- #1
Mddrill
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Homework Statement
Positive charge Q is distributed uniformly along the positive y-axis between y=0 and y=a . A negative point charge −q lies on the positive x-axis, a distance x from the origin
Calculate the y-component of the electric field produced by the charge distribution Q at points on the positive x-axis.
Homework Equations
E=K*Q/(r^2)
The Attempt at a Solution
I formed the equation dE_y=K*[(lambda)dy/(x^2+y^2)]*(y/(x^2+y^2)^(1/2) and trying to integrate
E_y=K(lambda)*[integral 0 to a][y/(x^2+y^2)^(3/2)]
using trig substitution y=xtan(theta) and dy=xsec^2(theta)d(theta)
to make this a little easier to read I will say K(lambda)=C
C[integral]{[(x^2)tan(theta)sec^2(theta)]/[(x^3)(1+tan^2(theta)^(3/2)]}d(theta)
=C[integral]{tan(theta)/xsec(theta)}
=C[integral]{sin(theta)/x} = C[-cos(theta)/x] cos(theta) = x/[(x^2 +y^2)^(1/2)]
=C[-1/[(x^2 +y^2)^(1/2)] from 0 to a
=(-K(lambda))/[(x^2 + a^2)^(1/2)] (lambda) = Q/a
=-(KQ)/[a(x^2+ a^2)^(1/2)]
This is the answer I came up with, but apparently its wrong. can anyone explain to me why this is wrong and what is the right way to do it?
Thank you
P.S. this is my first post, so idk if there's a "cleaner" way to write these equations, if there is please tell me. Sorry if they are hard to follow.