253-wrk-23 - HPHY 253 Workout 23 Electric Fields Dr. A. E....

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Unformatted text preview: HPHY 253 Workout 23 Electric Fields Dr. A. E. Bak Name 1. A & 4 : 50- & C particle lies at the location (in m ) of (1 : 00 ; 4 : 00) in the xy plane. Find the electric &eld that it produces at ( & 1 : 00 ; 7 : 00) . The source charge here is q = & 4 : 50 & C = & 4 : 50 ¡ 10 & 6 C : We have r = length vector from (1 : 00 ; 4 : 00) to ( & 1 : 00 ; 7 : 00) = ( & 1 : 00 & 1 : 00) e x + (7 : 00 & 4 : 00) e y = ( & 2 : 00 e x + 3 : 00 e y ) m ; having magnitude r = q ( & 2 : 00) 2 + (3 : 00) 2 ¢ 3 : 606 m : Therefore, we obtain E = k e q r r 3 = & 8 : 987551787 ¡ 10 9 ¡& & 4 : 50 ¡ 10 & 6 ¡ ( & 2 : 00 e x + 3 : 00 e y ) (3 : 606) 3 = (1730 e x & 2590 e y ) N= C as the electric &eld at ( & 1 : 00 ; 7 : 00) due to the & 4 : 50- & C point charge. 2. In the xy plane, a +6 : 00- & C particle lies at the location (in m ) of (0 : 300 ; 0) and a & 3 : 00- & C particle lies at (0 ; : 100) . Find the electric &eld that these charges together produce at the origin. £ The electric &eld due to source charge q 1 : Consider the point charge q 1 = +6 : 00 & C = +6 : 00 ¡ 10 & 6 C : We have r 1 = length vector from (0 : 300 ; 0) to (0 ; 0) = (0 & : 300) e x + (0 & 0) e y = & : 300 e x m ; having magnitude r 1 = j& : 300 j = 0 : 300 m : Therefore, we obtain E 1 = k e q 1 r 1 r 3 1 = & 8 : 987551787 ¡ 10 9 ¡& +6 : 00 ¡ 10 & 6 ¡ ( & : 300 e x ) (0 : 300) 3 = & 5 : 99 e x ¡ 10 5 N= C as the electric &eld at (0 ; 0) due to the +6 : 00 & C point charge. £ The electric &eld due to source charge q 2 : Consider the point charge q 2 = & 3 : 00 & C = & 3 : 00 ¡ 10 & 6 C : We have r 2 = length vector from (0 ; : 100) to (0 ; 0) = (0 & 0) e x + (0 & : 100) e y = & : 100 e y m ; having magnitude r 2 = j& : 100 j = 0 : 100 m : Therefore, we obtain E 2 = k e q 2 r 2 r 3 2...
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This note was uploaded on 05/01/2011 for the course HPHY 253 taught by Professor Bak during the Spring '09 term at Morehouse.

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253-wrk-23 - HPHY 253 Workout 23 Electric Fields Dr. A. E....

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