Electrical field near infinite, charged rod

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klm_spitifre
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Homework Statement


Calculate the electric field at point ##P## if the distance from the center of an infinitely long, charged rod to point ##P## is ##a = 0.6m##; the charge density equals ##\lambda = -CX^2##, ##C=10^{-3}C/m^3##. Show all steps in finding the equation of the field, then find the magnitude and direction of ##\vec{E}## at point ##P##.

Screen_Shot_2018_01_25_at_1_01_03_PM.png


2. The attempt at a solution
My professor actually solved the first part of this problem during lecture:
[tex] dq = \lambda dx\\<br /> R = \sqrt{a^2+x^2}\\<br /> d\vec{E} = \frac{K dq}{R^2} = \frac{K \lambda dx}{a^2+x^2}\\<br /> cos\theta = \frac{a}{R} = \frac{a}{\sqrt{a^2+x^2}}\\<br /> d\vec{E}_{x} = d\vec{E}cos\theta = \frac{K \lambda dx}{a^2+x^2} \cdot \frac{a}{\sqrt{a^2+x^2}} = \frac{K \lambda a dx}{(a^2+x^2)^{3/2}}\\<br /> \vec{E}_{x} = \int_{-\infty}^{+\infty} d\vec{E}_{x} = K a \lambda \int_{-\infty}^{+\infty} \frac{dx}{(a^2+x^2)^{3/2}} = \frac{2K\lambda}{a}[/tex]

Based off of my understanding of the problem, and the solution above, I'd wager the direction will be perpendicular to the rod in question. What throws me off is ##\lambda##. I get it represents the charge density (i.e. charge per unit distance), but the way it was defined in the original question throws me off. Namely, what is ##X##? Assuming the problem was solved correctly, and ##X## represents some constant, I should be able to plug ##C##, ##X##, ##K##, and ##a## into ##\frac{2K(-CX^2)}{a}## to get the magnitude -- right?
 

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Hi kim_spitifre and welcome to PF.

The problem was solved correctly. Note that ##x## is not a constant but a variable. It represents the position along the rod of a length element ##dx## from an origin chosen such that point P is at coordinates {0, a}. Thus an element of length ##dx## on the rod bears charge ##dq=\lambda~dx##.
 
Thanks for the welcome & response! :) So, dq = lambda * dx (seems) to make sense, but I think I'm missing the bigger picture. Given the final equation in the above solution, am I expected to calculate lambda before I can solve for the magnitude? If so, I'm not sure how...
 
klm_spitifre said:
am I expected to calculate lambda before I can solve for the magnitude? If so, I'm not sure how...
No, you cannot calculate λ, it is supposed to be given because you have an infinite rod. If the rod were of finite length L and had total charg Q, then λ = Q/L. Do you see the difference?