Determine charge at origin, based on charge density function

Click For Summary

Homework Help Overview

The discussion revolves around determining the charge at the origin based on a given charge density function. The problem involves concepts from electrostatics, particularly Gauss' law and the Dirac delta function, in a three-dimensional spherical coordinate system.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the Dirac delta function to isolate the charge density at the origin and question its dimensional implications in three-dimensional spherical coordinates. There is also exploration of applying Gauss' law to find the enclosed charge as the radius approaches zero.
  • Some participants suggest evaluating the total charge within a radius and then considering the limit as that radius approaches zero, while others express uncertainty about how to simplify the charge distribution function for integration.

Discussion Status

The discussion is ongoing, with participants sharing hints and suggestions on how to approach the problem without evaluating integrals directly. There are multiple lines of reasoning being explored, particularly around the application of Gauss' law and the implications of the charge density at the origin.

Contextual Notes

There are concerns about the behavior of the charge density function at r=0, and assumptions regarding the parameters involved, such as lambda and a, are being discussed. Participants are navigating the complexities of the problem while adhering to homework constraints.

xSilja
Messages
3
Reaction score
0

Homework Statement


upload_2015-9-23_18-30-1.png


a) and b) are no problem.

I need help to solve c) and d)

Homework Equations


c) Delta dirac function
Gauss' law

d) Gauss' law
## \int_V {\rho \, d\tau} = Q_{enclosed} ##

The Attempt at a Solution


By taking laplace on the potential I get:

## \rho(\mathbf{r}) = \frac{q_0}{4 \, \pi \, r} \, e^{-r/\lambda} \, \left( \frac{cos^2(\theta)}{\lambda^2} + \frac{2}{r^2} (1-3 \, cos^2(\theta)) \right)##

c) I got a hint that it was a good idea to use the dirac delta function along with the charge distribution.

But I'm not exactly sure why. As I understand it the dirac delta function "picks out" the value of a function at zero. So I'd get:

## \int {\rho(\mathbf{r}) \, \delta(r) \, dr} = \rho(0) ##

I realize that there must be a dimensional problem here, but I'm not sure how to use a delta function in 3D and spherical coordinates.
Also how will it help me to find the density at the origin? Can I apply Gauss' law here and let the radius go towards zero to get the charge in the origin?

d) I want to solve the integral

## Q = \int_0^\pi \int_0^{2 \, \pi} \int_a^{\infty} \, \rho(\mathbf{r}) \, r^2 \, \sin(\theta) \, dr \, d\theta \, d\phi ##

I tried evaluating this with Maple.
By assuming a>0 I get a complex function multiplied by infinity, which is not of much use.
If I also assume lambda>0 (as it says in the problem) I get rid of the infinity, but get exponential integrals instead.
I'm not sure how to move on from here. I suspect I need to modify my function for charge distribution by assumptions, to make it simpler.
 
Physics news on Phys.org
xSilja said:
c) I got a hint that it was a good idea to use the dirac delta function along with the charge distribution.
This assumes your charge density formula doesn't have problems with r=0. That might be, but I wouldn't rely on it.

Using the electric field, you can determine the total charge up to a radius a and then let a go to zero.
The opposite limit also works for (d) and does not need evaluating any actual integrals.
 
By best idea was also to use Gauss' law. ## \int_V \mathbf{\nabla} \cdot \mathbf{E} \, d\tau = \frac{1}{\epsilon_0} \, Q_{enclosed} ##
How can I avoid evaluating any actual integrals?
The charge density/ elctric field depends on both r and theta.
 
xSilja said:
How can I avoid evaluating any actual integrals?
Set them up, then find useful bounds on them.
 

Similar threads

Replies
5
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
14
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
10K