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Electric Field Produced by a Finite Charged Wire

  1. Jun 7, 2017 #1
    1. The problem statement, all variables and given/known data
    A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E⃗ at the point P, a distance d above the midpoint of the wire.

    What is the magnitude E of the electric field at point P? Throughout this part, express your answers in terms of the constant k, defined by k=1/4πϵ0. Express your answer in terms of L, λ, d, and k.

    15538_a.jpg


    2. Relevant equations

    E = k(q/r2)

    ##\int \frac {dx} {\sqrt {x^2 + d^2}} = \frac {x} {d^2 \sqrt {x^2 + d^2}}##


    3. The attempt at a solution

    ##E = E_x + E_y##

    ##E_x = 0## by symmetry.

    ##dE_y = dEcos(cos^{-1}(\frac {x} {\sqrt {x^2 + d^2}}) = dE \frac {x} {\sqrt {x^2 + d^2}}##

    so to find dE:

    ##E = k \frac {q} {r^2} ##

    ##dE = k \frac {dq} {\sqrt {x^2 + d^2}^2} ## , ##dq = \lambda dx##

    ##dE = k \frac {\lambda dx} {\sqrt {x^2 + d^2}^2}##

    then substitute this in the dE_y equation:

    ##dE_y = k \frac {\lambda dx} {x^2 + d^2} \frac {d} {\sqrt {x^2 + d^2}}##

    ## E_y = k\lambda d \int \frac {dx} {(x^2 + d^2) ^ {3/2}} ##

    then I used an integral table to for the integral:

    ##E_y = k\lambda d \frac {x} {d^2 \sqrt{x^2 + d^2}} ##

    I'm not sure what to do with the x's since the answer is supposed to be in terms of L, λ, d, and k.

    ##\int dq = \int \frac {\lambda} {dx} ##

    ##q = \lambda x##

    ##x = q / \lambda## , ##\lambda = q/2L##

    ##x = q (q / 2L) = 2L##

    So ##E = E_y = \frac {k\lambda 2L} {d \sqrt{4L^2 + d^2}} ##
    This is not right
     
  2. jcsd
  3. Jun 7, 2017 #2
    I think you are taking the wrong angle here.Try to look it again.Also I reccomend you to pick a point charge and we know the relationship that ##λ=dq/dx## from there as you proceed Theres only y component of the E-Field.Then after writing the integral turn it to in terms of dx and think about the boundries of your integral.In this way you should be able to do this very easily.
     
  4. Jun 7, 2017 #3
    it worked, I did the same integral but put limits -L and L like this:

    ##\int_{-L}^L \, \frac{dx} {(x^2 + d^2)^{3/2}} ##

    ## = [\frac {L} {d^2 \sqrt{(L^2 + d^2)}} - \frac {-L} {d^2 \sqrt{((-L)^2 + d^2)}}]##

    ## = \frac {2L} {d^2 \sqrt{(L^2 + d^2)}} ##

    final answer: ##E = k\lambda d ( \frac {2L} {d^2 \sqrt{(L^2 + d^2)}}) ##

    = ## k\lambda \frac {2L} {d \sqrt{(L^2 + d^2)}}##

    Thank you
     
  5. Jun 7, 2017 #4
    Your welcome :)
     
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