Physics Solution Manual for 1100 and 2101

Therefore the electric flux is e e cos a 150 nccos

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Unformatted text preview: also q = +6.6 µC directed upward (see the free-body diagram). Balancing this upward force are two downward forces: the weight mg of the sphere (where m is the mass of the sphere and g is the acceleration mg Fs due to gravity) and the force Fs exerted on the sphere by the spring (see the free-body diagram). We know that the spring exerts a downward force on the sphere because the equilibrium length Free-body diagram L = 0.059 m of the spring is shorter than its unstrained length L0 = 0.074 m. The magnitude Fs of the spring force is given by Fs = kx (Equation 10.2, without the minus sign), where k is the spring constant of the spring and x = L0 − L = 0.074 m – 0.059 m = 0.015 m is the distance by which the spring has been compressed. 980 ELECTRIC FORCES AND ELECTRIC FIELDS SOLUTION Solving FE = qE (Equation 18.2) for the magnitude E of the external electric field, we find that F (1) E= E q The sphere is in equilibrium, so the upward force (FE) must exactly balance the two downward forces (mg and Fs). Therefore, the magnitudes of the three forces are related by FE = mg + Fs (2) Substituting Equation (2) into Equation (1) yields E= mg + Fs (3) q Lastly, substituting Fs = kx (Equation 10.2, without the minus sign) into Equation (3), we obtain the desired electric field magnitude: ( )( ) 5.1× 10−3 kg 9.80 m/s 2 + ( 2.4 N/m )( 0.015 m ) mg + kx E= = = 1.3 × 104 N/C q 6.6 × 10−6 C 45. E2 SSM WWW REASONING The drawing shows the arrangement of the three charges. Let E q represent the electric field at the empty corner due to the –q charge. Furthermore, let E1 and E2 be the electric fields at the empty corner due to charges +q1 and +q2, respectively. +q1 Eq d5 –q E1 θ d +q 2 2d According to the Pythagorean theorem, the distance from the charge –q to the empty corner along the diagonal is given by (2d )2 + d 2 = 5d 2 = d 5 . The magnitude of each electric field is given by Equation 18.3, E = k q / r 2 . Thus, the magnitudes of each of the electric fields at the empty corner are given as follows: Eq = kq r 2 = kq (d 5 ) 2 = kq 5d 2 Chapter 18 Problems E1 = k q1 ( 2d ) 2 = k q1 and 4d 2 E2 = 981 k q2 d2 The angle θ that the diagonal makes with the horizontal is θ = tan −1 (d / 2d ) = 26.57° . Since the net electric field Enet at the empty corner is zero, the horizontal component of the net field must be zero, and we have E1 – Eq cos 26.57° = 0 k q1 or 4d 2 – k q cos 26.57° 5d 2 =0 Similarly, the vertical component of the net field must be zero, and we have E2 – Eq sin 26.57° = 0 or k q2 d2 – k q sin 26.57° 5d 2 =0 These last two expressions can be solved for the charge magnitudes q1 and q2 . SOLUTION Solving the last two expressions for q1 and q2 , we find that 4 5 q1 = q cos 26.57° = 0.716 q 1 5 q2 = q sin 26.57° = 0.0895 q ______________________________________________________________________________ 2 2 46. REASONING AND SOLUTION From kinematics, vy = v0y + 2ayy. Since the electron starts from rest, v0y = 0 m/s. The acceleration of the electron is given by ay = F eE = mm where e and m are the electron's charge magnitude and mass, respectively, and E is the magnitude of the electric field. The magnitude of the electric field between the plates of a parallel plate capacitor is E = σ/ε0, where σ is the magnitude of the charge per unit area on each plate. Thus, ay = eσ/(mε0). Combining this expression for a with the kinematics equation we have eσ v2 = 2 y y mε 0 982 ELECTRIC FORCES AND ELECTRIC FIELDS Solving for vy gives ( )( ) )( ) 2 1.60 ×10−19 C 1.8 ×10−7 C/m 2 1.5 ×10−2 m 2eσ y = = 1.0 ×107 m/s vy = −31 mε 0 8.85 ×10−12 C 2 / N ⋅ m 2 9.11×10 kg ______________________________________________________________________________ ( 47. REASONING The fact that the net electric field points upward along the vertical axis holds the key to this problem. The drawing at the right shows the fields from each charge, together with the horizontal components of each. The reason that the net field points upward is that these horizontal components point in opposite directions and cancel. Since they cancel, they must have equal magnitudes, a fact that will quickly lead us to a solution. ( ) E1 60.0º 30.0º E2 E2 sin 60.0º E1 sin 30.0º 30.0º 60.0º q2 q1 SOLUTION Setting the magnitudes of the horizontal components of the fields equal gives E2 sin 60.0° = E1 sin 30.0° The magnitude of the electric field created by a point charge is given by Equation 18.3. Using this expression for E1 and E2 and noting that each point charge is the same distance r from the center of the circle, we obtain k q2 r2 sin 60.0° = k q1 r2 sin 30.0° or q2 sin 60.0° = q1 sin 30.0° Solving for the ratio of the charge magnitudes gives q2 q1 = sin 30.0° = 0.577 sin 60.0° 48. REASONING a. The drawing at the right shows the electric fields at point P due to the two charges in the case that the second charge is positive. The presence of the q2 + +q1 d P d E2 E1 Chapter 18 Problems 983 second charge c...
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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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