{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solution37

solution37 - cherubin(mcc2645 – Homework 2 Electric Field...

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 Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: cherubin (mcc2645) – 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 (part 1 of 2) 10.0 points An electric field does 10 J of work on a . 0008 C charge. What is the voltage change? Correct answer: 12500 V. Explanation: Let : W = 10 J and q = 0 . 0008 C . Work is W = qV V = W q = 10 J . 0008 C = 12500 V . 002 (part 2 of 2) 10.0 points The same electric field does 20 J of work on a . 0016 C charge. What is the voltage change? Correct answer: 12500 V. Explanation: Let : W = 20 J and q = 0 . 0016 C . The voltage change is V = W q = 20 J . 0016 C = 12500 V . 003 10.0 points Two charges are located along the x-axis. One has a charge of 6 . 4 μ C, and the second has a charge of- 3 . 4 μ C. The Coulomb constant is k e = 8 . 98755 × 10 9 N m 2 / C 2 . The acceleration of gravity is 9 . 81 m / s 2 . If the electrical potential energy associated with the pair of charges is- 4 . 3 × 10- 2 J, what is the distance between the charges? Correct answer: 4 . 54936 m. Explanation: Let : q 1 = 6 . 4 μ C , q 2 =- 3 . 4 μ C , U electric =- 4 . 3 × 10- 2 J , and k e = 8 . 99 × 10 9 N · m 2 / C 2 . U electric = k e q 1 q 2 r r = k e q 1 q 2 U electric = (8 . 99 × 10 9 N · m 2 / C 2 ) × (6 . 4 × 10- 6 C)(- 3 . 4 × 10- 6 C)- 4 . 3 × 10- 2 J = 4 . 54936 m . 004 10.0 points A charge moves a distance of 2 . 3 cm in the direction of a uniform electric field having a magnitude of 222 N / C. The electrical potential energy of the charge decreases by 122 . 642 × 10- 19 J as it moves. Find the magnitude of the charge on the moving particle. (Hint: The electrical poten- tial energy depends on the distance moved in the direction of the field.) Correct answer: 2 . 40192 × 10- 18 C. Explanation: Let : Δ d = 2 . 3 cm , E = 222 N / C and Δ U electric =- 122 . 642 × 10- 19 J . cherubin (mcc2645) – Homework 2 Electric Field and Potential – Catala – (12054) 2 Δ U electric =- q E Δ d q =- Δ U electric E Δ d =-- 122 . 642 × 10- 19 J (222 N / C)(0 . 023 m) = 2 . 40192 × 10- 18 C . 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

{[ snackBarMessage ]}

Page1 / 6

solution37 - cherubin(mcc2645 – Homework 2 Electric Field...

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

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