What is the Electric Field at Point P Due to a Uniformly Charged Rod?

AI Thread Summary
The discussion revolves around calculating the electric field at point P due to a uniformly charged rod, with specific parameters provided. The user initially attempted to solve the problem using the electric field equation but arrived at an incorrect answer of 7.48 n/C instead of the correct 12.4 N/C. They realized that the mistake was in the integration process, particularly in handling the variable x and incorporating the correct trigonometric substitutions. After further clarification and guidance from another user, the original poster corrected their approach to successfully solve the problem. The conversation highlights the importance of accurate integration techniques in physics problems involving electric fields.
glennpagano44
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Homework Statement



Halliday and Resnick edition 7 chapter 22 number 29

In Fig 22-45, positive charge q = 7.81 pC is spread uniformly along a thin nonconducting rod of length L = .145m. The y distance between the point and rod is R = .06m. What is the magnitude and direction of the electric field produced at point P.

I will attempt to describe the figure:

There is a thin nonconducting rod in the x dimension with length L and a point P is directly above the center of the rod in the y dimension (R = .06m).

Homework Equations



E = (1/4\pi\epsilon)(q/r^{2})

The Attempt at a Solution



\intdE = 1/4\pi\epsilon\int\lambdaRdx/(R^{2}+x^{2})^{3/2}
I then took lamba and R out of the integral since they are constants
(4 is not rasied to (\pi\epsilon)

After taking the integral I got:

(\lambda/4\pi\epsilon)(x/R(R^{2}+x^{2})^{1/2}

When I plug in for the varibles I do not get the correct answer, I get 7.48 n/C but the correct answer is 12.4 N/C

I plug in L for x and R for R and for \lambda I use q/L
 
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Hi glennpagano44! :smile:

(have a lambda: λ and an epsilon: ε and a pi: π and a square-root: √ :wink:)
glennpagano44 said:
After taking the integral I got:

(\lambda/4\pi\epsilon)(x/R(R^{2}+x^{2})^{1/2}

No … how can you still have an x when you've just eliminated x by integrating over it?

And where did that extra R come from? :confused:
 
I went back to class today and I figured it out thanks alot. I just integrated it wrong, I had to do some trig subsitutions. (that is where the extra R came from)
 
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