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### a) Do this by dividing the charged rod into 2 equal length segments an.docx

1. a) Do this by dividing the charged rod into 2 equal length segments and treating each segment as a point change at the center of each segment. b) How could you improve your estimate? Solution A charge Q is uniformly distributed over a line of length L, linear charge density, lemda=total charge /total length =Q/L Q is equal to intergral of lemda dL from limit 0 to l. Then, we get E = (1/4?*epsilon -not) (lemda)dl/r 2 Let the rod subtend angle 2(theta) at P. Consider element dl at distance l having charge dq= lemda*dl l = x tan? dl = X sec 2 ? d? dE = k(lemda) dl/(x 2 +l 2 ) dE = k(lemda) X sec 2 ? d? /(x 2 +(X tan?) 2 ) dE = k(lemda) X sec 2 ? d?/(x 2 sec 2 ?) = k(lemda)d?/x Here, dE Cos ? along x-axis dE Sin? along y-axis which cancel each other above and below. a) If we divide a rod 2 equal length so l=L/2,then
2. E =E1+E2 = electric field due to ntegral from 0 to L/2 + electric field due to integral from L/2 to L E= K *lemda *sin?/x B) we get same electric field when integral from 0 to L.
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