- #1
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##.
2. The attempt at a solution
My professor actually solved the first part of this problem during lecture:
<br /> 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}<br />
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?
