hw3 - Physics 21 Fall, 2005 Solution to HW-3 21 P 45...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Physics 21 Fall, 2005 Solution to HW-3 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/ )...
View Full Document

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 .

Ask a homework question - tutors are online