This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: PHY2054 Spring 2009 1 P r o f . P r a d e e p K u m a r P r o f . P a u l A v e r y Prof. Yoonseok Lee Feb. 4, 2009 Exam 1 Solutions 1. A proton (+ e ) originally has a speed of v = 2.0 × 10 5 m/s as it goes through a plate as shown in the figure. It shoots through the tiny holes in the two plates across which 16 V | 100 V | 2 V of electric potential is applied. Find the speed as it leaves the second plate. Answer : 1.92 × 10 5 m/s | 1.44 × 10 5 m/s | 1.99 × 10 5 m/s Solution : The proton is slowed down by the electric field between the two plates, so the velocity will be reduced . Conservation of en- ergy yields 2 2 1 1 2 2 p i p f m v e V m v − Δ = , so 2 2 / f i p v v e V m = − Δ . 2. Five identical charges of +1 μ C are located on the 5 vertices of a regular hexagon of 1 m side leaving one vertex empty. A − 1 μ C | − 2 μ C | +2 μ C point charge is placed at the center of the hexagon. Find the magnitude and direction of the net Coulomb force on this charge. Answer : 9.0 × 10 − 3 N − x-direction | 1.8 × 10 − 2 N − x-direction | 1.8 × 10 − 2 N + x-direction Solution : Only the charge on the far left contributes to the net force on the charge in the center because its contribution is not cancelled by an opposing charge. If Q represents the charge at each hexagon vertex, q is the charge at the center and d is the length of a hexagon side (note that the distance from each hexagon point to the center is also d ), the force on q is 2 / kQq d along the + x direction if Qq is positive and along the − x direction if Qq is negative. 3. Wire A and Wire B are made out of copper and each wire carries the same current. If the di- ameter of A is twice | half | three times that of Wire B, how does the drift velocity v dA in Wire A compares to that in Wire B, v dB ? Answer : v dA = v dB / 4 | v dA = 4 v dB | v dA = v dB / 9 Solution : The current i is e d i An ev = , where 2 A r π = is the cross sectional area, n e is the free electron density and v d is the drift velocity. The answers follow by writing 2 / d e v i r e n π = and comparing the diameters. V v +++ +++ − − − − − − x y PHY2054 Spring 2009 2 4. The free electron density of copper is 8.5 × 10 28 m − 3 . A copper wire is connected to a 1.5 V battery and carries a current of 0.125 A | 0.015 A | 0.010 A. The electrons inside the wire move with the drift velocity of 2.5 × 10 − 4 m/s. What is the cross-sectional area of the wire?...
View Full Document
This note was uploaded on 07/16/2011 for the course PHY 2005 taught by Professor Lee during the Summer '08 term at University of Florida.
- Summer '08
- Applied Physics