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

AI Thread Summary
The discussion focuses on calculating the electric field at point P due to a uniformly charged rod centered at the origin. The user integrates from -L/2 to L/2, using the linear charge density and expressing the differential charge as dQ=lambda*dL. It is noted that the expression for R² should be corrected to R²=x²+y², where x is the horizontal distance along the rod. A suggestion is made to use a trigonometric substitution to simplify the integration process, which involves arctangent functions. Overall, the approach to solving the problem is confirmed to be reasonable with adjustments for clarity.
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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


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

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:
k*\lambda \int dL/R^{2}
from -L/2 to L/2 of course.

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

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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