(adsbygoogle = window.adsbygoogle || []).push({}); 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 line-charge (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

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.

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

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