Physics Solution Manual for 1100 and 2101

50 c 60 c thus the second charge is q2 60 c k

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Unformatted text preview: C = −6.2 × 107 N/C ) + (8.99 × 109 N ⋅ m2 /C2 )( 21 × 10−6 C) 2 (9.0 × 10−2 m ) 974 ELECTRIC FORCES AND ELECTRIC FIELDS The minus sign tells us that the net electric field points along the −x axis. b. The net electric field at x = +6.0 cm (point B) is E = E1 + E2: E = E1 + E2 = k q1 r12 + k q2 r22 (8.99 × 109 N ⋅ m2 /C2 )(8.5 × 10−6 C) + (8.99 × 109 N ⋅ m2 /C2 )( 21 × 10−6 C) = 2 2 (3.0 × 10−2 m ) (3.0 × 10−2 m ) = +2.9 × 108 N/C The plus sign tells us that the net electric field points along the +x axis. ______________________________________________________________________________ 38. REASONING At every position in space, the net electric field E is the vector sum of the external electric field Eext and the electric field Epoint created by the point charge at the origin: E = Eext + Epoint. The external electric field is uniform, which means that it has the same magnitude Eext = 4500 N/C at all locations and that it always points in the positive x direction (see the drawing). The magnitude Epoint of the electric field due to y Eext Epoint Epoint Epoint Eext the point charge at the origin is given by Epoint = Eext x kq (Equation 18.3), where r is the r2 distance between the origin and the location where the electric field is to be evaluated, q is the charge at the origin, and k = 8.99×109 N·m2/C2. All three locations given in the problem are a distance r = 0.15 m from the origin, so we have that Epoint = kq r2 (8.99 ×109 N ⋅ m 2 /C2 ) −8.0 ×10−9 C = 3200 N/C = ( 0.15 m )2 The direction of the electric field Epoint varies from location to location, but because the charge q is negative, Epoint is always directed towards the origin (see the drawing). Chapter 18 Problems 975 SOLUTION a. At x = −0.15 m, the electric field of the point charge and the external electric field both point in the positive x direction (see the drawing), so the magnitude E of the net electric field is the sum of the magnitudes of the individual electric fields: E = Eext + Epoint = 4500 N/C + 3200 N/C = 7700 N/C b. At x = +0.15 m, the electric field of the point charge is opposite the external electric field (see the drawing). Therefore, the magnitude E of the net electric field is the difference between the magnitudes of the individual electric fields: E = Eext − Epoint = 4500 N/C − 3200 N/C = 1300 N/C c. At y = +0.15 m, the electric fields Epoint and Eext are perpendicular (see the drawing). This makes them, in effect, the x-component (Eext) and y-component (Epoint) of the net electric field E. The magnitude E of the net electric field, then, is given by the Pythagorean theorem (Equation 1.7): 2 2 E = Eext + Epoint = ( 4500 N/C )2 + ( 3200 N/C )2 = 5500 N/C 39. SSM REASONING Since the charged droplet (charge = q) is suspended motionless in the electric field E, the net force on the droplet must be zero. There are two forces that act on the droplet, the force of gravity W = m g , and the electric force F = qE due to the electric field. Since the net force on the droplet is zero, we conclude that mg = q E . We can use this reasoning to determine the sign and the magnitude of the charge on the droplet. SOLUTION a. Since the net force on the droplet is zero, and the weight of magnitude W points downward, the electric force of magnitude F = q E must point upward. Since the electric field points upward, the excess charge on the droplet must be positive in order for the force F to point upward. F mg b. Using the expression mg = q E , we find that the magnitude of the excess charge on the droplet is mg (3.50 ×10 –9 kg)(9.80 m/s 2 ) q= = = 4.04 ×10 –12 C E 8480 N/C 976 ELECTRIC FORCES AND ELECTRIC FIELDS The charge on a proton is 1.60 × 10–19 C, so the excess number of protons is 1 proton ( 4.04 ×10–12 C ) 1.60 ×10–19 C = 2.53 ×107 protons ______________________________________________________________________________ 40. REASONING Since the proton and the electron have the same charge magnitude e, the electric force that each experiences has the same magnitude. The directions are different, however. The proton, being positive, experiences a force in the same direction as the electric field (due east). The electron, being negative, experiences a force in the opposite direction (due west). Newton’s second law indicates that the direction of the acceleration is the same as the direction of the net force, which, in this case, is the electric force. The proton’s acceleration is in the same direction (due east) as the electric field. The electron’s acceleration is in the opposite direction (due west) as the electric field. Newton’s second law indicates that the magnitude of the acceleration is equal to the magnitude of the electric force divided by the mass. Although the proton and electron experience the same force magnitude, they have different masses. Thus, they have accelerations of different magnitudes. SOLUTION According to Newton’s second law, Equation 4.2, the acceleration a of an object is equal to the net force divided by the object’s mass m. In this situ...
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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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