MathematicalMethodsSolutionsManual

MathematicalMethodsSolutionsManual - James B. Seaborn...

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James B. Seaborn University of Richmond , Springer SOLUTIONS MANUAL for Mathematical Methods for the Physical Sciences
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SOLUTIONS MANUAL for Mathematical Methods for the Physical Sciences
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Springer New York Berlin HeideLberg BarceLona Hong Kong London Milan Paris Singapore Tokyo
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James B. Seaborn University of Richmond , Springer SOLUTIONS MANUAL for Mathematical Methods for the Physical Sciences r
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Copyright © 2002 by Springer-Verlag New York. Inc. This material may be reproduced for testing or instructional purposes by people using the text. Printed in the United States of America.
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1 1 A Review 1. Let a = 0.075 m, b = 0.183 m, ql = 7.37 X 10- 9 C, and q2 = -3.58 X 10- 9 C. Ex = -kq1b 3 = -1,569 N/C (a 2 + b 2 )2 E y = -kqla _ kq2 = -5 085 N/C (a2 + b2)~ a 2 ' E = JE'; + E~ = 5,322 N/C ¢ = tan- 1 ~: = 252.8° The resultant electric field E is 5,322 N/C down to the left at 73° below the horizontal. 2. First, calculate the horizontal and vertical components of E (9 X 10 9 N . m 2 /C 2 )(3 x 10- 9 C) Ex = - (0.35 m)2 _ (9 x 10 9 N . m 2 /C 2 )(5 x 10- 9 C) (0.35 m) = -799.5 N/C ((0.35 m)2 + (0.15 m)2) 3/2 (9 X 10 9 N . m 2 /C 2 )(7 x 10- 9 C) (0.15 m)2 + (9 x 10 9 N . m 2 /C2)(5 x 10- 9 C) (0.15 m) = 1,322.2 N/C ((0.35 m)2 + (0.15 m)2) 3/2 E = J E'; + E~ = 1,545 N/C and The electric field is 1,545 N/C up to the left at 58.8° above the horizontal. 3. The point where E = 0 must lie on the line passing through the two charges and at a distance d to the left of the 9-f-LC charge. E = k(16 f-LC) _ k(9 f-LC) = 0 (d + 13 mm)2 tf2 . On rearranging, we get the quadratic equation for d, 7d 2 - 234d - 1,521 = (7d + 39)(d - 39) = O. Thus, the electric field is zero at the point 39 mm to the left of the 9-f-LC charge.
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1 A REVIEW 2 4. Let q1 = 8/-LC and q1 = -5/-LC. A line from the field point P to q1 makes an angle () with the horizontal line connecting the two charges. A line from P to q2 also makes an angle () with the line connecting the charges. Thus, Ex = E 1 cos () + E 2 cos () = (E 1 + E 2 ) cos () (9 X 10 9 N . m 2 /C 2 )(0.055 m) (8 0-6 C -6 C) = ----------- x 1 + 5 x 10 ((0.05 m)2 + (0.055 m)2) 3/2 = 1.567 X 10 7 N/C E y = E 1 sin() - E 2 sinB = (E 1 - E 2 )sin() (9 x 10 9 N . m 2 /C2)(0.05 m) (8 0-6 C -6 C) ---------- x 1 - 5 x 10 ((0.05 m)2 + (0.055 m)2) 3/2 = 0.329 X 10 7 N/C From these components, we find that the resultant electric field is 1.601 x 10 7 N/C up to the right at 11.8 0 above the horizontal. 5. Use Eq. (1.5) and solve for the ion density n, I 2.2 x 10- 6 Cis n= 2 = 2 (3e)7r(~) v (4.8 X 1O-19C)7rcoo~lm) (6.48 x 10 6 m/s) = 2.04 x 10 12 (ions/m 3 ) (.01 m/cm)3 = 2.04 x 10 6 ions/cm 3 . 6. The current I is 1= b.Q = [2(1.6 x 1O- 19 C)/ion](2.6 x 10 19 ions) = 0 39C/ b.t 60s .1 s.
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This note was uploaded on 05/24/2010 for the course PHYS 3113 taught by Professor Staff during the Spring '03 term at University of Central Florida.

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MathematicalMethodsSolutionsManual - James B. Seaborn...

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