HW05solutions - 27.10. Model: The rod is thin, so assume...

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27.10. Model: The rod is thin, so assume the charge lies along a line . Visualize: Solve: The force on charge q is rod . Fq E = GG From Example 27.3, the electric field a distance r from the center of a charged rod is () ( )( ) 92 2 9 5 rod 22 5 2 0 ˆ 9 . 01 0 N m /C 4 0 C 1 ˆˆ 1.406 10 N/C 4 / 2 0.04 m 0.04 m 0.05 m i Q Ei i rr L πε ×× == = × ++ G Thus, the force is ( )( ) 95 4 6.0 10 C 1.406 10 N/C 8.4 10 N Fi i −− × = × G More generally, ( ) 4 8.4 10 N, away from the rod F G .
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27.12. 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 27.5, the field of the left ring at z = 10 cm is () ( ) ( ) ( ) 92 2 9 4 1 1 3/2 22 0 9.0 10 N m /C 0.10 m 20 10 C 1.29 10 N/C 4 0.10 m 0.050 m z zQ E zR πε ×× == = × ⎡⎤ + + ⎣⎦ That is, ( ) 4 1 1.29 10 N/C, right . E G Ring 2 has the same quantity of charge and is at the same distance, so it will produce a field of the same strength. Because Q 2 is positive, 2 E G will point to the left. The net field at the midpoint between the two rings is 12 0 EEE =+= GGG N/C. (b) The field of the left ring at z = 0 cm is ( ) 1 0 z E = N/C. The field of the right ring at z = 20 cm to its left is ( ) ( ) ( ) 2 9 3 2 32 0.20 m 4.1 10 N/C 0.20 m 0.050 z E × + E EE ⇒= + = 0 N/C + (4.1 × 10 3 N/C, left) So the electric field strength is 4.1 × 10 3 N/C.
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27.18. Model: The electric field in a region of space between two charged circular disks is uniform. Solve: The electric field strength inside the capacitor is 0 E QA ε = . Thus, the area is ( )( ) () ( ) 91 9 2 42 12 2 2 5 0 3.0 10 1.6 10 C 2.71 10 m 4 8.85 10 C /N m 2.0 10 N/C 4 1.86 cm QD A E A D π ×× == = ⇒= = Assess: As long as the spacing is much less than the plate dimensions, the electric field is independent of the spacing and depends only on the diameter of the plates.
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27.24. Model: The infinite negatively charged plane produces a uniform electric field that is directed toward the plane. Visualize: Solve: From the kinematic equation of motion 22 10 02 a n d , vv a x F q E m a == + Δ = = 00 qE v mv ax mx q E −− == Δ = Δ Furthermore, the electric field of a plane of charge with surface charge density η is 0 2. E ε = Thus, () ( ) ( ) ( ) 2 27 6 12 2 2 2 19 6 2 1.67 10 kg 2.0 10 m/s 8.85 10 C /N m 0.185 m 1.60 10 C 2.0 10 C/m mv x q −× × × Δ= = = ×− ×
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27.28. Model: The electric field is that of three point charges q 1 , q 2 , and q 3 . Visualize: Please refer to Figure P27.28. Assume the charges are in the x-y plane. The 5.0 nC charge is q 1 , the 10 nC charge is q 3 , and the 5.0 nC charge is q 2 . The net electric field at the dot is net 1 2 3 E EEE =++ GG G G . The procedure will be to find the magnitudes of the electric fields, to write them in component form, and to add the components.
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This note was uploaded on 11/02/2009 for the course MASTERING PHYS taught by Professor All during the Spring '09 term at Kettering.

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HW05solutions - 27.10. Model: The rod is thin, so assume...

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