32. We assume q > 0. Using the notation λ = q / L we note that the (infinitesimal) charge 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 θ factors just discussed) the magnitude is ()()22 22 2 22 00 00 /2 2 32 22202200 0 22 2 2 00 22s i n442221 22 4 2 LLLL dq dx y E rx R x R qLR Rd x x Rx R xR qL q LR R LR LR θ πε πε πε πε πε πε
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This note was uploaded on 06/07/2010 for the course PHYS 344 taught by Professor Bb during the Spring '10 term at The Petroleum Institute.