# HW 3 - Kapoor(mk9499 – HW 03 – Turner –(60230 1 This...

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: Kapoor (mk9499) – HW 03 – Turner – (60230) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 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. E A E B = 8 1 . 2. E A E B = 4 1 . correct 3. E A E B = 1 1 . 4. E A E B = 2 1 . 5. E A E B = 1 2 . Explanation: Let : r B = 2 r A . The electric field strength E ∝ 1 r 2 , so E A E B = 1 r 2 A 1 r 2 B = r 2 B r 2 A = (2 r ) 2 r 2 = 4 . 002 (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. integraldisplay x 2 x 1 k e Q ( x 2- x 1 ) x 2 dx 2. integraldisplay x 2 x 1 k e Q ( x 2- x 1 )( x- x ) 2 dx correct 3. integraldisplay x 2 x 1 k e Q ( x 2- x ) x 2 dx 4. None of these 5. integraldisplay x 2 x 1 k e Q ( x 2- x )( x- x ) 2 dx 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 . 003 (part 2 of 2) 10.0 points If x 1 = 0 m, x 2 = 5 . 73 m and the charge Q = 4 . 24 μ C, what is the magnitude E of the electric field at x = 8 . 5 m? Correct answer: 1618 . 48 N / C. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , x 1 = 0 m , x 2 = 5 . 73 m , x = 8 . 5 m , and Q = 4 . 24 μ C . Kapoor (mk9499) – HW 03 – Turner – (60230) 2 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 . 73 m- 0 m) × (4 . 24 × 10- 6 C) × bracketleftbigg 1 8 . 5 m- 5 . 73 m- 1 8 . 5 m- 0 m bracketrightbigg = 1618 . 48 N / C ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 6

HW 3 - Kapoor(mk9499 – HW 03 – Turner –(60230 1 This...

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

View Full Document
Ask a homework question - tutors are online