Finding Electric Field for Spherical Charge Distribution?

Click For Summary
SUMMARY

The discussion focuses on calculating the electric field for a spherical charge distribution defined by a charge density ρ(r) = Ke^{-br} within the region 0 < r < a. The total charge Q is determined using the integral Q = ∫∫∫ ρ dV, leading to the expression dQ = 4πKr²e^{-br}dr for spherical coordinates. The correct approach to find the electric field E involves integrating the charge density rather than substituting Q directly into E = kQ/r², as the latter yields inconsistent results. The proper method emphasizes the use of spherical integration differentials to accurately evaluate the electric field.

PREREQUISITES
  • Understanding of electric fields and charge distributions
  • Familiarity with spherical coordinates and integration techniques
  • Knowledge of Gauss's Law and its applications
  • Basic concepts of electrostatics and charge density functions
NEXT STEPS
  • Study the application of Gauss's Law for spherical charge distributions
  • Learn about spherical coordinate integration techniques in electromagnetism
  • Explore the implications of charge density variations on electric fields
  • Investigate the relationship between charge density and electric potential
USEFUL FOR

Students and educators in physics, particularly those focusing on electromagnetism, as well as researchers working with electric fields and charge distributions in theoretical and applied contexts.

jesuslovesu
Messages
185
Reaction score
0

Homework Statement


There is a charge density rho that exists in a spherical region of space defined by 0 < r < a.
\rho (r) = Ke^{-br}
How do you find the electric field if a charge density varies as such?

The Attempt at a Solution



I found Q total = \int \int \int \rho dV
Now I need to find E.

My real question is can I just put Q (as a function of r) into E = kQ/r^2? Or do I need to reevaluate the integral using dq = \rho r^2 sin(\theta) dr d\theta d\phi

I get two different answers, (and I would have thought they should be the same) so which method is correct? I would have thought either would work.
 
Last edited:
Physics news on Phys.org
What do you mean? Q is the integral of the charge density over the volume. Also, I think you mean dr = rho r^2 sin(\theta) dr d\theta d\phi. What did you do for your integral?
 
Why sin(\theta)? rho depends only on r so dQ = 4\pi Kr^{2}e^{-br}dr
 
Oh whoops, I shouldn't have had rho in there, and I missed it when you had it. You were right about the dq I was questioning. dq= rho *spherical jacobian (i.e. spherical integration differentials), which is what you had.

Yes, dQ = 4\pi Kr^{2}e^{-br}dr

This is the way you want to go. I don't really understand what other way you would have gone about it.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
6
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
11
Views
1K
Replies
3
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
11
Views
1K
  • · Replies 17 ·
Replies
17
Views
3K