Electric Field w/ Continuous Charge Dist.

AI Thread Summary
A continuous line of charge with uniform linear charge density extends from x=x_0 to infinity, creating an electric field at the origin. The electric field is calculated by integrating the contributions from each infinitesimal charge element along the line, with limits of integration set from x_0 to infinity. The direction of the electric field is towards the left, indicating it is negative along the x-axis, due to the positive nature of the charge. The term involving infinity in the integral simplifies to zero, leading to the final expression for the electric field as E = -k_e λ/x_0. Understanding the limits of integration and the direction of the electric field is crucial for solving problems involving continuous charge distributions.
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Okay I'm really confused with continuous charge distributions, so I'm totally stuck...


A continuous line of charge lies along the x axis, extending from x=+x_0 to positive infinity. The line carries a uniform linear charge density lambda_0. What are the magnitude and direction of the electric field at the origin?


I have no idea where to approach or how to setup this problem up really.. What I tried to do is this..

Since, lambda = Q / L = dq / dx

E = int dE

= int k_e dq / r^2

= k_e int lambda dx / x^2

And then I'm completely stuck... Do I need to find lambda or just integrate? or...?


Any help is greatly appreciated. :smile:
Thanks.
 
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You're on the right track. Lambda is a constant, so put it out in front along with ke & integrate, (but first, determine your limits of integration).
 
Thanks for the quick response. I'm also confused on what my limits of integration are.. Do I go from {x_o to infinity}, or {0 to x_o}, or ...?

Okay so to continue on,

(k_e)(lambda) int{x_otoinf} x^-2 dx

(k_e)(lambda) int{x_otoinf} - 1 / x

(k_e)(lambda) [(-1 / x_o) - (1 / inf)] cos 0

[(-k_e lambda / x_o) + (-k_e lambda / inf)] i


Thanks for the help.:smile:
 
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Good.

Now, do you understand why it is {x_otoinf} or was that a lucky guess?

Where did cos 0 come from?

and what are you planning to do with (-k_e lambda / inf)?
 
Okay well the limits were partially a guess.. But I figured from x_o to infinity because it is a continuous line of charge along the x-axis but I wasn't sure because like you said I don't know what to do with (-k_e)(lambda)/inf.


I figured cos 0 because I thought it was a vector that lies along the x-axis but has no angle and no y component(j). So, cos 0 = 1 and the answer would have +i given in the notation. Is this wrong?
 
You're right about the limits. When you compute this integral, you are *summing* the electrical fields arising from the continuous line of tiny charges dq which runs from x_0 to infinity -- the location of the charges is what determines the limits of integration.

I would say the direction of the field is -i. Since it wasn't specified that the charge is negative, I would assume it is positive.
The field direction is away from a positive charge, and the origin is located to the left of the line of charge. So the field at the origin is directed towards the left.

As to the cosine, yes, cos 0 = 1 and yes, this field is along the x-axis and yes, there is no y component. It's just odd that the cosine just pops up out of nowhere. You're not doing a dot product here, and you didn't have any trig functions in the expression you are integrating. So it didn't arise out of anything in your computation and it doesn't really serve any purpose. You can explain your reasoning behind the direction of the field just as well without it, and in fact you'd have to give the same explanation to justify the cosine itself. So I'd say it doesn't belong there.

And as to (-k_e)(lambda)/inf
what is the value of x/ ∞ (where x = any finite number)?
 
could be zero. I'm confused on how that effects lambda though.
 
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My point is just that
-\frac{k_e \lambda}{\infty} = 0
so that entire term drops out, and the final result is just
E = - \frac{k_e \lambda}{x_0} \bold{i}
 
Ah okay.. I understand now.

Thanks for the help! :smile:
 
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