A Continuous Line of Charge Lies Along

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Magnitude and Direction of Electric Field at the origin? Please help

• Thread starter SilverGirl
• Start date Sep 28, 2008

Homework Statement

A continuous line of charge lies along the x axis, extending from x=+xo to positive infinity. The line carries charge with a uniform linear charge density lambda. What are the magnitude and direction of the electric field at the origin?

Homework Equations

E = lambda/2pieor for the magnitude of electric field produced by a uniformly charged infinite line.

The Attempt at a Solution

I have drawn a cylindrical Gaussian Surface around it. By the wording of the question, I am not sure where the line of charge is placed (does it start at the origin?). I have placed the line at the origin, with the cylinder starting at the origin. I have concluded (not sure if it is correct) that the field on the end of the cylinder is equal to zero, because the field can only pass through the center of the circle.

Am I close at all to the answer?

Answers and Replies

A continuous line of charge lies along the x axis, extending from x=+xo to positive infinity. The line carries charge with a uniform linear charge density lambda. What are the magnitude and direction of the electric field at the origin?
…
I have drawn a cylindrical Gaussian Surface around it. By the wording of the question, I am not sure where the line of charge is placed (does it start at the origin?) …

Hi SilverGirl!

(have a pi: π and an epsilon: ε and a lambda: λ and an infinity: ∞ )

No, the line starts at x₀ and goes to +∞.

hmm … no idea what you mean by a cylindrical Gaussian Surface.

Hint: slice the line into bits of length dx, and integrate from x₀ to +∞.

Is xo at any particular place on the x axis though? Maybe I didn't have to use the cylinder.

Is xo at any particular place on the x axis though? Maybe I didn't have to use the cylinder.

erm … it's at x₀ !

isn't that particular enough for you?

Not really..lol..because couldn't x0 be at the origin? Also, couldn't it be very very far away from the origin? Do you know of a way to do this without integrating?

lol..good point.

I am guessing the field is not 0 at the origin, even though the line of charge is not there.

Yes, you're right. It's not zero at the origin. It would be zero if the line extended from [tex]-\infty[/tex] to [tex]\infty[/tex]. Now tiny-tim's hint was to integrate it. You need to start with [tex]dE = \frac{dq}{4\pi \varepsilon_0 r(x)^2}[/tex], where r(x) is function of x which tells you the distance from any point on the line of charge to the origin.

You still need to express dq in some other way so that the integration can be done.

By symmetry, you should be able to determine what the E-field direction at O should be.

I've been using this problem for reference but I am so lost...I'm doing the same exact thing.

I am having trouble integrating. Could someone please walk me through it step by step with this equation?

I've been using this problem for reference but I am so lost...I'm doing the same exact thing.

I am having trouble integrating. Could someone please walk me through it step by step with this equation?

Hi Doc Z!

Show us how far you've got with the equation (and which equation? ), and where you're stuck, and then we'll know how to help.

I'm using E=2ke * lambda/r

I pretty much don't even know how to start integrating this.

I'm using E=2ke * lambda/r

I pretty much don't even know how to start integrating this.

Hint: slice the line into bits of length dx, and integrate from x₀ to +∞.

What do you get?

I found the answer to be ke * lambda/x0 in the -i direction but I don't understand how to get it. I don't remember integrals too well.

I found the answer to be ke * lambda/x0 in the -i direction but I don't understand how to get it. I don't remember integrals too well.

You know that E = k*Q/r² for a point charge.

So the field at the origin from any point over the range of charges is given by

E_(r) = k*Q/r²

But Q by the problem statement is

ΔQ = λ * Δr

Rewriting the equation:

E_(r) = k * λ * Δr/r²

Consequently then to sum up all the charge elements from r = x_o to r = ∞ you take the definite integral.

[tex] E_{(r)} = \int_{x_0}^{\infty} \frac{k * \lambda * dr}{r^2} [/tex]

Last edited: Oct 24, 2008

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