Electric field by infinite line of charge

AI Thread Summary
The discussion revolves around calculating the electric field due to an infinite line of charge with density λ along the x-axis. The formula used involves integrating the contributions of differential elements of charge, leading to the expression for dE. A participant expresses confusion about the integral resulting in zero due to the limits from -∞ to +∞. Another participant points out a potential oversight regarding the need for an additional term in the denominator of the equation. The conversation highlights the complexities of integrating over infinite limits in electrostatics.
Sam99
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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?
 
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Hi Sam99! Welcome to PF! :smile:

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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
 
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