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I would really appreciate a detailed answer!! :) thank you
PHYS208 Week 2 3. A thin rod with uniform charge +Q and length L is placed with one end at (0,b) and the other end at (0, -b)
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Answer #1

a] We know that electric field due to finite charged wire is given by:

E = k*lambda/a *(sin theta 1+ sin theta2) where theta1 and theta2 are angle subtended by ends with x axis.

= k*(Q/L)/a *(b/sqrt(a^2+b^2) + b/sqrt(a^2+b^2))

= k*(Q/L)/a *2b/sqrt(a^2+b^2)

= 2k*Q*b/[aL*sqrt(a^2+b^2)] where k is coulombs constant

b] Force = charge*E

= 2q*2k*Q*b/[aL*sqrt(a^2+b^2)]

= 4k*Q*q*b/[aL*sqrt(a^2+b^2)] where k is coulombs constant

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