the absence of any source of charge/current the Gauss' law states that the divergence of an electric field is zero. In this case although we are given a field it is given in the absence of any source, wouldn't that mean that density/permittivity then would also equate zero, finally giving the result that current charge density is zero? Am I right?
     
  6. Jan 19, 2014 #5

    hilbert2

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    Written in component form, the electric field is ##\vec{E}(\vec{r})=\frac{C \vec{r}}{\epsilon a^3}=\frac{Cx}{\epsilon a^3}\vec{i}+\frac{Cy}{\epsilon a^3}\vec{j}+\frac{Cz}{\epsilon a^3}\vec{k}##, where ##\vec{i},\vec{j},\vec{k}## are the unit vectors. The divergence is just the "dot product" of this with the vector ##\frac{\partial}{\partial x}\vec{i}+\frac{\partial}{\partial y}\vec{j}+\frac{\partial}{\partial z}\vec{k}##.

    The system corresponding to this kind of electric field is a sphere of uniform charge density, in the limit where the boundaries of the sphere are "infinitely far" from the origin of the coordinate system.
     
  7. Jan 19, 2014 #6
    Thanks for explaining that hilbert2. I calculated it and got the answer 3C/a^3. I know it was a noob question but I'm a bit new to this.
     
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