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E = − E3 + E4 = = ( − k q3
r32 + k q4
r42 )(
2
(5.0 × 10−2 m ) − 8.99 × 109 N ⋅ m 2 /C2 3.0 × 10−6 C ) + (8.99 × 109 N ⋅ m2 /C2 )(8.0 × 10−6 C)
2
( 7.0 × 10−2 m ) = +3.9 × 106 N/C
The plus sign indicates that the net electric field points along the +y direction .
______________________________________________________________________________
72. REASONING The unknown charges can
be determined using Coulomb’s law to
express the electrostatic force that each
unknown charge exerts on the 4.00 µC
charge. In applying this law, we will use
the fact that the net force points downward
in the drawing. This tells us that the
unknown charges are both negative and
have the same magnitude, as can be
understood with the help of the freebody
diagram for the 4.00 µC charge that is
shown at the right. The diagram shows +4.00 µC
30.0º F cos 30.0º F
F sin 30.0º
qA qB Chapter 18 Problems 1001 the attractive force F from each negative charge directed along the lines between the charges.
Only when each force has the same magnitude (which is the case when both unknown
charges have the same magnitude) will the resultant force point vertically downward. This
occurs because the horizontal components of the forces cancel, one pointing to the right and
the other to the left (see the diagram). The vertical components reinforce to give the
observed downward net force.
SOLUTION Since we know from the REASONING that the unknown charges have the
same magnitude, we can write Coulomb’s law as follows: ( 4.00 ×10−6 C ) qA
F =k
r2 ( 4.00 ×10−6 C ) qB
=k
r2 The magnitude of the net force acting on the 4.00 µC charge, then, is the sum of the
magnitudes of the two vertical components F cos 30.0º shown in the freebody diagram: ( 4.00 ×10−6 C) qA cos 30.0° + k ( 4.00 ×10−6 C ) qB cos 30.0°
ΣF = k
r2 r2 ( 4.00 ×10−6 C) qA cos 30.0°
= 2k
r2 Solving for the magnitude of the charge gives qA = = ( ( ΣF ) r 2 ) 2k 4.00 ×10−6 C cos 30.0° ( ( 405 N )( 0.0200 m )2 2 8.99 × 10 N ⋅ m / C
9 2 2 )( 4.00 ×10 −6 ) C cos 30.0° = 2.60 ×10−6 C Thus, we have qA = qB = −2.60 ×10−6 C . 73. REASONING The electric field is given by Equation 18.2 as the force F that acts on a test
charge q0, divided by q0. Although the force is not known, the acceleration and mass of the
charged object are given. Therefore, we can use Newton’s second law to determine the force
as the mass times the acceleration and then determine the magnitude of the field directly
from Equation 18.2. The force has the same direction as the acceleration. The direction of
the field, however, is in the direction opposite to that of the acceleration and force. This is 1002 ELECTRIC FORCES AND ELECTRIC FIELDS because the object carries a negative charge, while the field has the same direction as the
force acting on a positive test charge.
SOLUTION According to Equation 18.2, the magnitude of the electric field is E= F
q0 According to Newton’s second law, the net force acting on an object of mass m and
acceleration a is ΣF = ma. Here, the net force is the electrostatic force F, since that force
alone acts on the object. Thus, the magnitude of the electric field is ( )( ) 3.0 ×10−3 kg 2.5 × 103 m/s 2
F
ma
=
=
= 2.2 ×105 N/C
E=
−6
q0
q0
34 × 10 C
The direction of this field is opposite to the direction of the acceleration. Thus, the field
points along the −x axis . 74. REASONING The magnitude of the electric field between the plates of a parallel plate
capacitor is given by Equation 18.4 as E = σ
, where σ is the charge density for each plate
ε0 and ε0 is the permittivity of free space. It is the charge density that contains information
about the radii of the circular plates, for charge density is the charge per unit area. The area
of a circle is πr2. The second capacitor has a greater electric field, so its plates must have the
greater charge density. Since the charge on the plates is the same in each case, the plate area
and, hence, the plate radius, must be smaller for the second capacitor. As a result, we expect
that the ratio r2/r1 is less than one.
SOLUTION Using q to denote the magnitude of the charge on the capacitor plates and A = πr2 for the area of a circle, we can use Equation 18.4 to express the magnitude of the
field between the plates of a parallel plate capacitor as follows:
E= q
σ
=
ε 0 ε 0π r 2 Applying this result to each capacitor gives E1 = q ε π r2
14 0 3
241
First capacitor and E2 = q ε 0π r22
14243
Second capacitor Chapter 18 Problems 1003 Dividing the expression for E1 by the expression for E2 gives (
( )
) 2
E1 q / ε 0π r1
r22
=
=2
E2 q / ε π r 2
r1
02 Solving for the ratio r2/r1 gives r2
=
r1 E1
2.2 ×105 N/C
=
= 0.76
E2
3.8 × 105 N/C As expected, this ratio is less than one. 75. SSM REASONING AND SOLUTION Before the spheres have been charged, they exert
no forces on each other. After the spheres are charged, each sphere experiences a repulsive
force F due to the charge on the other sphere, according to Coulomb's law (Equation 18.1).
Therefore, since each sphere has the same charge, the magnitude...
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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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