- #1
tongpu
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Homework Statement
rod -a-------|---------a on x-axis, for x>-a has -Q charge, for X<a has +Q charge, find the electric field at point (x,0) x is positive
Homework Equations
lambda = Q/a linear charge density dQ = (lambda)(ds)
E=KQ/r^2
The Attempt at a Solution
i make dE=kdQ/(x-s)^2
integrate kdQ/(x-s)^2 between -a and a ( i can make it 0 to a and multiply by two, symmetry)
i get the result 2KQ/a[1/x^2 - 1/(x-a)^2] for E sub-x
now if i want to find x>>a i need o somehow change [1/x^2 - 1/(x-a)^2] to the 2nd taylor expansion term but how? And is my integration correct?