solution_pdf - Copy

# solution_pdf - Copy - vo(ytv59 – oldhomework 01 –...

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: vo (ytv59) – oldhomework 01 – Turner – (56705) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Three identical point charges hang from three strings, as shown. 45 ◦ 45 ◦ F g 12.0 cm 12.0 cm + + + +q +q +q 0.10 kg 0.10 kg 0.10 kg What is the value of q ? The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 , and the acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 7 . 92912 × 10 − 7 C. Explanation: Let : m = 0 . 10 kg , L = 12 . 0 cm , θ = 45 ◦ , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . r = 2 L sin θ = 2 L sin 45 ◦ = 2 L √ 2 2 = L √ 2 F T,x = F T sin θ F T,y = F T cos θ Each sphere is in equilibrium horizontally F electric − F T,x = 0 F electric − F T sin θ = 0 and vertically F T,y − F g = 0 F T cos θ − F g = 0 F T = F g cos θ . From the horizontal equilibrium, F electric = parenleftbigg F g cos θ parenrightbigg sin θ F electric = F g tan θ = F g (tan 45 ◦ ) = F g . For either of the end charges, F electric = k e q 2 r 2 + k e q 2 ( r 2 ) 2 = k e q 2 r 2 + 4 k e q 2 r 2 = 5 k e q 2 r 2 5 k e q 2 r 2 = mg . Thus | q | = radicalBigg r 2 mg 5 k e = radicalBigg ( L √ 2) 2 mg 5 k e = L · radicalbigg 2 mg 5 k e = (12 cm) parenleftbigg 1 m 100 cm parenrightbigg × radicalBigg 2(0 . 1 kg)(9 . 81 m / s 2 ) 5(8 . 98755 × 10 9 N · m 2 / C 2 ) = 7 . 92912 × 10 − 7 C . 002 10.0 points There are two identical small metal spheres with charges 82 . 4 μ C and − 50 . 9712 μ C. The distance between them is 6 cm. The spheres are placed in contact then set at their original distance. Calculate the magnitude of the force between the two spheres at the final position. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 616 . 502 N. Explanation: vo (ytv59) – oldhomework 01 – Turner – (56705) 2 Let : q A = 82 . 4 μ C = 8 . 24 × 10 − 5 C , q B = − 50 . 9712 μ C = − 5 . 09712 × 10 − 5 C , and d = 6 cm = 0 . 06 m . When the spheres are in contact, the charges will rearrange themselves until equi- librium is reached. Each sphere will then have half of the original total charge: q = q A + q B 2 = 8 . 24 × 10 − 5 C − 5 . 09712 × 10 − 5 C 2 = 1 . 57144 × 10 − 5 C . The force between the two spheres is F = k q 2 d 2 = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × (1 . 57144 × 10 − 5 C) 2 (0 . 06 m) 2 = 616 . 502 N . 003 10.0 points The diagram shows an isolated, positive charge Q , where point B is twice as far away from Q as point A . + Q A B 10 cm 20 cm The ratio of the electric field strength at point A to the electric field strength at point B is 1....
View Full Document

## This note was uploaded on 09/12/2010 for the course PHY 303L taught by Professor Turner during the Fall '10 term at Austin Community College.

### Page1 / 8

solution_pdf - Copy - vo(ytv59 – oldhomework 01 –...

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

View Full Document
Ask a homework question - tutors are online