641_PartUniversity Physics Solution

641_PartUniversity Physics Solution - 21-34 Chapter 21...

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21-34 Chapter 21 21.84. IDENTIFY: The electric field exerts equal and opposite forces on the two balls, causing them to swing away from each other. When the balls hang stationary, they are in equilibrium so the forces on them (electrical, gravitational, and tension in the strings) must balance. SET UP: (a) The force on the left ball is in the direction of the electric field, so it must be positive, while the force on the right ball is opposite to the electric field, so it must be negative. (b) Balancing horizontal and vertical forces gives qE = T sin θ /2 and mg = T cos /2. EXECUTE: Solving for the angle gives: = 2 arctan( qE/mg ). (c) As E , 2 arctan( ) = 2 ( π /2) = π = 180° EVALUATE: If the field were large enough, the gravitational force would not be important, so the strings would be horizontal. 21.85. IDENTIFY and SET UP: Use the density of copper to calculate the number of moles and then the number of atoms. Calculate the net charge and then use Coulomb's law to calculate the force. EXECUTE: (a) () 3 33 3 3 5 44 8.9 10 kg/m 1.00 10 m 3.728 10 kg mV r ρρ π −− ⎛⎞ == = × × = × ⎜⎟ ⎝⎠ ( ) ( ) 53 4 / 3.728 10 kg / 63.546 10 kg/mol 5.867 10 mol nmM × × 20 3.5 10 atoms A Nn N × (b) ( ) 20 22 29 3.5 10 1.015 10 e N = × electrons and protons ( )( )( ) 21 9 2 2 net 0.99900 0.100 10 1.602 10 C 1.015 10 1.6 C ee qe N e N =− = × × × = 2 2 10 2 2 1.6 C 2.3 10 N 1.00 m q Fk k r = × EVALUATE: The amount of positive and negative charge in even small objects is immense. If the charge of an electron and a proton weren't exactly equal, objects would have large net charges. 21.86. IDENTIFY: Apply constant acceleration equations to a drop to find the acceleration. Then use Fm a = to find the force and Fq E = to find q . SET UP: Let 2.0 cm D = be the horizontal distance the drop travels and 0.30 mm d = be its vertical displacement. Let + x be horizontal and in the direction from the nozzle toward the paper and let + y be vertical, in the direction of the deflection of the drop. 0 x a = and y aa = . EXECUTE: First, the mass of the drop: 63 3 11 4 (15.0 10 m) (1000 kg m ) 1.41 10 kg 3 ρ × . Next, the time of flight: (0.020 m) (20 m/s) 0.00100 s tD v = . 2 1 2 da t = . 4 2 22 ( 3 . 0 0 1 0 m ) 600 m s (0.001s) d a t × = . Then aFmq Em gives 2 11 13 4 (1.41 10 kg)(600 m s ) 1.06 10 C 8.00 10 N C qm aE × = × × . EVALUATE: Since q is positive the vertical deflection is in the direction of the electric field. 21.87. IDENTIFY: Eq. (21.3) gives the force exerted by the electric field. This force is constant since the electric field is uniform and gives the proton a constant acceleration. Apply the constant acceleration equations for the x - and y -components of the motion, just as for projectile motion. (a) SET UP: The electric field is upward so the electric force on the positively charged proton is upward and has magnitude F = eE . Use coordinates where positive y is downward. Then applying m = F a G G to the proton gives that 0 and / .
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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641_PartUniversity Physics Solution - 21-34 Chapter 21...

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