HW 4 - Kapoor(mk9499 – homework04 – Turner –(60230 1...

Info iconThis preview shows pages 1–2. 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: Kapoor (mk9499) – homework04 – Turner – (60230) 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 (part 1 of 2) 10.0 points A charge Q is distributed uniformly along the x axis from x 1 to x 2 . The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 Which of the following integrals is correct for the magnitude of the electric field at x on the x axis? Assume that x > x 2 > x 1 and k e = 1 4 πǫ . 1. None of these 2. integraldisplay x 2 x 1 k e Q ( x 2 − x 1 ) x 2 dx 3. integraldisplay x 2 x 1 k e Q ( x 2 − x ) x 2 dx 4. integraldisplay x 2 x 1 k e Q ( x 2 − x )( x − x ) 2 dx 5. integraldisplay x 2 x 1 k e Q ( x 2 − x 1 )( x − x ) 2 dx correct Explanation: For a continuous charge distribution, dE = k e dq r 2 . Recall that for a uniform linear charge dis- tribution, dq = λ dx = Q L dx = Q x 2 − x 1 dx . Furthermore, the point x is a distance ( x − x ) from a charge element dq . Hence integrating from x = x 1 to x = x 2 yields E = integraldisplay x 2 x 1 k e Q ( x 2 − x 1 )( x − x ) 2 dx . 002 (part 2 of 2) 10.0 points If x 1 = 0 m, x 2 = 5 . 59 m and the charge Q = 3 . 49 μ C, what is the magnitude E of the electric field at x = 8 . 2 m? Correct answer: 1465 . 59 N / C. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , x 1 = 0 m , x 2 = 5 . 59 m , x = 8 . 2 m , and Q = 3 . 49 μ C . E = integraldisplay x 2 x 1 k e Q ( x 2 − x 1 )( x − x ) 2 dx = k e Q ( x 2 − x 1 )( x − x ) vextendsingle vextendsingle vextendsingle vextendsingle x 2 x 1 = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) (5 . 59 m − 0 m) × (3 . 49 × 10 − 6 C) × bracketleftbigg 1 8 . 2 m − 5 . 59 m − 1 8 . 2 m − 0 m bracketrightbigg = 1465 . 59 N / C . 003 (part 1 of 4) 10.0 points Consider a disk of radius 3 . 1 cm, having a uniformly distributed charge of +5 . 7 μ C....
View Full Document

This note was uploaded on 06/28/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.

Page1 / 5

HW 4 - Kapoor(mk9499 – homework04 – Turner –(60230 1...

This preview shows document pages 1 - 2. Sign up to view the full document.

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