http://capa.mcgill.ca/res/mcgill/dcmcgill/oldproblems/newerlib/Graphics/Gtype51/prob19a.gif[/URL] 1. A charge Q = 8.10 ×10-4 C is distributed uniformly along a rod of length 2L, extending from y = -11.4 cm to y = +11.4 cm, as shown in the diagram 'on your assignment above. A charge q = 1.80 ×10-6 C, and the same sign as Q, is placed at (D,0), where D = 17.5 cm. Use integration to compute the total force on q in the x-direction It is already proven that : The magnitude of the force on charge q due to the small segment dy is dF=(kqQ/2Lr2)dy. 3. Given whats already proven that for a small dy: dF = (kqQ/2Lr2)dy so i integrated both sides and i got: ∫dF = k*q*Q / 2L ∫ 1/r^2 dy since r^2 = D^2+y^2 therefore the function becomes: ∫dF = k*q*Q / 2L ∫ 1/D^2+y^2 dy now what should i do next???? and the equation i derived, was it right at the first place??? THANKS!