Physics Solution Manual for 1100 and 2101

0 fbc cos 600 x b fba cos 600 fba chapter 18 problems

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Unformatted text preview: ombs (C). We will replace the prefix µ with 10−6 when calculating the distance r from Equation 18.1. SOLUTION Solving F = k r2 = k q1 q2 r2 q1 q2 F (Equation 18.1) for the distance r, we obtain or r= k q1 q2 F Chapter 18 Problems 949 Therefore, when the force magnitude F is 0.66 N, the distance between the charges must be r= k 9. q1 q2 F = (8.99 × 10 9 N ⋅ m /C 2 2 ) (8.4 × 10−6 C ) ( 5.6 ×10−6 C ) = 0.80 m 0.66 N SSM REASONING The number N of excess electrons on one of the objects is equal to the charge q on it divided by the charge of an electron (−e), or N = q/(−e). Since the charge on the object is negative, we can write q = − q , where q is the magnitude of the charge. The magnitude of the charge can be found from Coulomb’s law (Equation 18.1), which states that the magnitude F of the electrostatic force exerted on each object is given by F = k q q / r 2 , where r is the distance between them. SOLUTION The number N of excess electrons on one of the objects is N= q q −q = = −e −e e (1) To find the magnitude of the charge, we solve Coulomb’s law, F = k q q / r 2 , for q : q= F r2 k Substituting this result into Equation (1) gives ( 4.55 ×10 N )(1.80 ×10 m ) F r2 q k= 8.99 ×109 N ⋅ m 2 / C2 N= = =8 e e 1.60 × 10−19 C ______________________________________________________________________________ −21 −3 2 10. REASONING The gravitational force is an attractive force. To neutralize this force, the electrical force must be a repulsive force. Therefore, the charges must both be positive or both negative. Newton’s law of gravitation, Equation 4.3, states that the gravitational force depends inversely on the square of the distance between the earth and the moon. Coulomb’s law, Equation 18.1 states that the electrical force also depends inversely on the square of the distance. When these two forces are added together to give a zero net force, the distance can be algebraically eliminated. Thus, we do not need to know the distance between the two bodies. 950 ELECTRIC FORCES AND ELECTRIC FIELDS SOLUTION Since the repulsive electrical force neutralizes the attractive gravitational force, the magnitudes of the two forces are equal: kqq = 2 1 24 4r 3 Electrical force, Equation 18.1 GM e M m 2 14r 3 24 Gravitational force, Equation 4.3 Solving this equation for the magnitude q of the charge on either body, we find 2 −11 N ⋅ m 5.98 × 1024 kg 7.35 × 1022 kg 6.67 × 10 2 GM e M m kg q= = = 5.71 × 1013 C 2 k N⋅m 8.99 × 109 C2 ______________________________________________________________________________ ( 11. )( ) SSM WWW REASONING Initially, the two spheres are neutral. Since negative charge is removed from the sphere which loses electrons, it then carries a net positive charge. Furthermore, the neutral sphere to which the electrons are added is then negatively charged. Once the charge is transferred, there exists an electrostatic force on each of the two spheres, the magnitude of which is given by Coulomb's law (Equation 18.1), F = k q1 q2 / r 2 . SOLUTION a. Since each electron carries a charge of −1.60 ×10−19 C , the amount of negative charge removed from the first sphere is 1.60 ×10−19 C −6 3.0 ×1013 electrons 1 electron = 4.8 ×10 C ( ) Thus, the first sphere carries a charge +4.8 × 10–6 C, while the second sphere carries a charge −4.8 × 10–6 C. The magnitude of the electrostatic force that acts on each sphere is, therefore, F= k q1 q2 r2 (8.99 ×109 N ⋅ m2 /C2 ) ( 4.8 ×10−6 C) = ( 0.50 m )2 2 = 0.83 N b. Since the spheres carry charges of opposite sign, the force is attractive . ______________________________________________________________________________ Chapter 18 Problems 12. REASONING Let F2 and F1 represent the forces exerted on the charge q at the origin by the point charges q1 and q2, respectively. According to Equation 18.1, the magnitudes of these forces are given by F1 = k q1 q F2 = k and r12 951 y q2 q1 = −25 µC q2 q (1) r22 F1 where r1 is the distance between q1 and q, r2 is the distance q = +8.4 µC between q2 and q, and k = 8.99×109 N·m2/C2. The directions of the forces are determined by the signs of each charge-pair. The F2 sign of q1 is opposite that of q, so F1 is an attractive force, pointing in the positive y direction. The signs of q2 and q are both positive, so F2 is a repulsive force, pointing in the negative y direction (see the drawing). Because the net force F = F1 + F2 acting on q points in the positive y direction, the force F1 must have a greater magnitude than the force F2. Therefore, the magnitude F of the net electric force acting on q is equal to the magnitude of the attractive force F1 minus that of the repulsive force F2: F = F1 − F2 (2) SOLUTION Substituting Equations (1) into Equation (2) yields F = F1 − F2 = k q1 q r12 −k q2 q (3) r22 Solving Equation (3) for |q2|, we obtain k q2 q r22 =k q1 q r12 −F or q F q2 = r22 1 − r2 k q 1 Substituting the given values, we find that −25 × 10 −6 C 27 N = 1.8 × 10 −5 C q2 = ( 0.34...
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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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