Unformatted text preview: Physics 21 Fall, 2005 Solution to HW3 21 P 45 Suppose the charge Q on the ring of Fig. 21–27 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.) θ P dQ x y z a r The total charge Q is uniformly distributed over the semi circular ring of length πa , so the charge density λ is λ = Q πa , and the charge dQ on an arc length dl = a dθ is dQ = λ dl = Q π dθ. The field at P from a charge dQ at the point shown is d E = k dQ r 3 r = k dQ ( x ˆ i − a sin θ ˆ j + a cos θ ˆ k ) ( x 2 + a 2 ) 3 / 2 . We must integrate d E from θ = 0 to π , corresponding to the upper half circle where the charge is. By symmetry, the z component will cancel, so E = kQ π 1 ( x 2 + a 2 ) 3 / 2 Z π ³ x ˆ i − a sin θ ˆ j ´ dθ. The ˆ i integral is xπ , and the ˆ j integral is − 2 a , so we have E = kQ π ( xπ ˆ i − 2 a ˆ j ) ( x 2 + a 2 ) 3 / 2 = kQ ³ x ˆ i − (2 a/π )...
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 Fall '07
 Hickman
 Physics, Charge, Electric charge, 4L, 2L, 2 L, mg T, charge dq

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