{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Hw4_pdf - mcconnell(kam2342 – homework04 – Turner...

This preview shows pages 1–2. Sign up to view the full content.

mcconnell (kam2342) – 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 0 on the x axis? Assume that x 0 > x 2 > x 1 and k e = 1 4 πǫ 0 . 1. integraldisplay x 2 x 1 k e Q ( x 2 x 1 )( x 0 x ) 2 dx correct 2. integraldisplay x 2 x 1 k e Q ( x 2 x 0 )( x 0 x ) 2 dx 3. integraldisplay x 2 x 1 k e Q ( x 2 x 1 ) x 2 dx 4. None of these 5. integraldisplay x 2 x 1 k e Q ( x 2 x 0 ) 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 0 is a distance ( x 0 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 0 x ) 2 dx . 002 (part 2 of 2) 10.0 points If x 1 = 0 m, x 2 = 3 . 36 m and the charge Q = 2 . 44 μ C, what is the magnitude E of the electric field at x 0 = 10 . 7 m? Correct answer: 279 . 223 N / C. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , x 1 = 0 m , x 2 = 3 . 36 m , x 0 = 10 . 7 m , and Q = 2 . 44 μ C . E = integraldisplay x 2 x 1 k e Q ( x 2 x 1 )( x 0 x ) 2 dx = k e Q ( x 2 x 1 )( x 0 x ) vextendsingle vextendsingle vextendsingle vextendsingle x 2 x 1 = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) (3 . 36 m 0 m) × (2 . 44 × 10 6 C) × bracketleftbigg 1 10 . 7 m 3 . 36 m 1 10 . 7 m 0 m bracketrightbigg = 279 . 223 N / C . 003 (part 1 of 4) 10.0 points Consider a disk of radius 3 cm, having a uni- formly distributed charge of +5 μ C. The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Compute the magnitude of the electric field at a point on the axis and 2 . 6 mm from the center. Correct answer: 9 . 12393 × 10 7 N / C.

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.

{[ snackBarMessage ]}