Finding electrostatic potential from charge distribution

Click For Summary

Homework Help Overview

The discussion revolves around finding the electrostatic potential from a given charge distribution, focusing on the mathematical representation of the separation distance in the integral involved.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to express the separation distance |r-r'| in spherical coordinates but encounters difficulties with integration. Some participants question whether the potential must be found using the initial formula or if alternative methods, such as Gauss' law, could be employed.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to the problem. There is a suggestion to consider Gauss' law as a potentially simpler method, indicating a productive direction without reaching a consensus.

Contextual Notes

Participants are navigating the constraints of the problem, including the requirement to find the electrostatic potential and the challenges posed by the integral setup.

Plaetean
Messages
35
Reaction score
0
Note that posting images or image links does not count as filling out the homework template.

Homework Statement


Question

Homework Equations


Equation

The Attempt at a Solution


Attempt I am not sure how to write the |r-r'| in a way that allows me to actually solve the integral. I have tried writing |r-r'| in spherical co ords, but all I seem to be able to get is this as the separation distance, which seems impossible to integrate. Any help on whether I have gone about the problem the wrong way, or a way to write |r-r'| in a more manageable form would be very much appreciated
 
Physics news on Phys.org
Are you required to first find the potential using the formula you gave? Or, are you allowed to find E using some other equation (or law)?
 
No, perhaps I could do it just using Gauss' law instead, might be easier
 
Plaetean said:
No, perhaps I could do it just using Gauss' law instead, might be easier

Yes, give it a try.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 19 ·
Replies
19
Views
2K
Replies
5
Views
2K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K