Chapter 26 Solutions

Chapter 26 Solutions - Chapter 26 Solutions(solutions...

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Chapter 26 Solutions (solutions marked with * are graded problems) 26.3. Model: The electric field is that of the two 1 nC charges located on the y -axis. Visualize: Please refer to Figure Ex26.3. We denote the top 1 nC charge by q 1 and the bottom 1 nC charge by q 2 . The electric fields ( 1 E r and 2 E r ) of both the positive charges are directed away from their respective charges. With vector addition, they yield the net electric field net E r at the point P indicated by the dot. Solve: The electric fields from q 1 and q 2 are 1 1 2 0 1 1 , along -axis 4 q E x r !" # $ = + % % ( r ( )( ) ( ) 9 2 2 9 2 9.0 10 N m /C 1 10 C ˆ ˆ 3600 N/C 0.05 m i i ! " " = = 2 2 2 0 2 1 , above -axis 4 q E x r ! "# $ % = + ( ) r Because ( ) 1 tan 10 cm 5 cm, tan 2 63.43 . " = = = ° So, ( )( ) ( ) ( ) ( ) ( ) 9 2 2 9 2 2 2 9.0 10 N m /C 1 10 C ˆ ˆ ˆ ˆ cos63.43 sin63.43 322 644 N/C 0.10 m 0.05 m E i j i j ! " " = ° + ° = + + r The net electric field is thus ( ) net at P 1 2 ˆ ˆ 3922 644 N/C E E E i j = + = + r r r To find the angle this net vector makes with the x axis, we calculate 644 N/C tan 3922 N/C = ! 9.3 = ° Thus, the strength of the electric field at P is ( ) ( ) 2 2 net 3922 N/C 644 N/C E = + = 3975 N/C and net E r makes an angle of 9.3 ° above the + x -axis. Assess: Because of the inverse square dependence on distance, 2 1 E E < . Additionally, because the point P has no special symmetry relative to the charges, we expected the net field to be at an angle relative to the x -axis. 26.9. Model: The rod is thin, so assume the charge lies along a line . Visualize: Solve: The force on charge q is rod F qE = r r . From Example 26.3, the electric field a distance r from the center of a charged rod is
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( ) ( )( ) ( ) ( ) ( ) 9 2 2 9 5 rod 2 2 5 2 0 ˆ 9.0 10 N m /C 50 10 C 1 ˆ ˆ 1.757 10 N/C 4 /2 0.04 m 0.04 m 0.05 m i Q E i i r r L !" # $ $ = = = $ + + r Thus, the force is ( )( ) 9 5 4 ˆ ˆ 5 10 C 1.757 10 N/C 8.78 10 N F i i ! ! = ! " " = ! " r More generally, ( ) 4 8.78 10 N, toward the rod F ! = " r . 26.13. Model: Assume that the rings are thin and that the charge lies along circle of radius R . Visualize: Solve: (a) Let the rings be centered on the z -axis. According to Example 26.5, the field of the left ring at z = 10 cm is ( ) ( ) ( ) ( ) ( ) ( ) ( ) 9 2 2 9 4 1 1 3/2 3/2 2 2 2 2 0 9.0 10 N m /C 0.10 m 20 10 C 1.28 10 N/C 4 0.10 m 0.05 m z zQ E z R # $ $ = = = $ % + + ( ) * That is, ( ) 4 1 1.288 10 N/C, right E = ! r
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This note was uploaded on 04/03/2008 for the course PHYS 100 taught by Professor All during the Spring '08 term at Cal Poly.

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Chapter 26 Solutions - Chapter 26 Solutions(solutions...

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