homework 09-solutions

# homework 09-solutions - howard (cah3459) – homework 09...

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

Unformatted text preview: howard (cah3459) – homework 09 – Turner – (56705) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Initially, both metal spheres are neutral. In a charging process, 1 × 10 13 electrons are removed from one metal sphere and placed on a second sphere. Then the electrical poten- tial energy associated with the two spheres is found to be- . 061 J . The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 and the charge on an electron is 1 . 6 × 10- 19 C . What is the distance between the two spheres? Correct answer: 0 . 377183 m. Explanation: Let : n = 1 × 10 13 , q e =- 1 . 6 × 10- 19 C , U electric =- . 061 J , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . q 1 = nq e q 2 =- nq e U electric = k e q 1 q 2 r r = k e q 1 q 2 U electric = k e- [ nq e ] 2 U electric =- (8 . 98755 × 10 9 N · m 2 / C 2 ) · [(1 × 10 13 ) (1 . 6 × 10- 19 C)] 2- . 061 J = . 377183 m . 002 10.0 points Four charges are fixed at the corners of a square centered at the origin as follows: q at (- a, + a ); 2 q at (+ a, + a );- 3 q at (+ a,- a ); and 6 q at (- a,- a ). A fifth charge + q with mass m is placed at the origin and released from rest. Find the speed when it is a great distance from the origin, where the potential energy of the fifth charge due to the four point charges is negligible. 1. \$v = q 6 √ 3 k ma 2. \$v = q 6 √ 5 k ma 3. \$v = q 3 √ 5 k ma 4. \$v = q 3 √ 2 k ma 5. \$v = q 2 √ 2 k ma 6. \$v = q 3 √ 3 k ma 7. \$v = q 6 √ 2 k ma correct 8. \$v = q 2 √ 5 k ma 9. \$v = q 6 √ 6 k ma 10. \$v = q 3 √ 6 k ma Explanation: x a +6 q a- 3 q +2 q + q + q, m √ 2 a The initial energy of the charge is E i = K i + U i = U i howard (cah3459) – homework 09 – Turner – (56705) 2 = q k q √ 2 a + 2 k q √ 2 a + (- 3 q ) k √ 2 a + 6 k q √ 2 a = 6 k q 2 √ 2 a . The final energy is E f = 1 2 mv 2 . From energy conversation, we have E i = E f 6 k q 2 √ 2 a = 1 2 mv 2 v = q 6 √ 2 k ma . 003 10.0 points A charged particle is connected to a string that is is tied to the pivot point P . The particle, string, and pivot point all lie on a horizontal table (consequently the figure below is viewed from above the table). The particle is initially released from rest when the string makes an angle 37 ◦ with a uniform electric field in the horizontal plane (shown in the figure). 37 ◦ 2 . 2 m 462 V / m P . 01 kg 3 μ C initial release ω parallel P Determine the speed of the particle when the string is parallel to the electric field. Correct answer: 0 . 350429 m / s. Explanation: Let : L = 2 . 2 m , q = 3 μ C = 3 × 10- 6 C , m = 0 . 01 kg , and θ = 37 ◦ ....
View Full Document

## This note was uploaded on 10/11/2010 for the course PHY 56715 taught by Professor Turner during the Spring '10 term at University of Texas.

### Page1 / 9

homework 09-solutions - howard (cah3459) – homework 09...

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

View Full Document
Ask a homework question - tutors are online