Find charge density of an electric field

In summary, the electric field in a certain region of space is given by E(\vec{}r)=C\vec{}r/ε0a3, where a has dimension length and C is a constant. The charge density is given by?
  • #1
JonNash
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Homework Statement



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?

Homework Equations



∇.E=ρ/ε

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?
 
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  • #2
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.
 
  • #3
vanhees71 said:
I can't read your electric field well. Does it mean
[tex]\vec{E}(\vec{r})=\frac{C \vec{r}}{\epsilon a^3}?[/tex]
Yes, that's the equation, new here, having trouble with writing complex equations.
vanhees71 said:
Then there's no problem at the origin, but it's well-defined everywhere.
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?

vanhees71 said:
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.

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
 
  • #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?
 
  • #5
JonNash said:
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.

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.
 
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  • #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.
 

What is charge density?

Charge density is a measure of the amount of electric charge per unit volume in a given region. It is typically represented by the symbol ρ (rho) and is measured in units of coulombs per cubic meter (C/m³).

How is charge density related to an electric field?

The electric field is directly proportional to the charge density. This means that as the charge density increases, the electric field also increases, and vice versa. The electric field is given by the equation E = ρ/ε₀, where ε₀ is the permittivity of free space.

How can I find the charge density of an electric field?

To find the charge density of an electric field, you will need to know the electric field strength and the permittivity of free space. You can then use the equation ρ = ε₀E to calculate the charge density. Make sure to use consistent units for all values.

What are the units of charge density?

The units of charge density are coulombs per cubic meter (C/m³) in the SI system. In other systems of measurement, it may be represented in different units, such as coulombs per liter (C/L) or coulombs per cubic centimeter (C/cm³).

How does charge density affect the behavior of an electric field?

The charge density plays a crucial role in determining the behavior of an electric field. A higher charge density will result in a stronger electric field, which can lead to stronger forces between charged particles and larger voltage differences. Conversely, a lower charge density will result in a weaker electric field and correspondingly weaker effects.

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