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foxjwill
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Homework Statement
Find the magnitude of the total electric field strength, [tex]E[/tex], at a distance [tex]x[/tex] produced by a wire of length [tex]l[/tex] and uniform linear charge density [tex]\lambda[/tex] such that [tex]x \gg l[/tex].
Homework Equations
The Attempt at a Solution
This how we went through it in class:
First, we realized that the electric field in the y direction would be zero. So, working in the x direction,
[tex]dE = \frac{kdQ}{\sqrt{x^2 + y^2}}\cos{\theta}[/tex]
Since [tex]\lambda[/tex] is a uniform, it will be the same no matter how small of a portion of the wire is taken, so we can say that [tex]\lambda dy = dQ[/tex]. Also, based on the graphic and some ingenious trig,
[tex]\cos{\theta} = \frac{x}{\sqrt{x^2+y^2}}[/tex].
So,
[tex]dE = \frac{xk\lambda dy}{(x^2+y^2)^{3/2}}[/tex].
Integrating from [tex]-l[/tex] to [tex]l[/tex],
[tex]
E = {\int^l _{-l} \frac{xk\lambda}{(x^2+y^2)^{3/2}} dy}= \frac{2k\lamda l}{x\sqrt{x^2+l^2}}
[/tex]
But the question has that [tex]x\gg l[/tex]. How does this work? According to our teacher (it's her first year teaching this particular course, and for some reason the previous teachers all decided, among other things, not to tell her what kind of questions to expect), the answer should come out to be
[tex]E = \frac{2k\lambda}{x^2}[/tex].
What's stumping us is how we can just get rid of [tex]l[/tex]. If it was just on the bottom, we could find the limit as [tex]l \rightarrow 0[/tex]. But as it is now, doing so would result in a value of 0, which is wrong.
I was thinking possibly of doing something with differentials,
(e.g.
[tex]dE = \frac{-2k\lamda (l dx (2x^2 + l^2) - x^3 dl)}{x^2(x^2+l^2)^{3/2}}[/tex]
[tex]\lim_{l \rightarrow 0} dE = \frac{2k \lambda dl}{x^2}[/tex]
but then what do we do with [tex]dl[/tex]?)
or maybe taking the limit as [tex](x-l) \rightarrow \inf [/tex] (which I'm not entirely sure how to do).