[symbolic] Linear uniform charge density (E field at a point)

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Homework Help Overview

The discussion revolves around calculating the electric field at a point P due to a uniformly charged rod centered at the origin. The problem involves integrating to find the electric field using charge density and distance relationships.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration limits and the expression for differential charge, with some questioning the use of variables and the form of the distance R in the integral.

Discussion Status

There is an ongoing exploration of variable substitution and integration techniques. Suggestions have been made to clarify variable definitions and improve the setup of the integral, indicating a productive direction in the discussion.

Contextual Notes

Participants note potential confusion regarding the variable L and its usage in different contexts within the problem. The integration process is acknowledged to involve complex expressions, including trigonometric substitutions.

syntroniks
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Homework Statement


Center a rod of length L at (0,0) with the length going horizontally.

Take a point P at (0,y).

Find the electric field at P.

Homework Equations


[tex]\lambda[/tex]=Q/L
[tex]E= \int k*dQ/R^{2}[/tex]

The Attempt at a Solution


I am integrating from -L/2 to L/2
Since Q=lambda*L, I guess differentially dQ=lambda*dL.

Substituting that into the integral, it becomes:
[tex]k*\lambda \int dL/R^{2}[/tex]
from -L/2 to L/2 of course.

R is pretty messy so I'll just write what I came up with for [tex]R^{2}[/tex]:
[tex]R^{2}=(L^{2}/4)+y^{2}[/tex]

So... Doesn't this seem pretty reasonable? I just want to be double sure that this is OK.
 
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There may be a confusion with L, used in two different ways here.
I suggest you change dQ=lambda*dL to dQ=lambda*dx, where x is a distance along the x axis. This x runs from -L/2 to L/2.
I think you'll find that R² = x² + y².
Looks like one of those trig substitution integrals.
 
Thankfully for this particular problem I get to do the integration by software... kinda. Turns out to involve arctangent and a relative mess of symbols.

Thanks for the suggestion about variables, it is definitely more clear that way.
 
Looks like the substitution x = y*tan A really simplifies it!
And the A is a real angle in the problem.
 

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