Physics Solution Manual for 1100 and 2101

2e vb va m 2 16 10 19 c 25 000 v 94

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Unformatted text preview: F of this force is F= k q1 q2 r2 = (8.99 × 109 N ⋅ m 2 /C2 )(1.60 × 10−6 C) 2 = 2.30 N (0.100 m) 2 The repulsive force on each sphere compresses the spring to which it is attached. The magnitude of this repulsive force is related to the amount of compression by Equation 10.1: FxApplied = kx . Setting FxAppled = F and solving for k, we find that F 2.30 N = = 92.0 N/m x 0.0250 m ______________________________________________________________________________ k= 76. REASONING AND SOLUTION In order for the net force on any charge to be directed inward toward the center of the square, the charges must be placed with alternate + and – signs on each successive corner. The magnitude of the force on any charge due to an adjacent charge located at a distance r is F= kq r2 2 (8.99 × 109 N ⋅ m2 / C2 )( 2.0 × 10−6 C )2 = 0.40 N = ( 0.30 m )2 The forces due to two adjacent charges are perpendicular to one another and produce a resultant force that has a magnitude of Fadjacent = 2 F 2 = 2 ( 0.40 N ) = 0.57 N 2 1004 ELECTRIC FORCES AND ELECTRIC FIELDS The magnitude of the force due to the diagonal charge that is located at a distance of r 2 is Fdiagonal = kq 2 (r 2 ) 2 = kq 2r 2 2 = 0.40 N = 0.20 N 2 since the diagonal distance is r 2 . The force Fdiagonal is directed opposite to Fadjacent (since the diagonal charges are of the same sign). Therefore, the net force acting on any of the charges is directed inward and has a magnitude Fnet = Fadjacent – Fdiagonal = 0.57 N – 0.20 N = 0.37 N ______________________________________________________________________________ CHAPTER 19 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL ANSWERS TO FOCUS ON CONCEPTS QUESTIONS ____________________________________________________________________________________________ 1. (d) The force on the positive charge is in the same direction as the electric field, and the displacement of the charge is opposite to the direction of the force. Therefore, the work is negative (see Section 6.1). The electric potential energy at point B differs from that at A by EPEB − EPEA = −WAB (Equation 19.1). Since WAB is negative, EPEB is greater than EPEA. 2. (d) The electric potential energy EPE is related to the electric potential V by EPE = q0V (Equation 19.3). So, even though the electric potentials at two locations are the same, the electric potential energies are different since the charges placed at these locations are different. 3. VB − VA = −5.0 × 102 V 4. (a) The change in the proton’s electric potential energy (EPEB − EPEA) in going from A to B is related to the change in the potential (VB − VA) by Equation 19.4 as EPEB − EPEA = (+e)( VB − VA), where +e is the charge on the proton. On the other hand, the change in the electron’s electric potential energy (EPEA − EPEB) in going from B to A is related to the change in the potential (VA − VB) by EPEA − EPEB = (−e)( VA − VB), where −e is the charge on the electron. Comparing the right-hand sides of these two equations shows that the change in the proton’s electric potential energy is the same as the change in the electron’s electric potential energy. 5. (c) According to Equation 19.6, the potential produced by the charge q is V = kq/r. The smaller the value of r, the greater is the potential. The potential, however, does not depend on the charge (q0 or 2q0) placed at points P1 or P2. See Section 19.3. 6. (e) The total electric potential at the origin is the algebraic sum of the potentials due to all the charges. Since each potential is of the form V = kq/r (Equation 19.6) and r is same for each charge, the total electric potential is proportional to the sum of the charges. The sum of the charges in A (+4q) equals the sum in C (+4q), which is greater than the sum in B (+2q). 7. (b) The electric potential energy of the two charges in the top drawing is EPE = q0V, where V = kq/r is the electric potential produced by the charge at the origin (see Equation 19.6). Thus, EPE = kqq0/r. In a similar fashion, the electric potential energy for the charges in the bottom drawing is kq(2q0)/(2r) = kqq0/r, which is the same as that in the top drawing. 8. VB − VA = 6.7 × 103 V 1006 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL 9. (a) The electric potential energy EPE of two charges q1 and q2 is EPE = kq1q2/r, where r is the distance between them (see Section 19.3). Since the distance r is the same for all four pairs of charges, the electric potential energy is proportional to the product q1q2 of the 2 charges. The products of the charges in A and C are the same (+12q ), and the products of the charges in B and D are the same (−12q2). 10. (c) The electric field E is related to the electric potential difference ∆V by E = −∆V/∆s (Equation 19.7), where ∆s is the displacement of one point in the region relative to another point. If the potential is the same everywhere, then ∆V = 0 V, so E is zero everywhere. ∆V (Equation 19.7). ∆s Since ∆s is the same for all three capacitors, E is proportional to the potential difference ∆V = Vright − Vleft between the right and left plates. ∆V for capacitor C (300 V) is greater than that for capacitor B (250 V), which is greater than that for capacitor...
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