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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.
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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.
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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 lefthand 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.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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