{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW3Sol - Platt David – Homework 3 – Due 4:00 am –...

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: Platt, David – Homework 3 – Due: Sep 12 2005, 4:00 am – Inst: Ken Shih 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Note for numeric problems: 1,As far as a magnitude value is concerned, it always means the absolute value. Be careful of the sign of your answer. 2,Please check student’s instruction concerning the significant figure and precision etc in case you do the right way but get the ”wrong” answer. 001 (part 1 of 1) 10 points A rod 9 . 7 cm long is uniformly charged and has a total charge of- 27 μ C. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Determine the magnitude of the electric field along the axis of the rod at a point 18 . 2357 cm from the center of the rod. Correct answer: 7 . 85274 × 10 6 N / C. Explanation: Let : ‘ = 9 . 7 cm , Q =- 27 μ C , r = 18 . 2357 cm , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . For a rod of length ‘ and linear charge density (charge per unit length) λ , the field at a dis- tance d from the end of the rod along the axis is E = k e Z d + ‘ d λ x 2 dx = k e- λ x fl fl fl fl d + ‘ d = k e λ‘ d ( ‘ + d ) , where dq = λdx . The linear charge density (if the total charge is Q ) is λ = Q ‘ so that E = k e Q ‘ ‘ d ( ‘ + d ) = k e Q d ( ‘ + d ) . In this problem, we have the following situa- tion (the distance r from the center is given): d l r r The distance d is d = r- ‘ 2 = 18 . 2357 cm- 9 . 7 cm 2 = 0 . 133857 m , and the magnitude of the electric field is E = k e Q d ( ‘ + d ) = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × |- 2 . 7 × 10- 5 C | (0 . 133857 m)(0 . 097 m + 0 . 133857 m) = 7 . 85274 × 10 6 N / C . Now, the direction must be toward the rod, since the charge distribution is negative (a positive test charge would be attracted). So the sign should be positive, according to the convention stated in the problem. 002 (part 1 of 1) 10 points A circular arc has a uniform linear charge density of 9 nC / m. The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . 1 3 6 ◦ 2 . 4 m x y Platt, David – Homework 3 – Due: Sep 12 2005, 4:00 am – Inst: Ken Shih 2 What is the magnitude of the electric field at the center of the circle along which the arc lies? Correct answer: 62 . 4983 N / C. Explanation: Let : λ = 9 nC / m = 9 × 10- 9 C / m , Δ θ = 136 ◦ , and r = 2 . 4 m . θ is defined as the angle in the counter- clockwise direction from the positive x axis as shown in the figure below....
View Full Document

{[ snackBarMessage ]}

Page1 / 6

HW3Sol - Platt David – Homework 3 – Due 4:00 am –...

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