Electric field by infinite line of charge

Click For Summary

Homework Help Overview

The problem involves calculating the electric field generated by an infinitely long line of charge with a given linear charge density λ. The electric field is to be determined at a specific point in space defined by coordinates X = x(x') + y(y') + z(z').

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the electric field equation and the integration limits. There is a focus on the expression for the differential electric field dE and whether the terms in the denominator are correctly represented. Some participants question the treatment of the limits of integration and the resulting calculations.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's attempts. There is a suggestion to reconsider the terms in the denominator of the electric field expression, indicating a productive exploration of the mathematical setup.

Contextual Notes

Participants are navigating through the implications of integrating over an infinite range and the assumptions made regarding the coordinate system and the representation of distances in the electric field formula.

Sam99
Messages
2
Reaction score
0

Homework Statement


Given an infinitely long line of charge density λ extended along the x-axis, what is the electric field at a point X = x(x')+y(y')+z(z') (in space)?

Homework Equations



E = kq / r^2, dq = (lamda)dx

The Attempt at a Solution



dE = kλ ∫ [x(x')+y(y')+z(z')- x(x')] dx / [(y^2) +(z^2)]^(3/2) (integral from -∞ to +∞)

I end up getting 0 because -∞ + ∞ equals 0?
 
Last edited:
Physics news on Phys.org
Welcome to PF!

Hi Sam99! Welcome to PF! :smile:

(try using the X2 button just above the Reply box :wink:)
Sam99 said:
dE = kλ ∫ [x(x')+y(y')+z(z')- x(x')] dx / [(y2) +(z2)]3/2 (integral from -∞ to +∞)

shouldn't there be an x2 on the bottom also? :redface:
 
Thanks!
I don't think so because I'm using (x-x')/ |x-x'|^3.
would that be right?
 
yes, but there's still a difference in x, which has to go on the bottom
 

Similar threads

Replies
9
Views
2K
Replies
64
Views
6K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
11
Views
4K
  • · Replies 28 ·
Replies
28
Views
1K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 36 ·
2
Replies
36
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K