Physics Solution Manual for 1100 and 2101

Taking east as the positive x direction the electric

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Unformatted text preview: ation there is only one force, the electric force F, so it is the net force. According to Equation 18.2, the magnitude of the electric force is equal to the product of the magnitude of the charge and the magnitude of the electric field, or F = q0 E. Thus, the magnitude of the acceleration can be written as a= F q0 E = m m The magnitude of the acceleration of the proton is a= q0 E m (1.60 × 10−19 C ) (8.0 × 104 N /C ) = = 1.67 × 10 −27 kg 7.7 × 1012 m /s 2 The magnitude of the acceleration of the electron is a= q0 E (1.60 × 10−19 C) (8.0 × 104 N /C) = = 1.4 × 1016 m /s 2 m 9.11 × 10−31 kg ______________________________________________________________________________ Chapter 18 Problems 41. REASONING AND SOLUTION Figure 1 at the right shows the configuration given in text Figure 18.20a. The electric field at the center of the rectangle is the resultant of the electric fields at the center due to each of the four charges. As discussed in Conceptual Example 11, the magnitudes of the electric field at the center due to each of the four charges are equal. However, the fields produced by the charges in corners 1 and 3 are in opposite directions. Since they have the same magnitudes, they combine to give zero resultant. 977 -q +q 11 2 44 3 +q +q The fields produced by the charges in corners 2 and 4 point in Figure 1 the same direction (toward corner 2). Thus, EC = EC2 + EC4, where EC is the magnitude of the electric field at the center of the rectangle, and EC2 and EC4 are the magnitudes of the electric field at the center due to the charges in corners 2 and 4 respectively. Since both EC2 and EC4 have the same magnitude, we have EC = 2 EC2. The distance r, from any of the charges to the center of the rectangle, can be found using the Pythagorean theorem (see Figure 2 at the right): 11 d 5.00 cm d = (3.00 cm) +(5.00 cm) = 5.83 cm 2 2 2 d Therefore, r = = 2.92 cm = 2.92 × 10−2 m 2 θ 44 3 3.00 cm Figure 2 The electric field at the center has a magnitude of EC = 2 EC 2 = 2k q2 r2 = 2(8.99 × 109 N ⋅ m 2 /C 2 )(8.60 × 10−12 C) = 1.81 × 102 N/C −2 2 (2.92 × 10 m) Figure 3 at the right shows the configuration given in text Figure 18.20b. All four charges contribute a non-zero component to the electric field at the center of the rectangle. As discussed in Conceptual Example 11, the contribution from the charges in corners 2 and 4 point toward corner 2 and the contribution from the charges in corners 1 and 3 point toward corner 1. Notice also, the magnitudes of E24 and E13 are equal, and, from the first part of this problem, we know that E24 = E13 = 1.81 × 102 N/C -q -q 11 2 E 13 E 24 C 44 +q Figure 3 3 +q 978 ELECTRIC FORCES AND ELECTRIC FIELDS The electric field at the center of the rectangle is the vector sum of E24 and E13. The x components of E24 and E13 are equal in magnitude and opposite in direction; hence (E13)x – (E24)x = 0 Therefore, EC = ( E13 ) y + ( E24 ) y = 2( E13 ) y = 2( E13 ) sin θ From Figure 2, we have that sinθ = and 5.00 cm 5.00 cm = = 0.858 d 5.83 cm ( ) EC = 2 ( E13 ) sin θ = 2 1.81×102 N/C ( 0.858 ) = 3.11× 102 N/C ______________________________________________________________________________ 42. REASONING AND SOLUTION The magnitude of the force on q1 due to q2 is given by Coulomb's law: k q1 q2 F12 = (1) r122 The magnitude of the force on q1 due to the electric field of the capacitor is given by σ F1C = q1 EC = q1 ε 0 (2) Equating the right hand sides of Equations (1) and (2) above gives k q1 q2 r12 2 σ = q1 ε 0 Solving for r12 gives r12 = ε 0 k q2 σ [8.85 ×10−12 C2 /(N ⋅ m 2 )](8.99 ×109 N ⋅ m 2 /C 2 )(5.00 ×10− 6 C) = 5.53 ×10 –2 m −4 2 (1.30 ×10 C/m ) ______________________________________________________________________________ = Chapter 18 Problems 979 43. REASONING The magnitude E of the electric field is the magnitude F of the electric force exerted on a small test charge divided by the magnitude of the charge: E = F/ q . According to Newton’s second law, Equation 4.2, the net force acting on an object is equal to its mass m times its acceleration a. Since there is only one force acting on the object, it is the net force. Thus, the magnitude of the electric field can be written as E= F ma = q q The acceleration is related to the initial and final velocities, v0 and v, and the time t through v − v0 Equation 2.4, as a = . Substituting this expression for a into the one above for E gives t v − v0 m ma t = m ( v − v0 ) E= = q q qt SOLUTION The magnitude E of the electric field is E= m ( v − v0 ) qt (9.0 × 10−5 kg ) ( 2.0 × 103 m /s − 0 m /s ) = = ( 7.5 × 10−6 C ) ( 0.96 s ) 2.5 × 104 N /C ______________________________________________________________________________ 44. REASONING The external electric field E exerts a force FE = qE (Equation 18.2) on the sphere, where q = +6.6 µC is the net charge FE of the sphere. The external electric field E is directed upward, so the force FE it exerts on the positively charged sphere is...
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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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