Deriving Electric Field at Origin for Infinite Line Charge

In summary, the problem involves an infinitely long line charge with a varying linear charge density that is inversely proportional to distance from the origin. The expression for the electric field at the origin is derived by setting up an integral of dq/x^2 and simplifying it to dq/x^3. This results in an electric field of λ0/(4(pi)ε0) at the origin point.
  • #1
MrMaterial
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0

Homework Statement


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→+∞)

Homework Equations



E = q/r^2

I think by "L" the professor mean x0.


The Attempt at a Solution



First thing I wanted to do was to draw the situation.

JZPX8.jpg


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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  • #2


MrMaterial said:
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,
[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 x3, not x2.
 
  • #3


haruspex said:
More accurately,
[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 x3, not x2.

ok this is the bit that confuses me.

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


MrMaterial said:
[itex](\lambda0*x0)/x[/itex] is a density, and i need a charge, so i need to multiply by x.
For an element dx with density ρ the charge is ρ.dx. I.e. the multiplication by length is already there in the integration process.
 
  • #5


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! :eek:

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 got to do more integral practice
 

1. What is an infinite line charge?

An infinite line charge is a hypothetical charge distribution that extends infinitely in one direction and has a constant charge density along its length.

2. How is the electric field at the origin derived for an infinite line charge?

The electric field at the origin for an infinite line charge can be derived using Coulomb's Law, which states that the electric field is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance between the charge and the point of interest.

3. What is the formula for calculating the electric field at the origin for an infinite line charge?

The formula for calculating the electric field at the origin for an infinite line charge is E = λ/2πε0r, where λ is the charge density, ε0 is the permittivity of free space, and r is the distance from the charge to the origin.

4. Can the electric field at the origin for an infinite line charge be negative?

Yes, the electric field at the origin for an infinite line charge can be negative if the charge density is negative. This indicates that the direction of the electric field is towards the origin, rather than away from it.

5. How does the electric field at the origin for an infinite line charge compare to that of a point charge?

The electric field at the origin for an infinite line charge is always perpendicular to the charge, whereas the electric field for a point charge can be in any direction. Additionally, the magnitude of the electric field for an infinite line charge decreases as the distance increases, whereas the magnitude of the electric field for a point charge decreases as the inverse square of the distance.

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