Physics Solution Manual for 1100 and 2101

2 107 n c 12 107 n c 2 2 17 107 n c part

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Unformatted text preview: utions cancel. SOLUTION Using Coulomb’s law for each contribution to the net force, we calculate the ratio of the greatest to the smallest net force as follows: ( ΣF ) C = ( ΣF ) B k k q 2 d2 q +k 2 d2 −k q 2 d2 q +k 2 d2 +k q 2 ( 2d ) 2 q 2 = 1+1+ 1 4 1 4 = 9.0 ( 2d ) 2 23. REASONING The kinetic energy of the orbiting electron is KE = 1 mv 2 (Equation 6.2), 2 where m and v are its mass and speed, respectively. We can obtain the speed by noting that the electron experiences a centripetal force whose magnitude Fc is given by Fc = mv 2 / r (Equation 5.3), where r is the radius of the orbit. The centripetal force is provided almost exclusively by the electrostatic force of attraction F between the electron and the protons, so Fc = F. The electrostatic force points toward the center of the circle and its magnitude is given by Coulomb’s law as F = k q1 q2 / r 2 (Equation 18.1), where q1 and q2 are the magnitudes of the charges. SOLUTION The kinetic energy of the electron is KE = 1 mv 2 2 (6.2) Solving the centripetal-force expression, Fc = mv 2 / r (Equation 5.3), for the speed v, and substituting the result into Equation 6.2 gives rF KE = 1 mv 2 = 1 m c = 1 r Fc 2 2 2 m (1) The centripetal force is provided almost entirely by the electrostatic force, so Fc = F, where F is the magnitude of the electrostatic force of attraction between the electron and the three protons. This force is given by Coulomb’s law, F = k q1 q2 / r 2 (Equation 18.1). Substituting Coulomb’s law into Equation 1 yields k q q k q1 q2 KE = 1 r Fc = 1 r F = 1 r 12 2 = 2 2 2 2r r 962 ELECTRIC FORCES AND ELECTRIC FIELDS Setting q1 = −e and q2 = +3e , we have KE = = k −e +3e 2r ( 8.99 ×109 N ⋅ m 2 / C2 ) −1.60 × 10−19 C +3 ×1.60 × 10−19 C = 1.96 ×10−17 J 2(1.76 ×10 m) ______________________________________________________________________________ −11 24. REASONING When the airplane and the other end of the guideline carry point charges +q and −q, the airplane is subject to an attractive electric force of magnitude qq q −q q2 F =k 1 22 =k = k 2 (Equation 18.1), where k = 8.99 ×109 N ⋅ m2 / C2 and r is the r r2 r 2 mv2 length of the guideline. This electric force provides part of the centripetal force Fc2 = r (Equation 5.3) necessary to keep the airplane (mass = m) flying along its circular path at the higher speed v2 (which is associated with the greater kinetic energy KE2). The remainder of the centripetal force is provided by the maximum tension Tmax in the guideline, so we have that 2 mv2 F + Tmax = Fc2 = (1) r When the airplane is neutral and flying at the slower speed v1 (which is associated with the smaller kinetic energy KE1), there is no electrical force, so the centripetal force Fc1 = 2 mv1 r acting on the airplane is due solely to the maximum tension Tmax in the guideline: Tmax = Fc1 = 2 mv1 r (2) We note that the kinetic energy of the airplane is given by KE = 1 mv 2 (Equation 6.2), so 2 that the quantities in the numerators of Equations (1) and (2) are proportional to the kinetic energies KE2 and KE1 of the airplane: 2 mv2 = 2 ( KE 2 ) SOLUTION Solving F = k sides, we obtain and 2 mv1 = 2 ( KE1 ) (3) q2 (Equation 18.1) for q2 and taking the square root of both 2 r Chapter 18 Problems q2 = Fr 2 k Fr 2 k q= or 963 (4) Substituting Equations (3) into Equations (1) and (2) yields F + Tmax = 2 ( KE 2 ) r Tmax = and 2 ( KE1 ) (5) r Substituting the second of Equations (5) into the first of Equations (5), we obtain F+ 2 ( KE1 ) r = 2 ( KE 2 ) r or F= 2 ( KE 2 ) r − 2 ( KE1 ) r = 2 ( KE 2 − KE1 ) r (6) Substituting Equation (6) into Equation (4), we find that q= Fr 2 = k 2 ( KE 2 − KE1 ) r 2 = k r 2 ( KE 2 − KE1 ) r k Therefore, the magnitude q of the charge on the airplane is q= 25. 2 ( 51.8 J − 50.0 J ) ( 3.0 m ) = 3.5 × 10 −5 C 8.99 × 109 N ⋅ m 2 /C 2 SSM REASONING Consider the drawing qU q1 FAU at the right. It is given that the charges qA, d FA2 q1, and q2 are each positive. Therefore, the 4d qA charges q1 and q2 each exert a repulsive q2 = +3.0 µC force on the charge qA. As the drawing θ shows, these forces have magnitudes FA1 (vertically downward) and FA2 (horizontally FA1 to the left). The unknown charge placed at the empty corner of the rectangle is qU, and it exerts a force on qA that has a magnitude FAU. In order that the net force acting on qA point in the vertical direction, the horizontal component of FAU must cancel out the horizontal force FA2. Therefore, FAU must point as shown in the drawing, which means that it is an attractive force and qU must be negative, since qA is positive. 964 ELECTRIC FORCES AND ELECTRIC FIELDS SOLUTION The basis for our solution is the fact that the horizontal component of FAU must cancel out the horizontal force FA2. The magnitudes of these forces can be expressed using Coulomb’s law F = k q q′ / r 2 , where r is the distance between the charges q and q′ . Thus, we have FAU = k qA qU FA2 = and ( 4d )2 + d 2 k qA q2 ( 4d ) 2 where we have used the fact that the distance between...
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