A charge Q = 5.50 x 10-04 C is distributed uniformly along a rod of length 2L, e

Question

A charge Q = 5.50 x 10-04 C is distributed uniformly along a rod of length 2L, extending from y = -21.0 cm to y = +21.0 cm, as shown in the diagram above. A charge q = 7.25 x 10-06 C, and the same sign as Q, is placed at (D,0), where D = 13.5 cm. Use integration to compute the total force on q in the x-direction. My attempt... Im pretty sure the integral i need to use is dF= (kqQ/2Lr^2)dy which is the magnitude of the force on charge q due to the small segment dy... Then after integrating this or trying I ended up with F = kqQ/Dsqrt(D^2+L^2)..I am getting 375 N and this isnt right can you please help me figure out the right answer?

Explanation / Answer

Choose a small piece of rod "dy" located at some "y". That piece has charge "dq" which exerts force "dF" on "Q" by Coulomb's Law; dF = kQdq/r^2 , (where r^2 = y^2 + D^2) Because dF is a vector you can not simply add the force due to all the dq's. So You have to get components first. And you're right the y-component will be zero from symmetry. The x-component is then; dFx = dFCos() = dF(D/r) = kQDdq/r^3 You next write "dq" in terms of metrical quantities via a density. The rod is uniformly charged so its density is q/2L , which is constant for any piece of rod so it is also dq/dy . Equating these gives; dq = (q/2L)dy Put this into the dFx eq, then rewrite "r" in terms of y & D then you can integrate over "y" from -L to +L.

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