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Unformatted text preview: Physics 21 Fall, 2005 Solution to HW3 21 P 45 Suppose the charge Q on the ring of Fig. 2127 was all distributed uniformly on only the upper helf of the ring, and no charge was on the lower half. Determine the electric E at P . (Take y vertically upward.) P dQ x y z a r The total charge Q is uniformly distributed over the semi circular ring of length a , so the charge density is = Q a , and the charge dQ on an arc length dl = a d is dQ = dl = Q d. The field at P from a charge dQ at the point shown is d E = k dQ r 3 r = k dQ ( x i a sin j + a cos k ) ( x 2 + a 2 ) 3 / 2 . We must integrate d E from = 0 to , corresponding to the upper half circle where the charge is. By symmetry, the z component will cancel, so E = kQ 1 ( x 2 + a 2 ) 3 / 2 Z x i a sin j d. The i integral is x , and the j integral is 2 a , so we have E = kQ ( x i 2 a j ) ( x 2 + a 2 ) 3 / 2 = kQ x i (2 a/ )...
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This note was uploaded on 08/06/2008 for the course PHYS 21 taught by Professor Hickman during the Fall '07 term at Lehigh University .
 Fall '07
 Hickman
 Physics, Charge

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