solution4

solution4 - fullwood (tf5349) Homework 2 Electric Field and...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: fullwood (tf5349) Homework 2 Electric Field and Potential Catala (12054) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A positron (a particle with a charge of + e and a mass equal to that of an electron) that is accelerated from rest between two points at a fixed potential difference acquires a speed of 9 . 3 10 7 m / s. What speed is achieved by a proton acceler- ated from rest between the same two points? (Disregard relativistic effects.) Correct answer: 2 . 17005 10 6 m / s. Explanation: Let : v f,positron = 9 . 3 10 7 m / s , m positron = 9 . 109 10- 31 kg , m pr = 1 . 673 10- 27 kg , and q pr = q positron = 1 . 60 10- 19 C . Since K i = 0 J , K f = U 1 2 mv 2 f = q V . V = m positron ( v f,positron ) 2 2 q positron = (9 . 109 10- 31 kg) (9 . 3 10 7 m / s) 2 2 (1 . 60 10- 19 C) = 24619 . 9 V . Thus v f,pr = radicalBigg 2 U electric m pr = radicalBigg 2 ( q pr V ) m pr = radicalBigg 2 (1 . 60 10- 19 C) (24619 . 9 V) 1 . 673 10- 27 kg = 2 . 17005 10 6 m / s . 002 (part 1 of 2) 10.0 points An electric field does 9 J of work on a 0 . 0009 C charge. What is the voltage change? Correct answer: 10000 V. Explanation: Let : W = 9 J and q = 0 . 0009 C . Work is W = qV V = W q = 9 J . 0009 C = 10000 V . 003 (part 2 of 2) 10.0 points The same electric field does 18 J of work on a . 0018 C charge. What is the voltage change? Correct answer: 10000 V. Explanation: Let : W = 18 J and q = 0 . 0018 C . The voltage change is V = W q = 18 J . 0018 C = 10000 V . 004 10.0 points The magnitude of a uniform electric field be- tween the two plates is about 2 10 5 N / C. If the distance between these plates is . 1 cm, find the potential difference between the plates. Correct answer: 200 V. fullwood (tf5349) Homework 2 Electric Field and Potential Catala (12054) 2 Explanation: Let : E = 2 10 5 N / C and d = 0 . 1 cm = 0 . 001 m . The magnitude of the potential difference is V = E d = (2 10 5 N / C) (0 . 001 m) = 200 V . 005 10.0 points The Coulomb constant is 8 . 98755 10 9 N m 2 / C 2 and the acceleration of gravity is 9 . 81 m / s 2 ....
View Full Document

This note was uploaded on 06/10/2010 for the course PHY 100 taught by Professor Pro.kiroshima during the Spring '10 term at Adelphi.

Page1 / 6

solution4 - fullwood (tf5349) Homework 2 Electric Field and...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online