{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

641_PartUniversity Physics Solution

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

This preview shows pages 1–2. Sign up to view the full content.

21-34 Chapter 21 21.84. I DENTIFY : 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. S ET U P : (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 θ /2. E XECUTE : Solving for the angle θ gives: θ = 2 arctan( qE/mg ). (c) As E , θ 2 arctan( ) = 2 ( π /2) = π = 180° E VALUATE : If the field were large enough, the gravitational force would not be important, so the strings would be horizontal. 21.85. I DENTIFY and S ET U P : 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. E XECUTE : (a) ( ) ( ) 3 3 3 3 3 5 4 4 8.9 10 kg/m 1.00 10 m 3.728 10 kg 3 3 m V r ρ ρ π π = = = × × = × ( ) ( ) 5 3 4 / 3.728 10 kg / 63.546 10 kg/mol 5.867 10 mol n m M = = × × = × 20 3.5 10 atoms A N nN = = × (b) ( ) ( ) 20 22 29 3.5 10 1.015 10 e N = × = × electrons and protons ( ) ( )( )( ) 2 19 22 net 0.99900 0.100 10 1.602 10 C 1.015 10 1.6 C e e q eN eN = = × × × = ( ) ( ) 2 2 10 2 2 1.6 C 2.3 10 N 1.00 m q F k k r = = = × E VALUATE : 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. I DENTIFY : Apply constant acceleration equations to a drop to find the acceleration. Then use F ma = to find the force and F q E = to find q . S ET U P : 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 a a = . E XECUTE : First, the mass of the drop: 6 3 3 11 4 (15.0 10 m) (1000 kg m ) 1.41 10 kg 3 m V π ρ × = = = × . Next, the time of flight: (0.020 m) (20 m/s) 0.00100 s t D v = = = . 2 1 2 d at = . 4 2 2 2 2 2(3.00 10 m) 600 m s (0.001s) d a t × = = = . Then a F m qE m = = gives 2 11 13 4 (1.41 10 kg)(600 m s ) 1.06 10 C 8.00 10 N C q ma E × = = = × × . E VALUATE : Since q is positive the vertical deflection is in the direction of the electric field. 21.87. I DENTIFY : 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) S ET U P : 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 / .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online