Physics Solution Manual for 1100 and 2101

10 c 0 a d applies as well solution the capacitance

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Unformatted text preview: = − (Vpositive − Vnegative ) (Equation 19.7a). Note that ∆s and ∆s (Vpositive − Vnegative) are positive numbers, so the electric field is a negative number, denoting that it points to the left in the drawing: Chapter 19 Problems − − − Electric field 1035 + + Electron − F − −x + +x + Capacitor plates SOLUTION The total energy of the electron is conserved, so its total energy at the positive plate is equal to its total energy at the negative plate: KE positive + EPE positive = KE negative + EPE negative 1444 24444 4 3 1444 24444 4 3 Total energy at positive plate Total energy at negative plate Since the electron starts from rest at the negative plate, KEnegative = 0 J. Thus, the kinetic energy of the electron at the positive plate is KEpositive = −(EPEpositive − EPEnegative). We know from Equation 19.4 in the REASONING section that EPEpositive − EPEnegative = (−e)(Vpositive − Vnegative), so the kinetic energy can be written as KEpositive = −(EPEpositive − EPEnegative) = e (Vpositive − Vnegative) Since the potential difference is related to the electric field E and the displacement ∆s by Vpositive − Vnegative = − E ∆s (Equation 19.7a), we have that ( ) KE positive = e Vpositive − Vnegative = e( − E ∆s) ( )( ) = 1.60 × 10−19 C − −2.1×106 V/m ( +0.012 m ) = 4.0 × 10−15 J In arriving at this result, we have used the fact that the electric field is negative, since it points to the left in the drawing. 1036 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL 39. SSM REASONING AND SOLUTION Let point A be on the x-axis where the potential is 515 V. Let point B be on the x-axis where the potential is 495 V. From Equation 19.7a, the electric field is E=− V − VA ∆V 495 V − 515 V =− B =− = − 1.7 × 103 V/m −3 ∆s ∆s −2 6.0 × 10 m ( ) The magnitude of the electric field is 1.7 × 103 V/m . Since the electric field is negative, it points to the left , from the high toward the low potential. 40. REASONING The outside force is the only nonconservative force acting on the particle. Therefore, the work Wnc done by the outside force changes the total mechanical energy E = KE + EPE of the particle according to Wnc = ∆KE + ∆EPE (Equation 6.7b), where ∆KE = KE B − KE A is the difference between the kinetic energy of the particle at equipotential surface B and that at equipotential surface A, and ∆EPE = EPE B − EPE A is the corresponding difference between the electric potential energies of the particle. The kinetic energy of the particle is found from its mass m and speed v via KE = 1 mv 2 2 (Equation 6.2): 2 2 (1) KE A = 1 mvA and KE B = 1 mvB 2 2 The electric potential energy of the particle is given by EPE = qV (Equation 19.3), where q is the charge of the particle, and V is the potential at an equipotential surface. Therefore, we have that EPE A = qVA and EPE B = qVB (2) SOLUTION Substituting ∆KE = KE B − KE A and ∆EPE = EPE B − EPE A into the relation Wnc = ∆KE + ∆EPE (Equation 6.7b), we obtain Wnc = ∆KE + ∆EPE = ( KEB − KEA ) + ( EPEB − EPEA ) Substituting Equations (1) and (2) into Equation (3) yields Wnc = 2 2 2 2 ( 1 mvB − 1 mvA ) + ( qVB − qVA ) = 1 m ( vB − vA ) + q (VB − VA ) 2 2 2 Therefore, the work done by the outside force in moving the particle from A to B is (3) Chapter 19 Problems 1037 2 2 Wnc = 1 ( 5.00 ×10−2 kg ) ( 3.00 m/s ) − ( 2.00 m/s ) + ( 4.00 ×10−5 C ) ( 7850 V − 5650 V ) 2 = 0.213 J 41. REASONING a. The potential difference VB − VA between points A and B is related to the work WAB done by the electric force when a charge q0 moves from A to B; VB − VA = −WAB /q0 (Equation 19.4). Since the displacement from A to B is perpendicular to the electric force, no work is done, so WAB = 0 J (see Section 6.1). b. The potential difference ∆V = VC − VB between B and C is related to the electric field E and the displacement ∆s by ∆V = −E ∆s (Equation 19.7a). Since E and ∆s are known, we can use this expression to find ∆V. c. Since the electric force is a conservative force, it does not matter which path we take between points A and C to find the potential difference VA − VC. Therefore, we can retrace our steps in parts a and b to find this potential difference. C +y 10.0 cm 8.0 cm +x 6.0 cm B A E E SOLUTION a. As discussed in the Reasoning section, the displacement from A to B is perpendicular to the electric force, so no work is done. Thus the potential difference is VB − VA = −WAB 0 J = = 0V q0 q0 b. We have that VC − VB = −E ∆s (Equation 19.7a). Since the electric field points in the negative y direction, E = −3600 N/C. The displacement from B to C is positive, so ∆s = + 0.080 m. Thus, the potential difference is 1038 ELECTRIC POTENTIAL ENERGY AND THE ELECTRIC POTENTIAL VC − VB = − E ∆s = − ( −3600 N/C ) ( +0.080 m ) = +290 V c. To determine the potential difference VA − VC , we may take any path from C to A that is convenient. We wil...
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