ch22_p29 - 29. We assume q > 0. Using the notation = q/L...

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on an element dx of the rod contains charge dq = λ dx . By symmetry, we conclude that all horizontal field components (due to the dq ’s) cancel and we need only sum (integrate) the vertical components. Symmetry also allows us to integrate these contributions over only half the rod (0 x L /2) and then simply double the result. In that regard we note that sin θ = R / r where 22 rx R =+ . (a) Using Eq. 22-3 (with the 2 and sin factors just discussed) the magnitude is () 2 00 2 2 32 2 0 0 2 2 2 2s i n 44 21 4 2 LL L L dq dx y E R xR qLR Rd x x Rx R qL q LR R LR πε §· λ == ¨¸ + ©¹ + ªº λ «» + + ¬¼ + + ³³ ³ G where the integral may be evaluated by elementary means or looked up in Appendix E
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