This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: lim (kl9356) – Homework 2 – Weathers – (22202) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Four point charges are placed at the four cor ners of a square, where each side has a length a . The upper two charges have identical pos itive charge, and the two lower charges have charges of the same magnitude as the first two but opposite sign. That is, q 1 = q 2 = q and q 3 = q 4 = − q where q > 0. b b b b b q 4 q 2 q 3 q 1 P j i Determine the direction of the electric field at the center, point P. 1. − i 2. 3. j 4. 1 √ 2 ( i − j ) 5. 1 √ 2 ( i + j ) 6. − 1 √ 2 ( i + j ) 7. − j correct 8. − 1 √ 2 ( i − j ) 9. i Explanation: The direction is already clear: all the x components cancel, and the lower charges at tract and the top ones repel, so the answer is − j . 002 (part 1 of 2) 10.0 points Three positive charges are arranged as shown at the corners of a rectangle. The Coulomb constant is 8 . 98755 × 10 9 N m 2 / C 2 . + + + . 411 m 1 . 233 m 8 . 06 nC 4 . 03 nC 3 . 29 nC Find the magnitude of the electric field at the fourth corner of the rectangle. Correct answer: 229 . 147 N / C. Explanation: r 1 r 2 θ ϕ Q 2 Q 1 Q 3 E 1 E 2 E 3 Let : q 1 = 4 . 03 nC = 4 . 03 × 10 − 9 C , q 2 = 8 . 06 nC = 8 . 06 × 10 − 9 C , q 3 = 3 . 29 nC = 3 . 29 × 10 − 9 C , r 2 = 1 . 233 m , r 1 = 0 . 411 m , and k e = 8 . 98755 × 10 9 N m 2 / C 2 . The length of the diagonal is r 3 = radicalBig r 2 1 + r 2 2 = radicalBig (1 . 233 m) 2 + (0 . 411 m) 2 = 1 . 2997 m . The electric field produced by the charge q 1 is along the positive yaxis, so E 1 ,x = 0 E 1 ,y = E 1 = k c q 1 r 2 1 . The electric field produced by the charge q 2 is along the negative xaxis, so E 2 ,x = − E 2 = − k c q 2 r 2 2 E 2 ,y = 0 . lim (kl9356) – Homework 2 – Weathers – (22202) 2 The electric field produced by the charge q 3 is upward and to the left along the diagonal of the rectangle, and is directed away from the charge since q 3 is positive, so E 3 = k e q 3 r 2 3 . Consider the direction of E 3 cos ϕ = r 2 r 3 and sin ϕ = r 1 r 3 Thus E 3 ,x = − k e q 3 r 2 3 sin ϕ = − k e q 3 r 2 r 3 3 E 3 ,y = k C q 3 r 2 3 cos ϕ = k e q 3 r 1 r 3 3 , so that E net,x = E 2 ,x + E 3 ,x = − k e q 2 r 2 2 − k e q 3 r 2 r 3 3 = − ( 8 . 98755 × 10 9 N m 2 / C 2 ) × 8 . 06 × 10 − 9 C (1 . 233 m) 2 − ( 8 . 98755 × 10 9 N m 2 / C 2 ) × (3 . 29 × 10 − 9 C) (1 . 233 m) (1 . 2997 m) 3 = − 64 . 255 N / C , and E net,y = E 1 ,y + E 3 ,y = k e q 1 r 2 1 + k e q 3 r 1 r 3 3 = ( 8 . 98755 × 10 9 N m 2 / C 2 ) × 4 . 03 × 10 − 9 C (0 . 411 m) 2 + ( 8 . 98755 × 10 9 N m 2 / C 2 ) × (3 . 29 × 10 − 9 C) (0 . 411 m) (1 . 2997 m) 3 = 219 . 954 N / C , so that E net = radicalBig E net,x 2 + E net, 4 2 E net = bracketleftbig ( − 64 . 255 N / C) 2 +(219 . 954 N / C) 2 bracketrightbig 1 2 = 229 . 147 N / C ....
View
Full
Document
This note was uploaded on 09/07/2010 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.
 Spring '00
 Littler
 Charge, Magnetism, Work

Click to edit the document details