Physics Solution Manual for 1100 and 2101

According to equation 193 the electric potential

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Unformatted text preview: e origin is 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 free-body 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 free-body 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.

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