problems with answers for pretest

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Problems with Answers for Pre Test Problem: 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. We find the magnitudes of the individual forces on the charge at the upper right corner: F 1 = F 2 = kQQ/L 2 = kQ 2 /L 2 = (9.0 10 9 N · m 2 /C 2 )(6.00 10 –3 C) 2 /(0.100 m) 2 = 3.24 10 7 N. F 3 = kQQ/(L2) 2 = kQ 2 /2L 2 = (9.0 10 9 N · m 2 /C 2 )(6.00 10 –3 C) 2 /2(0.100 m) 2 = 1.62 10 7 N. The directions of the forces are determined from the signs of the charges and are indicated on the diagram. For the forces on the upper-right charge, we see that the net force will be along the diagonal. For the net force, we have F = F 1 cos 45° + F 2 cos 45° + F 3 = 2(3.24 10 7 N) cos 45° + 1.62 10 7 N = 6.20 10 7 N along the diagonal, or away from the center of the square. From the symmetry, each of the other forces will have the same magnitude and a direction away from the center: The net force on each charge is 6.20 10 7 N away from the center of the square. Note that the sum for the three charges is zero. F 1 + F 2 F 3 Q x y L + + + L Q Q Q

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Problems With Answers

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Page 1: Problems With Answers for PreTest

Problems with Answers for Pre TestProblem: 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.

We find the magnitudes of the individual forces on the charge at the upper right corner:

F1

+

F2

F3

Q

x

y

L

++

+

L

QQ

Q

F1 = F2 = kQQ/L2 = kQ2/L2

= (9.0109 N · m2/C2)(6.0010–3 C)2/(0.100 m)2 = 3.24107 N.

F3 = kQQ/(L√2)2 = kQ2/2L2 = (9.0109 N · m2/C2)(6.0010–3 C)2/2(0.100 m)2 = 1.62107 N.

The directions of the forces are determined from the signs of the charges and are indicated on the diagram. For the forces on the upper-right charge, we see that the net force will be along the diagonal. For the net force, we have

F = F1 cos 45° + F2 cos 45° + F3 = 2(3.24107 N) cos 45° + 1.62107 N= 6.20107 N along the diagonal, or away from the center of the square.

From the symmetry, each of the other forces will have the same magnitude and a direction away from the center: The net force on each charge is 6.20107 N away from the center of the square.Note that the sum for the three charges is zero.