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Finding the Space Charge Density

  • Thread starter weathergal
  • Start date
  • #1
I am taking a lightning class for my atmospheric science masters degree. However its been awhile and Im having trouble with remembering how to do the physics portion of it. I hope someone can help me!

Homework Statement



I have an electric field [itex]\vec{}E[/itex]=E[itex]_{}x[/itex][itex]\widehat{}X[/itex]+E[itex]_{}y[/itex][itex]\widehat{}Y[/itex]+E[itex]_{}z[/itex][itex]\widehat{}Z[/itex] Find the divergence and curl of this electric field. I have done this with no problem. The second part is where Im confused. I have to find the space density charge of [itex]\vec{}E[/itex]=E(r)[itex]\dot{}r\widehat{}[/itex]

Homework Equations



I have an equation for space charge density but am unsure how to derive it. The space charge equation I have is ([itex]\rho[/itex]/[itex]\epsilon[/itex])=delta dot E

Can some one give me some pointers on even how to find a space charge denstiy?\
Thanks

The Attempt at a Solution

 

Answers and Replies

  • #2
diazona
Homework Helper
2,175
6
Do you really have to derive the equation? Based on what you've posted, it seems that you only have to use it.
[tex]\frac{\rho}{\epsilon_0} = \vec{\nabla}\cdot \vec{E}[/tex]
You've said that you know how to calculate the divergence and curl of a vector field. How can you apply that to this problem?
 
  • #3
760
69
The equation in 2. is Gauss's Law for the Electric Field in differential form.

It's not delta, but nabla, the symbol for del, the vector differential operator. In this case nabla dot E means divergance of the electric field.
 
  • #4
Well this is a two part question. The first part is to find the curl and divergence. Part B is if E=E(r)[itex]\widehat{r}[/itex] what is the space charge. I just firgure that part a is used in part b. I guess im wrong. I think the answer to this would be just

([itex]\rho[/itex]/[itex]\epsilon[/itex])=dE(r)/dr
 
  • #5
diazona
Homework Helper
2,175
6
The first part is to find the curl and divergence. Part B is if E=E(r)[itex]\widehat{r}[/itex] what is the space charge. I just firgure that part a is used in part b. I guess im wrong.
No, you're correct. Part a is used in part b. Or, at least, you could use part a in part b, though it's not the only way.

Have you learned anything about calculating the divergence in non-Cartesian coordinate systems? If so, you can use that knowledge for part b. If not, consider this: how can you express the vector field you're given in part b, [itex]E_r\hat{r}[/itex], in terms of the Cartesian unit vectors [itex]\hat{x},\hat{y},\hat{z}[/itex]?
I think the answer to this would be just

([itex]\rho[/itex]/[itex]\epsilon[/itex])=dE(r)/dr
Not quite, but you're thinking along the right lines.
 

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