# Field resulting from line charge

1. May 28, 2006

### DFWDraco76

A continuous line of charge lies along the x axis, extending from $$-x_{0}{$$ to $$-\infty$$. The line carries a uniform linear charge density $$\lambda_{0}$$. What is E, the magnitude of the electric field, at the origin?

Use the symbols $$\lambda_{0}, x_{0}$$ and k (for the Coulomb constant) to enter your answer.

Ok so I'm 99% sure I've got the correct answer, but for some reason the *#!\$ website we have to use to enter the answers is not accepting it. So before I blow a gasket over it I wanted to make sure the answer is correct.

Since we're looking for the magnitude of the field it's easier for me to flip it over to the +x axis.

Basically it comes down to integrating E =
$$\int_x_{0}^\infty \frac{k\lambda_{0}}{(x=x_{0})^2$$, which, if I've done it correctly, comes to $$\frac{k\lambda_{0}}{2x_{0}}$$.

Ah I can't get the latex right. integral from x_not to infinity of (k lambda_not) / (x_not + x)^2 dx.

Is that correct?

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On a side note - I'm new to the site and am still learning how to use latex. why is it that when you click "preview post" you can't see the result of the latex code? Unless I'm missing something, I have to actually post before I can see if I've made a mistake. Is that right?

Last edited: May 28, 2006
2. May 29, 2006

### maverick280857

I can't see your TeX...so I'll just describe the correct way of solving this. The answer seems correct by the way.

Consider an elemental charge dq at a distance x from the origin. Clearly $dq=\lambda dx$. The field due to this dq is directed towards the +x axis (assuming its all positve charge) and is given by

$$dE = \frac{k \lambda dx}{x^2}$$

Now all you have to do is integrate from $x=-x_{0}$ to $x=-\infty$. This is probably what you did...I just threw in the solution just in case you have a doubt.

Cheers
Vivek