Homework03

Homework03 - johnson (rj6247) hw 3 Opyrchal (121014) This...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
johnson (rj6247) – hw 3 – Opyrchal – (121014) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points An electric feld oF magnitude 8000 N / C and directed upward perpendicular to the Earth’s surFace exists on a day when a thunderstorm is brewing. A truck that can be approximated as a rectangle 4 . 9 m by 4 . 6 m is traveling along a road that is inclined 11 relative to the ground. Determine the electric ±ux through the bot- tom oF the truck. Correct answer: 1 . 77007 × 10 5 N · m 2 / C. Explanation: Let : E = 8000 N / C , = 4 . 9 m , w = 4 . 6 m , and θ = 11 . By Gauss’ law, Φ = v E · v A The ±ux through the bottom oF the car is Φ = E A cos θ = E ℓ w cos θ = (8000 N / C) (4 . 9 m) × (4 . 6 m) cos(11 ) = 1 . 77007 × 10 5 N · m 2 / C . 002 10.0 points A nonconducting wall carries a uniForm charge density oF 13 . 14 μ C / cm 2 . What is the electric feld 4 . 3 cm in Front oF the wall? Correct answer: 7 . 42022 × 10 9 N / C. Explanation: Let : σ = 13 . 14 μ C / cm 2 and r = 4 . 3 cm . The electric feld oF an infnite plane oF surFace charge density σ is E = σ 2 ǫ 0 = 13 . 14 μ C / cm 2 2 (8 . 85419 × 10 12 C 2 / N · m 2 ) × p 1 C 10 6 μ C P · p 100 cm 1 m P 2 = 7 . 42022 × 10 9 N / C . 003 (part 1 oF 2) 10.0 points A thin spherical shell oF radius 4 . 93 m has a total charge oF 8 . 11 C distributed uniFormly over its surFace. Let : k e = 8 . 988 × 10 9 N · m 2 / C 2 . 4 . 93 m v E + + ²ind the electric feld v E 11 . 7 m From the center oF the shell (outside the shell). Correct answer: 5 . 32491 × 10 8 N / C. Explanation: Let : a = 4 . 93 m , Q = 8 . 11 C , and r = 11 . 7 m . IF we construct a spherical Gaussian surFace oF radius r > a , concentric with the shell, then the charge inside this surFace is Q . ThereFore the feld at a point outside the shell is equiva- lent to that oF a point charge Q at the center. ²or r > a , E 1 = k Q r 2 = (8 . 988 × 10 9 N · m 2 / C 2 )(8 . 11 C) (11 . 7 m) 2 = 5 . 32491 × 10 8 N / C .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
johnson (rj6247) – hw 3 – Opyrchal – (121014) 2 004 (part 2 of 2) 10.0 points Find the electric ±eld E 2 3 . 13 m from the center of the shell (inside the shell).
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/11/2010 for the course PHYS 121 taught by Professor Opyrchal during the Spring '08 term at NJIT.

Page1 / 4

Homework03 - johnson (rj6247) hw 3 Opyrchal (121014) This...

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

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