Chapter 27

# Chapter 27 - 27.1. Model: The electric field is that of the...

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27.1. Model: The electric field is that of the two charges placed on the y -axis. Visualize: Please refer to Figure EX27.1. We denote the upper charge by q 1 and the lower charge by q 2 . Because both the charges are positive, their electric fields at P are directed away from the charges. Solve: The electric field from q 1 is 1 1 2 01 1 , below -axis 4 q Ex r θ πε ⎛⎞ =+ ⎜⎟ ⎝⎠ G ( )( ) () 92 2 9 22 9.0 10 N m /C 3.0 10 C ˆˆ cos sin 0.050 m 0.050 m ij ×× =− + Because tan 5 cm 5 cm 1, == the angle 45 . = ° Hence, 1 11 5400 N/C E G Similarly, the electric field from q 2 is 2 2 2 02 1 , above -axis 5400 N/C 4 q E xi j r = + G net at P 1 2 E EE ⇒= + GG G 3 1 2 5400 N/C 7.6 10 N/C 2 ii × Thus, the strength of the electric field is 7.6 × 10 3 N/C and its direction is horizontal. Assess: Because the charges are located symmetrically on either side of the x -axis and are of equal value, the y - components of their fields will cancel when added.

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27.2. Model: The electric field at the point is found by superposition of the fields due to the two charges located on the y -axis. Visualize: Please refer to Figure EX27.2. The electric field due to the positive charge q 1 at the point is away from q 1 . On the other hand, the electric field due to the negative charge q 2 at the point is toward q 2 . These two electric fields are then added vertically to obtain the net electric field at the point. Solve: The electric field from q 1 is ( )( ) () 92 2 9 1 1 22 2 01 9.0 10 N m /C 3.0 10 C 1 ˆˆ , below + -axis cos sin 4 0.050 m 0.050 m q E xi j r θ θθ πε ⎛⎞ ×× ⎜⎟ == + ⎝⎠ G Because tan 5 cm 5 cm, 45 . ° So, 1 11 5400 N/C E ij =+ G Similarly, the electric field from q 2 is 2 2 2 02 1 , below -axis 4 q Ex r =− G 5400 N/C 3 net 1 2 1 2 5400 N/C 7637 N/C 7.6 10 N/C 2 EE E j j ⇒= + = = = × GG G Thus, the strength of the electric field is 7.6 × 10 3 N/C and its direction is vertically downward. Assess: A quick visualization of the components of the two electric fields shows that the horizontal components cancel.
27.3. Model: The electric field is that due to superposition of the fields of the two 3.0 nC charges located on the y -axis. Visualize: Please refer to Figure EX27.3. We denote the top 3.0 nC charge by q 1 and the bottom 3.0 nC charge by q 2 . The electric fields 12 ( and ) E E GG of both the positive charges are directed away from their respective charges. With vector addition, they yield the net electric field net E G at the point P indicated by the dot. Solve: The electric fields from q 1 and q 2 are 1 1 2 01 1 , along -axis 4 q Ex r πε ⎛⎞ =+ ⎜⎟ ⎝⎠ G ( )( ) () 92 2 9 2 9.0 10 N m /C 3.0 10 C ˆˆ 10,800 N/C 0.05 m ii ×× == 2 2 2 02 1 , above -axis 4 q r θ G Because 1 tan 10 cm 5 cm, tan 2 63.43 . θθ = ° So, ( )( ) ( ) ( ) 2 9 2 22 ˆ ˆ cos63.43 sin63.43 966 1127 N/C 0.10 m 0.050 m Ei j i j + ° = + + G The net electric field is thus ( ) net at P 1 2 11,766 1127 N/C EE E i j =+= + G To find the angle this net vector makes with the x -axis, we calculate 1127 N/C tan 11,766 N/C φ = 5.5 = ° Thus, the strength of the electric field at P is ( ) net 11,766 N/C 1127 N/C E = 11,820 N/C = 1.18 × 10 3 N/C and net E G makes an angle of 5.5 ° above the + x -axis.

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## This note was uploaded on 01/30/2011 for the course PHYS 131 taught by Professor E.salik during the Winter '10 term at Cal Poly Pomona.

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Chapter 27 - 27.1. Model: The electric field is that of the...

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