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fishturtle1
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Homework Statement
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.
Homework Equations
E = k(q/r2)
##\int \frac {dx} {\sqrt {x^2 + d^2}} = \frac {x} {d^2 \sqrt {x^2 + d^2}}##
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