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25 Chapter Solutions (* indicates a graded problem) 25.1. Model: Use the charge model. Solve: (a) In the process of charging by rubbing, electrons are removed from one material and transferred to the other because they are relatively free to move. Protons, on the other hand, are tightly bound in nuclei. So, electrons have been removed from the glass rod to make it positively charged. (b) Because each electron has a charge of 1.60 " 10 !19 C , the number of electrons removed is 5 " 10 !9 C = 3.13 " 1010 1.60 " 10 !19 C 25.7. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve: The first step shows two neutral metal spheres touching each other. In the second step, the negative rod repels the negative charges which will retreat as far as possible from the top of the left sphere. Note that the two spheres are touching and the net charge on these two spheres is still zero. While the rod is there on top of the left sphere, the right sphere is moved away from the left sphere. Because the right sphere has an excess negative charge by the same amount as the left sphere has an excess positive charge, the separated right sphere is negatively charged as shown in the third step. As the two spheres are moved apart further and the negatively charged rod is moved away from the spheres, the charges on the two spheres redistribute uniformly over the entire sphere surface. Thus, we have oppositely charged the two spheres. 25.14. Model: Charges A, B, and C are point charges. r Visualize: Please refer to Figure Ex25.14. Charge A experiences an electric force FB on A due to charge B and an electric r r r force FC on A due to charge C. The force FB on A is directed to the right and the force FC on A is directed to the left. Solve: Coulomb's law yields: FB on A = (9 " 10 N m /C )(1 " 10 C)(1 " 10 (1 " 10 m ) 9 2 2 !9 !2 2 9 !9 C ) = 9.0 " 10 !5 N FC on A The net force on A is (9 " 10 = N m 2 /C2 1 " 10 !9 C 4 " 10 !9 C )( )( ( 2 " 10 !2 m ) 2 ) = 9.0 " 10 !5 N r r r ^ ^ Fon A = FB on A + FC on A = 9.0 " 10 !5 N i + 9.0 " 10 !5 N !i = 0 N ( ) ( )( ) 25.16. Model: Assume the glass bead, the proton, and the electron are point charges. Visualize: Solve: Coulomb's law gives Fbead on electron = Fbead on proton = (a) Newton's second law is F = ma, so (9 " 10 9 N m 2 /C2 20 " 10 !9 C 1.60 " 10 !19 C !2 )( (1.0 " 10 )( m ) 2 ) = 2.88 " 10 !13 N aproton = In vector form Fbead on proton mproton = 2.88 " 10 !13 N = 1.72 " 1014 m/s2 1.67 " 10 !27 kg r aproton = 1.72 ! 1014 m/s2 , away from bead ( ) (b) Similarly, aelectron = Fbead on electron 2.88 " 10 !13 N = = 3.16 " 1017 m/s2 melectron 9.11 " 10 !31 kg r Thus aelectron = 3.16 ! 1017 m/s2 , toward bead . ( ) 25.25. Model: A field is the agent that exerts an electric force on a charge. Visualize: r Solve: Newton's second law on the plastic ball is ! Fnet Fon q = w # q E = mg # E = Because Fon q ( ) =F y r on q r " w . To balance the weight with the electric force, 1.0 " 10 !3 kg (9.8 N/kg ) mg = = 3.27 " 106 N/C q 3.0 " 10 !9 C ( ) must be upward and the charge is negative, the electric field at the location of the plastic ball must be r pointing downward. Thus E = 3.27 ! 10 6 N/C, downward . r r r r r r Assess: F = qE means the sign of the charge q determines the direction of F or E . For positive q, E and F are pointing r r in the same direction. But E and F point in opposite directions when q is negative. ( ) *25.33. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve: Plastic is an insulator and does not transfer charge from one sphere to the other. The charge of metal sphere A is (1.0 " 10 )(!1.60 " 10 12 !19 C = !160 nC and the charge of metal sphere B is 0 C. ) *25.41. Model: The charges are point charges. r Visualize: r Solve: The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1 . We have r " K q1 q2 # F2 on 1 = $ , toward q2 % 2 $ r % & ' " 9.0 ( 109 N m 2 /C2 10 ( 10 !9 C 5 ( 10 !9 C # =$ , toward q2 % 2 $ % $ % 1.0 ( 10 !2 m & ' = 4.5 ( 10 !3 N, toward q = 4.5 ( 10 !3 ^ N j ( )( )( ) ( ) ( 2 ) r " K q1 q3 # F3 on 1 = $ , toward q3 % $ r2 % & ' " 9.0 ( 109 N m 2 /C2 10 ( 10 !9 C 15 ( 10 !9 C # =$ , toward q3 % 2 $ % $ % 3.0 ( 10 !2 m & ' !3 !3 ^ = 1.5 ( 10 N, toward q = !1.5 ( 10 i N ( )( )( ) ( ) ( 3 ) r r r ^ " Fon 1 = F2 on 1 + F3 on 1 = !1.5 # 10 !3 i + 4.5 # 10 !3 ^ N j ( ) The magnitude and direction of the resultant force vector are Fon 1 = (!1.5 " 10 !3 N ) + (4.5 " 10 2 !3 N ) 2 = 4.74 " 10 !3 N tan ! = 4.5 # 10 "3 N ! ! = tan "1 (3) = 71.6 above the " x axis 1.5 # 10 "3 N 25.50. Model: The charged particles are point charges. Visualize: Please refer to Figure P25.50. Solve: The charge q2 is in static equilibrium, so the net electric field at the location of q2 is zero. We have r r r Enet = Eq1 + E#2 nC = 2 $ 10 #9 C q1 1 ^ + 1 ^ i #i = 0 N/C 4!" 0 (0.2 m )2 4!" 0 (0.10 m )2 ( ) ( ) ( ) ^ We have used the sign to indicate that a positive charge on q1 leads to an electric field along +i and a negative charge ^ ^ on q leads to an electric field along !i . Because the above equation can only be satisfied if we use +i , we infer that the 1 charge q 1 is a positive charge. Thus, 2 " 10 (+i^ )! (0.10 m )C (i^ )= 0 N/C ! q = 8 nC 0.2 m ) ( 2 2 1 q1 !9 25.60. Model: The charged spheres are point charges. Visualize: Each sphere is in static equilibrium when the string makes an angle of 20 with the vertical. The three forces acting on each sphere are the electric force, the weight of the sphere, and the tension force. r r r r r Solve: In the static equilibrium, Newton's first law is Fnet = T + w + Fe = 0. In component form, (Fnet )x = Tx + wx + (Fe )x = 0 N "T sin! + 0 N + T sin! = (Fnet )y = Ty + wy + (Fe )y = 0 N T cos! " mg + 0 N = 0 N Kq2 =0N d2 Kq2 + Kq2 = 2 2 d (2L sin! ) T cos! = + mg Dividing the two equations and solving for q, q= 4sin2 ! tan! L2 mg = K 4 sin2 20 tan20 (1.0 m ) 3.0 # 10 "3 kg (9.8 N/kg ) 9.0 # 10 N m /C 9 2 2 ( ) 2 ( ) = 746 nC 25.61. Model: The electric field is that of a positive point charge located at the origin. Visualize: Please refer to Figure P25.61. Place the 10 nC charge at the origin. Solve: The electric field is 9 2 2 #9 % r $ 1 q % $ 9.0 & 10 N m /C 10 & 10 C E=' , away from q ( = ' , away from q ( 2 2 ( r ) 4!" 0 r * ' ) * ( )( ) $ 90.0 N m 2 /C % =' , away from q ( 2 r ) * At each of the three points, r # E1 = % % % ( r # E2 = % % % ( $ , away from q & = 1.0 ' 105 ^ N/C j 2 & & 3.0 ' 10 "2 m ) $ 90.0 N m 2 /C ^ , away from q & = 3.6 ' 10 4 N/C cos! i + sin! ^ j 2 & "2 & 5.0 ' 10 m ) ^ 3j ^ = 3.6 ' 10 4 N/C 4 i + 5 ^ = 2.88 ' 10 4i + 2.16 ' 10 4 ^ N/C j 5 90.0 N m 2 /C ( ( ) ( ) )( ) ( )( ) ) ( ) $ r # 90.0 N m 2 /C ^ E3 = % , away from q & = 5.63 ' 10 4i N/C 2 % & "2 % 4.0 ' 10 m & ( ) ( *25.69. Model: The charged ball attached to the string is the point charge. Visualize: The charged ball is in static equilibrium in the external electric field when the string makes an angle with the vertical. The three forces acting on the charge are the electric force due to the electric field, the weight of the ball, and the tension force. r r r r r Solve: In static equilibrium, Newton's second law for the charged ball is Fnet = T + w + F3 = 0 . In component form, (Fnet )x = Tx + 0 N + qE = 0 N (Fnet )y = Ty ! mg + 0 N = 0 N These two equations become T sin! = qE and T cos! = mg . Dividing the equations gives tan! = 25 # 10 "9 C (200,000 N/C ) qE = = 0.255 $ ! = 14.3 mg 2.0 # 10 "3 kg (9.8 N/kg ) ( ( ) )

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