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Electric Field of Uniformly Charged Rod at bisector

  1. Apr 12, 2017 #1
    1. The problem statement, all variables and given/known data
    (Taken from Serway and Jewett Chapter 23, Q44, 9th Edition)
    A thin rod of length ## l ## and uniform charge per unit length ##λ## lies along the x axis as shown in the image attached.
    (a) Show that the electric field at P, a distance d from the rod along its perpendicular bisector, has no x component and is given by ## E=2k_eλ sinθ_0/d.##
    b) What If? Using your result to part (a), show that the field of a rod of infinite length is ##E=2k_eλ/d.##

    2. Relevant equations
    ##dq=λ dl##
    ##\vec E=k_e∫ \frac {dq}{r^2} \hat {\mathbf r} ##
    Pythagoras i.e. ## r^2 = d^2 + (l/2)^2.##


    3. The attempt at a solution
    Starting out with replacing the infinitesimal ##dq## in integral with ## λ dl## to give an expression of ##\vec E=k_e λ ∫ \frac {dl} {r^2}\hat {\mathbf r}## (I took the ##λ## outside the integral since it is uniform and thus I assumed it was a constant). From there I could only find one way to continue.
    That way involved using Pythagoras saying that ##r^2 = {d^2 + \frac{l^2} {4}} ## which I then put back into the integral but found that I could not make any more progress with this. (Maybe some standard integral would work here but I do not know it).
    Where I stopped:
    ##\vec E= k_e λ ∫ \frac {dl}{ {d^2} +\frac {l^2}{4}}\hat {\mathbf r}##.
    What I would like to know is how to get rid of the vector by turning it into a scalar and ultimately make progress with this question.
    For part (b) I noticed that ##sinθ## disappears and becomes 1. Thus ##θ## must become ##90°## to allow this to happen. However, I struggle to deduce the other reasoning necessary for reaching the answer. Thus I would like help with this.

    I thank you for any help offered.
     

    Attached Files:

  2. jcsd
  3. Apr 12, 2017 #2
    Hi there! A question for you: What happened to your rhat? Is it possible that rhat also depends on l in some way?
     
  4. Apr 12, 2017 #3
    Well I know that ##\hat {\mathbf r}=\frac{\mathbf r}{\left|r\right|}## but I don't know how ##\hat{\mathbf r}## is related to ##l##. However I guess there must be a way to connect ##\hat{\mathbf r}## and ##l## as implied by your question but it escapes me at the moment.
     
  5. Apr 12, 2017 #4

    gneill

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    Staff: Mentor

    The side lengths of triangle POx are all related to d and θ by trig relationships (where 0 ≤ θ ≤ θo, and 0 ≤ x ≤ l/2). For example, the hypotenuse is equal to d/cos(θ). You can express x in terms of d and θ as well, and hence find how dx relates to dθ. How dq relates to dθ follows immediately.

    With some simplification you should end up being able to express your Coulomb's Law integration in a particularly simple form.
     
  6. Apr 12, 2017 #5
    True (there are many ways to do it) but this doesn't answer OP's question about vectors:

    So, referring to ##\hat {\mathbf r}=\frac{\mathbf r}{\left|r\right|}##, you already told us what ##\left|r\right|## was from the Pythagorean theorem. Now, ##\mathbf r## is a vector that points from the charge element to the point where you're trying to find the field. Try to write this in terms of variables given. All of this needs to be in the integral, as you correctly wrote.

    In the end, you can use trig knowledge to simplify and convert to ##\theta##, as gneill suggested.
     
  7. Apr 13, 2017 #6
    I'm still struggling with ##\mathbf r## and what variables it corresponds to.
    To try to solve this I looked at the textbook examples(Ex 23.7 The Electric Field due to a Charged Rod) and in it they say that vectors do not need to be handled due to the sum of the rod's contributions all being in one direction.
    However this confuses me how the vector notation can drop off like this and I would appreciate it if someone could assist with helping me to understand this.
     
  8. Apr 14, 2017 #7
    Absolutely, I never like how people explain that part away. The math will give you the correct answer if you follow it through. The vector ##\mathbf r## points from the charge element that you're integrating to the point where you're trying to calculate the field. If you consider a small piece of the rod to the left of the midpoint, then this vector will be ##\mathbf r = l \hat x + d \hat y##, where I'm using your notation with ##l## as the integration variable (let's then use the notation ##L## for the fixed length of the rod).

    So if you now plug this into your integral, you'll find that you can separate your integral into two directions. What is the expression that you have? Can you tell that one of the integrals will be zero when the bounds are ##-L/2## to ##L/2##?
     
  9. Apr 15, 2017 #8
    I have the following expression for the integral after using ##r=l~\hat x + d~\hat y## :
    ##\vec E= k_e~λ∫\frac {dl}{r^2} \frac {l~\hat x + d~\hat y}{\left| r\right|}##

    Then using L/2 and -L/2 as bounds and separating the x and y component integrals I noticed that the x integral equals zero.
    However, the y integral I had much more trouble with. I tried trigonometry and differentiation tied together i.e. ##dl=2r~cosθ~dθ## and ##\frac d r = cosθ## and used them in the integral. I continued with simplifying the integral until I reached this point :
    ## \vec E=2k_e λ\int_ 0^{θ_0} \frac {cos^2 θ} {\left|r\right|} dθ~\hat y##
    From there I could not reach the correct answer given in the question.
    What conceptual or "silly" mistake did I make?
     
  10. Apr 16, 2017 #9
    You're doing good, and you correctly noticed that the ##\hat x## integral zeros out. I think there is one oversight, which is ##\left|r\right|=r## in your notation. In other words, the integral you're trying to compute is ##\vec E= k_e~λ∫\frac {d}{r^3}dl \hat y##. It is a little tricky to see the next relation that you need, which is ##dl=r~dθ/cosθ##. I can post a sketch showing why this is true, if you like.

    After inputting these parts, your integral should be: ## \vec E=\frac{2k_e λ}{d}\int_ 0^{θ_0}cosθ dθ~\hat y##. Then you're home free.
     
  11. Apr 17, 2017 #10
    That would be great to see a sketch showing why ##dl= r~dθ/cosθ## is true as I don't know how it is derived.
    With respect to the question, I keep getting ## \frac {k_e λ} {d} sinθ_o## as my answer which is annoying as I am only off by a factor of 2. I used the trig relation that I mentioned in an earlier post ## \frac d r= cosθ## to get to my answer.
    I forgot to mention that I don't understand how ## \vec r = l \hat x + d\hat y## is derived so if you could show me how you got to this fact that would be great.
     
  12. Apr 18, 2017 #11
    Here's a quick diagram, hopefully it is clear:
    dx_theta.png
    Regarding the factor of 2, it comes from the limits of integration.

    Regarding how ##\vec r=l~\hat x + d~\hat y## is derived, it points from where the charges are to the point where you're trying to calculate the field (try drawing it).
     
  13. Apr 20, 2017 #12
    Where the factor of 2 comes from still puzzles me as in one of your posts the 2 is already outside the integral before you integrate the function.Therefore how can it come from the limits of integration?
    Also,I sketched the ##\vec r## equation and attempted to understand it geometrically but I can't seem to get ##\vec r## to equal ## l \hat x + d \hat y##.
    What I seem to get depends on where I start the drawing i.e. at the end of the rod( on the left side) but that gave me a vector of ## \frac l 2 \hat x + d\hat y## and when I sketched at the end of the rod on the right hand side I obtained##\frac{-l}{2} \hat x + d\hat y##.
    Also does the ## \vec r ## equation hold for all infinitesimal points on the rod? What I mean is wouldn't the component of ## \hat x## become smaller as we moved closer to the O point on the diagram?
    Sorry for the many questions, I am still not understanding this problem entirely.
    Thank you for the help so far plasmon_shmasmon and sorry for the late reply.
     
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