Physics Solution Manual for 1100 and 2101

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Unformatted text preview: sides of this result gives 1 4 = 2 2 2 L L + ( 2.00 m ) L2 + ( 2.00 m ) = 4 L2 2 or Solving for L, we obtain 3L2 = ( 2.00 m ) 2 or L= 2.00 m = 1.15 m 3 2 1022 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL 21. REASONING The potential V created by a point charge q at a spot that is located at a kq distance r is given by Equation 19.6 as V = , where q can be either a positive or negative r quantity, depending on the nature of the charge. We will apply this expression to obtain the potential created at the empty corner by each of the three charges fixed to the square. The total potential at the empty corner is the sum of these three contributions. Setting this sum equal to zero will allow us to obtain the unknown charge. SOLUTION The drawing at the right shows the three charges fixed to the corners of the square. The length of each side of the square is denoted by L. Note that the distance between the unknown charge q and the empty corner is L. Note also that the distance between one of the 1.8-µC charges and the empty corner is r = L , but that the distance between the other 1.8-µC charge and the empty corner is +1.8 µC L L L +1.8 µC q L r = L + L = 2 L , according to the Pythagorean theorem. Using Equation 19.6 to express the potential created by the unknown charge q and by each of the 1.8-µC charges, we find that the total potential at the empty corner is 2 2 Vtotal ( )( ) −6 −6 kq k +1.8 ×10 C k +1.8 ×10 C = + + =0 L L 2L In this result the constant k and the length L can be eliminated algebraically, leading to the following result for q: q + 1.8 ×10−6 C + 1.8 ×10−6 C =0 2 or ( ) 1 −6 q = −1.8 ×10−6 C 1 + = −3.1×10 C 2 22. REASONING The radius R of the circle is the distance between each charge and the center of the circle, where we know the total electric potential V. The total electric potential is the kq sum of the contributions V1, V2, and V3 made by the three charges. We will use V = r 9 22 (Equation 19.6), where k = 8.99×10 N·m /C and r = R to determine each of the three contributions to the total electric potential. Summing the contributions and equating the result to V = −2100 V, we will obtain the radius R of the circle. kq (Equation 19.6), the total potential V at the center of the circle is r k ( q1 + q2 + q3 ) kq kq kq (1) V = V1 + V2 + V3 = 1 + 2 + 3 = R R R R SOLUTION From V = Solving Equation (1) for R, we find that Chapter 19 Problems R= 1023 k ( q1 + q2 + q3 ) V (8.99 ×109 N ⋅ m 2 /C2 )( −5.8 ×10−9 C − 9.0 ×10−9 C+7.3 ×10−9 C ) = 0.032 m = −2100 V 23. SSM REASONING The only force acting on the moving charge is the conservative electric force. Therefore, the sum of the kinetic energy KE and the electric potential energy EPE is the same at points A and B: 1 2 2 2 mvA + EPE A = 1 mvB + EPE B 2 Since the particle comes to rest at B, v B = 0 . Combining Equations 19.3 and 19.6, we have kq EPE A = qVA = q 1 d and kq EPE B = qVB = q 1 r where d is the initial distance between the fixed charge and the moving charged particle, and r is the distance between the charged particles after the moving charge has stopped. Therefore, the expression for the conservation of energy becomes 1 2 2 mvA + kqq1 d = kqq1 r This expression can be solved for r. Once r is known, the distance that the charged particle moves can be determined. SOLUTION Solving the expression above for r gives 1024 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL r= = kqq1 1 mv 2 + kqq1 A 2 d (8.99 ×109 N ⋅ m 2 / C2 )(–8.00 ×10− 6 C)(−3.00 ×10− 6 C) 9 2 2 −6 −6 1 (7.20 × 10 –3 kg)(65.0 m/s) 2 + (8.99 × 10 N ⋅ m / C )( – 8.00 × 10 C)( −3.00 × 10 C) 2 0.0450 m = 0.0108 m Therefore, the charge moves a distance of 0 .0450 m – 0.0108 m = 0.0342 m . 24. REASONING The speed vB of the test charge at position B, the center of the square, is related q 2 to its kinetic energy by KE B = 1 mvB 2 (Equation 6.2). No forces other than the conservative electric force act on the test charge, so the total mechanical energy of the test charge is conserved as it accelerates from its initial position A at one corner of the square to its final position B. There is no rotational motion and no elastic potential energy, and the test charge is at rest at position A, so the conservation of energy takes the following form: 1 mv 2 + EPE B B 2 {{ Electric Kinetic potential energy energy at B at B = 1 m ( 0 m/s ) + EPE A 24 14 244 3{ 2 Kinetic energy at A q0 rA A rB q0 vB B rA = 0.480 m rB q= +7.20 µC or 1 2 2 mvB + EPE B = EPE A (1) Electric potential energy at A In Equation (1), m is the mass of the test charge. At both position A and position B, the electric potential energy EPE of the test charge is determined by EPE = q0V (Equation 19.3), where V is the total electric potential due to both of the charges q. We will kq use V = (Equation 19.6), where k = 8.99×109 N·m2/C2 to obtain the total electric r potential at positions A and B. SOLUTION At positions A and B, the total electric potential is the sum of the electric potentials due to the charges q. Since the charges are identical, and both are equidistant from positions A and B, the total potentials VA and VB are twice the potentials due to either charge kq alone. Therefore, from V = (Equation 19.6), we h...
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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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