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

by MrMaterial
Tags: charge, density, electric, field, infinite, line, parallel
 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 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 =$\int$$\frac{(\Delta q)}{\Delta x\stackrel{2}{}} dx$ $\rightarrow$ $\frac{1}{4\pi\epsilon\stackrel{}{0}}$$\int$$\frac{((\lambda0*x0)/xi)*xi}{xi\stackrel{2}{}}dx$ this seems to simplify to E0 = ∫(λ0*x0)/x^2 dx since λ0*x0 is a constant, it turns into [B]E0 = $\frac{(\lambda0*x0)}{4\pi\epsilon\stackrel{}{0}}$$\int$$\frac{dx}{x\stackrel{2}{}}$ 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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P: 9,832
 Quote by MrMaterial 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 infinitely long line-charge (L→+∞) I think by "L" the professor mean x0.
No, L is the length. You can consider (if it helps) that the wire extends from x = x0 to x0+L, then let L tend to infinity.
 I figure that every xi piece of the line charge makes a certain E field at the origin point which is x0 away.
No, it would be xi away.
 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 = ∫Δq/Δx^2 dx → 1/(4(pi)ε0)*∫(((λ0*x0)/x)*x)/x^2 dx
More accurately,
$E0 = ∫dq/x^2 = \frac{1}{4\pi\epsilon_0}\int_{x_0}^{\infty}\frac{λ_0x_0 dx}{x}\frac{1}{x^2}$
Note the denominator is x3, not x2.
P: 28
 Quote by haruspex More accurately, $E0 = ∫dq/x^2 = \frac{1}{4\pi\epsilon_0}\int_{x_0}^{\infty}\frac{λ_0x_0 dx}{x}\frac{1}{x^2}$ Note the denominator is x3, not x2.
ok this is the bit that confuses me.

here's what I did: E = $\int$$\frac{((\lambda0*x0)/x)*x}{x^2}$ because $(\lambda0*x0)/x$ 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 = $(\lambda0*x0)/xi$

can you please explain where i went wrong; how did you decide that dq was $(\lambda0*x0)/x$, not $((\lambda0*x0)/x)*x$

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 Quote by MrMaterial $(\lambda0*x0)/x$ is a density, and i need a charge, so i need to multiply by x.