Need help with Proper integral of a uniformly charged rod

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SUMMARY

The discussion focuses on calculating the x-component of the electric field generated by a uniformly charged rod positioned along the x-axis from x=0 to x=L. The relevant equations include Coulomb's law for linear charge distributions and the integral formulation for electric field components. The user attempts to set up the integral using the linear charge density λ and the infinitesimal length element dl, leading to the expression for the electric field. The correct approach involves integrating the expression for the electric field over the length of the rod while considering the geometry of the system.

PREREQUISITES
  • Understanding of Coulomb's law and electric fields
  • Familiarity with calculus, specifically integration techniques
  • Knowledge of linear charge density and its application
  • Basic concepts of vector calculus in three dimensions
NEXT STEPS
  • Study the derivation of electric fields from continuous charge distributions
  • Learn about the application of integration in electrostatics problems
  • Explore the concept of line integrals in vector calculus
  • Review the use of coordinate systems in physics, particularly Cartesian coordinates
USEFUL FOR

Students of electromagnetism, physics enthusiasts, and anyone studying electric fields and charge distributions will benefit from this discussion.

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Homework Statement



A uniformly charged rod is places along the x-axis from x=0 to x= L. carefully set up, but do not solve the proper integral to determine the x component of the electric field at the point (L,a)

Homework Equations


E=u[tex] \int \frac{dq}{(r^2)}[/tex]

for a line dq=λdl

The Attempt at a Solution


i know for a line dq=λdl
i think its something along the lines of
Ey=Kλy=[tex] \int \frac{dx}{(x^2y^2)^(3/2)}[/tex] from 0 to L

I have no idea if this is right. I am using my notes to guide me.
 
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Coulomb's law for a linear charge distribution is:

[tex]\textbf{E}(\textbf{r})=\frac{1}{4\pi\epsilon_0}\int \lambda(\textbf{r}')\frac{\textbf{r}-\textbf{r}'}{|\textbf{r}-\textbf{r}'|^3}dl'[/tex]

Where [itex]dl'[/itex] is an infinitesimal length of the source, located at [itex]\textbf{r}'[/itex], and the integration is over the entire line of charge.

Use [itex]\textbf{r}=x\mathbf{\hat{x}}+y\mathbf{\hat{y}}+z\mathbf{\hat{z}}[/itex] and [itex]\textbf{r}'=x'\mathbf{\hat{x}}+y'\mathbf{\hat{y}}+z'\mathbf{\hat{z}}[/itex] to find the x-component.
 

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