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Distribution of Charge along a straight line

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data

    A straight nonconducting plastic wire 8.50 cm long carries a charge density of +175nC/m distributed uniformly along its length. It is lying on a horizontal table top
    (a) Find the magnitude and direction of the electric field this wire produces at a point 7.00cm directly above its midpoint.

    Q = +175nC/m (charge density)
    L = .085m (length of charge)
    D = .060m (Distance from charge)
    r^2 = x^2 + H^2

    2myvr02.jpg

    2. Relevant equations

    dQ = (Q / L) dx (Total charge over Total distance)
    dEy = kdQ/r^2 sin θ
    sin θ = D/r
    dEx = 0
    k = electric constant
    3. The attempt at a solution

    Ey = kQD/L(x^2 + H^2) ∫ 1 / (x^2 + D^2) ^(3/2) Integrand from 0 to L
    = KQD/(D^2(D^2 + X^2)^(1/2))

    Answer I got was 2.5 x 10^5 N/C

    This is wrong because the charge is a density C/m. I tried to multiply that answer with the total distance to cancel the m but answer still came out wrong. What did I do wrong here?
     
    Last edited: Oct 9, 2011
  2. jcsd
  3. Oct 9, 2011 #2
    If you don't mind, I'll first make sure I've got your procedure right by translating it into TeX (because for some reason my eyes don't deal well with ASCII equations and I want to make sure I'm helping you find the right errors.

    So I'm seeing

    [itex]E_{y} = \int^{x = L/2}_{x = -L/2} \frac{\lambda D dx}{4\pi\epsilon_{0}(x^{2}+D^{2})^{3/2}} = \frac{\lambda D}{4\pi\epsilon_{0}} \int^{x = L/2}_{x = -L/2} \frac{dx}{(x^{2}+D^{2})^{3/2}}[/itex]

    Is this what you had? Because if it is, I'm pretty sure you're on the right track. If not, can you see how I derived it? If you don't see it and it's not what you got, just let me know so I can go back and help you derive it step-by-step.

    Anyway, once you've got that you can solve the integral and evaluate it at its proper bounds and it should yield the correct answer.

    (You can simplify things substantially by taking advantage of the symmetry of the situation by setting the midpoint of the charged line segment at x = 0, thereby putting your bounds of integration to |x| <= L/2. Note also that by symmetry all of the horizontal components of the electric field vector will cancel each other.)
     
  4. Oct 9, 2011 #3
    This is what I did

    Ey = kQHdx
    ||||| D || (x2+ H2)3/2

    My integral is from 0 to D.
    Since Q is a charge density and not the actual charge it self, what do I have to do to convert it so it fits the equation of E?

    Ignore the white lines, that my bad attempt to make the equation look clear for you guys.
    Btw, how do you type like that?
     
  5. Oct 9, 2011 #4
    What are you setting D equal to? The distance between x and the midpoint of the rod? Because from the picture you gave, when you integrate from 0 to D you're integrating an x-dependence over an interval that includes a y-value. o_O;
     
  6. Oct 10, 2011 #5
    Sorry I meant from 0 to L, I wrote one thing and drew another.

    I'm still confused with the given charge density, how to I turn that into a charge?
     
  7. Oct 10, 2011 #6
    Okay, so your linear density of charge (which you've been representing as Q and I've been representing as [itex]\lambda[/itex]) is, by definition, the charge per unit length through a differential element dx of the charged rod. Recall that the electric field is given by

    [itex]E_{y} = \int\frac{dq sin\theta}{4\pi\epsilon_{0}r^{2}}[/itex].

    Because your differential charge dq is given by

    [itex]dq = \lambda dx[/itex]

    and the distance r between an arbitrary point x along the rod and the point (L/2, D) is given by

    [itex]r^{2} = (x - L/2)^{2} + D^{2}[/itex]

    and [itex]sin\theta[/itex] is given by

    [itex]sin\theta = \frac{D}{r} = \frac{D}{\sqrt{(x - \frac{L}{2})^{2} + D^{2}}}[/itex]

    the integral

    [itex]E_{y} = \int\frac{dq sin\theta}{4\pi\epsilon_{0}r^{2}}[/itex]

    becomes

    [itex]E_{y} = \int^{L}_{0}\frac{\lambda D dx}{4\pi\epsilon_{0}( (x - \frac{L}{2})^{2} + D^{2})^{3/2}}[/itex]

    which by symmetry about the midpoint [itex]\frac{L}{2}[/itex] actually just becomes

    [itex]E_{y} = 2\int^{\frac{L}{2}}_{0}\frac{\lambda D dx}{4\pi\epsilon_{0}( (x - \frac{L}{2})^{2} + D^{2})^{3/2}} = \frac{\lambda D}{2\pi\epsilon_{0}}\int^{\frac{L}{2}}_{0}( (x-\frac{L}{2})^{2} + D^{2})^{\frac{-3}{2}}dx[/itex]

    So the tl;dr version would simply be this:
    Using the notation

    [itex]\lambda[/itex] (or [itex]Q[/itex], depending on preference) [itex]= \frac{q}{L}[/itex]

    we see that [itex] q = \lambda L (= QL)[/itex]. We can also conclude from the analytical definition of linear charge density that the differential charge (dq) through a differential length element (dx) is given by

    [itex]Q = \lambda = \frac{dq}{dx} \Rightarrow dq = \lambda dx[/itex] or [itex] Q dx[/itex] depending on the notation you prefer.

    You can verify this by integrating [itex]\Sigma q = \int^{L}_{0}\lambda dx[/itex] which yields [itex]\Sigma q = \lambda L (= QL)[/itex] which is what we should expect because the linear charge density is constant/uniform (or in this case simply x-invariant) -- and it does in fact conform to our previous derivation.

    This confirms the argument that you can use your linear charge density to impute the differential charge via the relationship [itex]dq = \lambda dx[/itex].
     
  8. Oct 11, 2011 #7
    Ah, I see. Thank you very much!
     
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