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chapt2 - PHYS 133 WEEK 2 CHAPTER 26 26.4 Model The electric...

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PHYS 133, WEEK 2 CHAPTER 26 26.4. Model: The electric field is that of the two charges located on the y -axis. Visualize: Please refer to Figure Ex26.4. We denote the positive charge by q 1 and the negative charge by q 2 . The electric field r E 1 of the positive charge q 1 is directed away from q 1 , but the field r E 2 is toward the negative charge q 2 . We will add r E 1 and r E 2 vectorially to find the strength and the direction of the net electric field vector. Solve: The electric fields from q 1 and q 2 are r E 1 = 1 4 !" 0 q 1 r 1 2 , away from q 1 along + x axis # $% & ( = 9.0 ) 10 9 N m 2 / C 2 ( ) 1 ) 10 * 9 C ( ) 0.05 m ( ) 2 ˆ i = 3600 ˆ i N / C r E 2 = 1 4 0 q 2 r 2 2 , # below$ x -axis % & ( ) * From the geometry of the figure, tan ! = 10 cm 5 cm " = 63.43 ° ! r E 2 = 9.0 " 10 9 N m 2 / C 2 ( ) 1 " 10 # 9 C ( ) 0.10 m ( ) 2 + 0.05 m ( ) 2 # cos63.43 ° ˆ i # sin63.43 ° ˆ j ( ) = # 322 ˆ i + 644 ˆ j ( ) N / C ! r E net = r E 1 + r E 2 = 3278 ˆ i " 644 ˆ j ( ) N / C ! E net = 3278 N / C ( ) 2 + " 644 N / C ( ) 2 = 3340 N/C To find the angle this net vector makes with the horizontal, we calculate tan = E net ( ) y E net x ( ) = 644 N / C 3278 N / C " = 11.1 ° Thus, the strength of the net electric field at P is 3341 N/C and r E net makes an angle of 11.1 ° below 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 r F = q r E rod . From Example 26.3, the electric field a distance r from the center of a charged rod is r E rod = 1 4 0 Q r r 2 + L / 2 ( ) 2 ˆ i = 9.0 # 10 9 N m 2 / C 2 ( ) 50 # 10 $9 C ( ) ˆ i 0.04 m ( ) 0.04 m ( ) 2 + 0.05 m ( ) 5 = 1.757 # 10 5 ˆ i N / C Thus, the force is This preview has intentionally blurred sections. Sign up to view the full version. View Full Document r F = ! 5 " 10 ! 9 C ( ) 1.757 " 10 5 N / C ( ) ˆ i = ! 8.78 " 10 ! 4 ˆ i N More generally, r F = 8.78 ! 10 " 4 N, toward the rod ( ) . 26.14. Model: Each disk is a uniformly charged disk. When the disk is charged negatively, the on-axis electric field of the disk points toward the disk. The electric field points away from the disk for a positively charged disk. Visualize: Solve: (a) The surface charge density on the disk is ! = Q A = Q " R 2 = 50 # 10$ 9 C 0.05 m ( ) 2 = 6.366 # 10 $6 C / m 2 From Equation 26.22, the electric field of the left disk at z = 0.10 m is E 1 ( ) z = 2 0 1 # 1 1 + R 2 z 2$ % & & ( ) ) = # 6.366 * 10 # 6 C / m 2 2 8.85 * 10 # 12 C 2 / N m 2 ( ) 1 # 1 1 + 0.05 m 0.10 m ( ) 2 \$ % & & & ( ) ) ) = # 38,000 N / C In other words, r E 1 = 38,000 N / C, left ( ) . Similarly, the electric field of the right disk at z = 0.10 m (to its left) is
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chapt2 - PHYS 133 WEEK 2 CHAPTER 26 26.4 Model The electric...

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