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Homework Help: Much greater than

  1. Jan 5, 2008 #1
    1. The problem statement, all variables and given/known data
    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].


    2. Relevant equations

    3. 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].


    [tex]dE = \frac{xk\lambda dy}{(x^2+y^2)^{3/2}}[/tex].

    Integrating from [tex]-l[/tex] to [tex]l[/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}}

    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,
    [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).
  2. jcsd
  3. Jan 5, 2008 #2


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    As far as I can tell, the correct answer is
    [tex]E = \frac{2kl\lambda}{x^2}[/tex].

    That makes sense because when [tex]x>>l[/tex] you are essentially looking at a finite lump of charge from a distance. Bigger [tex]l[/tex] means more charge so more E.
  4. Jan 5, 2008 #3
    The correct answer is

    [tex]E = \frac{2kl\lambda}{x^2}[/tex]

    From [tex]x\gg l[/tex] you have [tex]\frac{l}{x}\ll 1\Rightarrow \frac{l}{x}\rightarrow 0[/tex]. So write the denominator as

    [tex]x\,\sqrt{x^2+l^2}=x\,\sqrt{x^2\left(1+(\frac{l}{x})^2\right)} \rightarrow x^2 \quad \text{for} \quad \frac{l}{x}\rightarrow 0 \quad \text{and} \quad x>0 [/tex]
  5. Jan 5, 2008 #4
    But if it's negligible compared to x, wouldn't changing it slightly not have much affect on E?

    I was just thinking it through on the lines of differentials. Above, I wrote that

    [tex]\lim_{l \rightarrow 0} dE = \frac{2k \lambda dl}{x^2}[/tex].

    However, since [tex]x \gg l[/tex], wouldn't that mean that [tex]dx \gg dl[/tex]? So, that would mean that

    [tex]\lim_{l \rightarrow 0} dE = \frac{2k \lambda}{x^2} {\lim_l \rightarrow 0} dl = 0[/tex]

    and not [tex]\frac{2k \lambda dl}{x^2}[/tex].

    Therefore, we can say that, overall, [tex]l[/tex] has negligible affect on [tex]E[/tex], and

    [tex]E = \frac{2k\lambda}{x^2}[/tex].

    Is this right?
  6. Jan 5, 2008 #5
    HallsofIvy, the "k" is just the Coulomb's constant! Nothing peculiar :smile:
  7. Jan 5, 2008 #6
    I found the actual question. It's similar to what I have above (which I'm still confused about; I don't really understand why you can keep the [tex]l[/tex] on top, but not on the bottom.), but not the same.

    A very long, straight wire has charge per unit length 2.10e-10 C/m. At what distance from the wire is the electric field magnitude equal to 3.00 N/C?

    The answer is 1.26 m, which can be gotten by plugging values into the formula:
    [tex]E = \frac{2k\lambda}{x^2}[/tex],
    where [tex]E[/tex] = electric field strength, [tex]k = \frac{1}{4 \pi \epsilon_0} \texttt{m/F}[/tex], [tex]\lambda[/tex] = linear charge density, and [tex]x[/tex] = distance.

    Again, I got to

    [tex]E = \frac{2kl\lambda}{x\sqrt{x^2+l^2}}[/tex].

    Now, since this time [tex] l \gg x[/tex], what do I do?
    Last edited: Jan 5, 2008
  8. Jan 5, 2008 #7
    Look at post #3
  9. Jan 5, 2008 #8
    Hmm. I think what's tripping me up is the definition of "much greater than." Is there a rigorous definition?

    I'm going to have to think of a better way to formulate my confusion.
  10. Jan 5, 2008 #9
    Since now you have
    you have to integrate in the domain [tex](-\infty,+\infty)[/tex] so the electric field is

    [tex]E = \frac{2\,k\,\lambda}{x}[/tex]
  11. Jan 5, 2008 #10
    Oh. That makes sense.

    I'd still like to know if there's a rigorous definition for "much greater than."
  12. Nov 17, 2009 #11
    As would I. Can anyone provide something? I am working with RF propagation, and far field region is defined as d>>wavelength. 10x? 20x? 100x?
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