Textbook Notes (363,260)
Economics (1)
1000 (1)
Chapter

# hw01-2.doc

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School
British Columbia Institute of Technology
Department
Economics
Course
1000
Professor
Nicole Neverson
Semester
Fall

Description
Homework 1 Ch16: P 12, 13, 15, 17, 19 12. (II) Particles of charge + 75 , + 48,and − 85 μ C are placed in a line (Fig. 16–49). The center one is 0.35 m from each of the others. Calculate the net force on each charge due to the other two. Solution Let the right be the positive direction on the line of charges. Use the fact that like charges repel and unlike charges attract to determine the direction of the forces. In the following expressions, k =8.988×10 N × m C 2. (75μC )(μC ) ( 75 μC)(5 μ C) F +75 −k 2 +k 2 14−7.2N ≈1.− 10×N (0.35m ) ( 0.70m) F = k (75μC )(μC ) +k( 48 μ)(85 μ C) 563.5N 5.6 10×N +48 (0.35m ) (0.35m )2 (85μC )(μC ) ( 85 μ)(48 μ C) 2 F −85 −k 2 −k 2 41−6.3N ≈4.− 10×N (0.70m ) ( 0.35m) 13. (II) Three positive particles of equal charge, + 11 .0 μ C, are located at the corners of an equilateral triangle of side 15.0 cm (Fig. 16–50). Calculate the magnitude and direction of the net force on each particle. Solution The forces on each charge lie along a line connecting the charges. Let the variable d represent the length of a side of the triangle, and let the variable Q represent the charge at each corner. Since the triangle is equilateral, each angle is 60 . 2 2 2 r r F = k Q → F =k Q cos60 , F =k Q sin60o F13 F12 12 2 12x 2 12y 2 d d d Q1 Q 2 Q 2 Q 2 d d F = k → F = −k cos60 , F k sin60o 13 d2 13x d2 13y d2 Q 2 Q3 Q 2 o Q2 F1x F 12x+F 13x=0 F1y =F12y +F13y =2 k 2 sin60 =3k 2 d d d −6 2 Q 2 (11.0×10 C ) F1= F1x F 1y3k = 3 8.(88×10 N×m C 2 2) = 83.7N d2 (0.150m )2 r The direction of F1 is in the y-direction . Also notice that it lies along the bisector of the opposite side of the triangle. Thus the force on the lower left charge is of magnitude.7N , and o will point30 below the − x axis . Finally, the force on the lower right charge is of magnitude 83.7N , and will poin30 below the + x axis . 14. (II) A charge of 6.00 mC is placed at each corner of a square 0.100 m on a side. Determine the magnitude and direction of the force on each charge. 15. (II) Repeat Problem 14 for the case when two of the positive charges, on opposite corners, are replaced by negative charges of the same magnitude (Fig. 16–51). Solution Determine the force on the upper right charge, and then the symmetry of the configuration says that the force on the lower left charge is the opposite of the force on the upper right charge. Likewise, determine the force on the lower right charge, and then the symmetry of the configuration says that the force on the upper left charge is the opposite of the force on the lower right charge. r The force at the upper right corner of the square is the vector sum of the r F F 42 forces due to the other three charges. Let the variablerepresent the 0.100 Q 41 Q m length of a side of the square, and let the variablerepresent the 6.00 1 r 4 F mC charge at each corner. 43 2 2 d F = k Q → F = −kQ , F 0 41 d 2 41x d 2 41y Q 2 Q 2 2 Q 2 2 Q 2 Q 2 Q3 F 42k → F 42x =k cos45o =k , 42y =k 2d 2 2d2 4d 2 4d 2 2 2 Q Q F 43k 2 → F 43x =0 , 43y = −k 2 d d Add the x and y components together to find the total force, noting tha4xF 4y. 2 2 2 Q 2Q Q  2  F 4xF 41x+ F42x+F 43x= −k 2 +k 2 0+ =k 2 1− + ÷ =F4y d 4d d  4  2 2 Q 2 2 Q 2 1 F 4 F4x F 4y = F4x 2 = k 2 −1 + 2 =k 2 2 − ÷ d 4 d  2 6.00×10 C3 2 = 8.988×10 N × m C 2 ( )  2 − 1  =2.96 1×0 N ( ) 2  2 ÷ (0.100m )   −1F4y o
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