Physics Solution Manual for 1100 and 2101

Since both ec2 and ec4 have the same magnitude we

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Unformatted text preview: _____________________________________________________________________ 30. REASONING a. The magnitude of the electric field is obtained by dividing the magnitude of the force (obtained from the meter) by the magnitude of the charge. Since the charge is positive, the direction of the electric field is the same as the direction of the force. b. As in part (a), the magnitude of the electric field is obtained by dividing the magnitude of the force by the magnitude of the charge. Since the charge is negative, however, the direction of the force (as indicated by the meter) is opposite to the direction of the electric field. Thus, the direction of the electric field is opposite to that of the force. SOLUTION a. According to Equation 18.2, the magnitude of the electric field is E= F 40.0 µ N = = 2.0 N /C q 20.0 µ C As mentioned in the REASONING, the direction of the electric field is the same as the direction of the force, or due east . 968 ELECTRIC FORCES AND ELECTRIC FIELDS b. The magnitude of the electric field is E= F 20.0 µ N = = 2.0 N /C q 10.0 µ C Since the charge is negative, the direction of the electric field is opposite to the direction of the force, or due east . Thus, the electric fields in parts (a) and (b) are the same. ______________________________________________________________________________ 31. SSM WWW REASONING The electric field created by a point charge is inversely proportional to the square of the distance from the charge, according to Equation 18.3. Therefore, we expect the distance r2 to be greater than the distance r1, since the field is smaller at r2 than at r1. The ratio r2/r1, then, should be greater than one. SOLUTION Applying Equation 18.3 to each position relative to the charge, we have E1 = kq and r12 E2 = kq 2 r2 Dividing the expression for E1 by the expression for E2 gives E1 E2 = k q / r12 k q / r22 = r22 r12 Solving for the ratio r2/r1 gives r2 = r1 E1 248 N/C = = 1.37 E2 132 N/C As expected, this ratio is greater than one. Chapter 18 Problems 969 32. REASONING AND SOLUTION The electric field lines must originate on the - 4 q +5 q positive charges and terminate on the negative charges. They cannot cross one another. Furthermore, the number of field lines beginning or ending on any charge must be proportional to the magnitude of the charge. If 10 electric field lines leave the +5q charge, then six lines must originate from the +3q charge, and eight lines must end on each –4q charge. The drawing shows -4 q the electric field lines that meet these +3 q criteria. ______________________________________________________________________________ 33. REASONING Each charge creates an electric field at the center of the square, and the four fields must be added as vectors to obtain the net field. Since the charges all have the same magnitude and since each corner is equidistant from the center of the square, the magnitudes kq of the four individual fields are identical. Each is given by Equation 18.3 as E = 2 . The r directions of the various contributions are not the same, however. The field created by a positive charge points away from the charge, while the field created by a negative charge points toward the charge. SOLUTION The drawing at the right shows each of the field contributions at the center of the square (see black dot). Each is directed along a diagonal of the square. Note that ED and EB point in opposite directions and, therefore, cancel, since they have the same magnitude. In contrast EA and EC point in the same direction toward corner A and, therefore, combine to give a net field that is twice the magnitude of EA or EC. In other words, the net field at the center of the square is given by the following vector equation: +C B+ ED EA EC EB A− Σ E = E A + E B + EC + E D = E A + E B + EC − E B = E A + EC = 2 E A Using Equation 18.3, we find that the magnitude of the net field is Σ E = 2 EA = 2 kq r2 +D 970 ELECTRIC FORCES AND ELECTRIC FIELDS In this result r is the distance from a corner to the center of the square, which is one half of the diagonal distance d. Using L for the length of a side of the square and taking advantage of the Pythagorean theorem, we have r = 1 d = 1 L2 + L2 . With this substitution for r, the 2 2 magnitude of the net field becomes ΣE = 2 ( kq 1 2 L2 + L2 ) 2 = 4k q L2 = ( )( 4 8.99 ×109 N ⋅ m 2 / C2 2.4 ×10−12 C ( 0.040 m )2 ) = 54 N/C 34. REASONING Part (a) of the drawing given in the text. The electric field produced by a charge points away from a positive charge and toward a negative charge. Therefore, the electric field E+2 produced by the +2.0 µC charge points away from it, and the electric fields E−3 and E−5 produced by the −3.0 µC and −5.0 µC charges point toward them (see the left-hand side of the following drawing). The magnitude of the electric field produced by a point charge is given by Equation 18.3 as E = k q /r2. Since the distance from each charge to the origin is the same, the magnitude of the electric field is proportional only to th...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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