How to Calculate Charge Density on an Irregular Conductor?

Click For Summary

Homework Help Overview

The discussion revolves around calculating the local surface charge density on an irregularly shaped conductor, given varying electric field strengths on its surface.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between electric field strength and charge density, with some attempting to connect radius of curvature to charge density. Questions arise regarding the adequacy of the provided information for calculating exact values.

Discussion Status

Some participants have offered insights into the relationship between charge density and radius of curvature, while others express uncertainty about how to proceed with the given data. There is acknowledgment of previous misunderstandings, and a potential equation relating electric field and charge density has been identified.

Contextual Notes

Participants note that the problem is constrained by the information provided, specifically the electric field values, and question whether this is sufficient to derive exact charge density values.

benndamann33
Messages
20
Reaction score
0
Question: The electric field on the surface of an irregularly shaped conductor varies from 56.0 kN/c to 28.0kN/c. calculate the local surface charge density at the point on the surface where the radius of curvature of the surface is (a) greatest and (b) smallest.

I am stuck as far as how to get an exact number. The only thing I concluded was that at the largest radius of curvature, charge density is smallest, and I called this lamda. Because electric field ranges to double the smallest value, I concluded at the smallest radius of curvature the charge density is 2 lamda. Is this totally incorrect?
 
Physics news on Phys.org
The charge density is proportional to 1/(radius of curvature at that particular point).
 
I understand that, but I'm just given the information above, which is simply an electric field, so I couldn't find the exact charge density any more than what I have above, could I?
 
Ok correct my mistake I totally read that wrong. The equation relating the potential and the electric field just outside the surface is:

[tex]v=aE=\frac{a\sigma}{\epsilon_0}[/tex]

where a is the radius of curvature at a particular point and sigma is the charge density.
 
thanks man, I was overlooking that for some reason, appreciate it.
 
sorry about the original post just didn't have my head screwed on.
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 15 ·
Replies
15
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
Replies
9
Views
2K
Replies
5
Views
2K
Replies
1
Views
1K