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The Electric Field parallel to an infinite line charge (with change in charge density 
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#1
Nov1212, 09:09 PM

P: 28

1. The problem statement, all variables and given/known data
A line charge starts at x = +x0 and extends to positive infinity. The linear charge density varies inversely with distance from the origin, λ(x)=(λ0*x0)/x derive the expression for the electric field at the origin, E0, due to this infinetly long linecharge (L→+∞) 2. Relevant equations E = q/r^2 I think by "L" the professor mean x0. 3. The attempt at a solution First thing I wanted to do was to draw the situation. so the line charge i drew is a bit thick, but i made it big so it would be easier to show you guys how I am doing it. I figure that every xi piece of the line charge makes a certain E field at the origin point which is x0 away. This is the way i usually solve these types of problems. However, this is the first time a varying charge density has entered the equation for me. Not only is the distance of the charge varying, but the amount of charge per xi is varying as well. so every xi yields a certain q, which is a certain distance away from the origin which can be summed up with an integral from x0 to +∞. The q, or Δq, yielded would be xi*λi (distance * charge per distance) which will give me a charge value. E0 =[itex]\int[/itex][itex]\frac{(\Delta q)}{\Delta x\stackrel{2}{}} dx[/itex] [itex]\rightarrow[/itex] [itex]\frac{1}{4\pi\epsilon\stackrel{}{0}}[/itex][itex]\int[/itex][itex]\frac{((\lambda0*x0)/xi)*xi}{xi\stackrel{2}{}}dx[/itex] this seems to simplify to E0 = ∫(λ0*x0)/x^2 dx since λ0*x0 is a constant, it turns into [B]E0 = [itex]\frac{(\lambda0*x0)}{4\pi\epsilon\stackrel{}{0}}[/itex][itex]\int[/itex][itex]\frac{dx}{x\stackrel{2}{}}[/itex] does this seem right to you guys? I get an evaluation of (λ0*x0)/(4(pi)ε0*x0) then the x0's cancel and i get a straight constant of λ0/(4(pi)ε0) as the Efield at the origin point. 


#2
Nov1212, 10:59 PM

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[itex]E0 = ∫dq/x^2 = \frac{1}{4\pi\epsilon_0}\int_{x_0}^{\infty}\frac{λ_0x_0 dx}{x}\frac{1}{x^2}[/itex] Note the denominator is x^{3}, not x^{2}. 


#3
Nov1312, 12:05 AM

P: 28

here's what I did: E = [itex]\int[/itex][itex]\frac{((\lambda0*x0)/x)*x}{x^2}[/itex] because [itex](\lambda0*x0)/x[/itex] is a density, and i need a charge, so i need to multiply by x. (x here meaning the xi) And to me that makes sense because that would be the charge value from the small piece of the line charge. looking at my diagram, I have length xi and charge density λi. λi = [itex](\lambda0*x0)/xi[/itex] can you please explain where i went wrong; how did you decide that dq was [itex](\lambda0*x0)/x[/itex], not [itex]((\lambda0*x0)/x)*x[/itex] 


#4
Nov1312, 12:44 AM

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The Electric Field parallel to an infinite line charge (with change in charge density



#5
Nov1312, 01:30 AM

P: 28

ok so i confused the density*dx with density*x. I'm going to have to think about that, for some reason i associated the same x with both the density and the distance. That scares me!
I am starting to remember that dx = Δx, but it seemed like there was another Δx for the distance. ugh, didn't think i'd forget how to set up an integral. anyway thanks for helping! Looks like i gotta do more integral practice 


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