# hw01-2.doc

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British Columbia Institute of Technology

Economics

1000

Nicole Neverson

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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