solution3_pdf

solution3_pdf - taherisefat (mt23852) – hw 3 – Opyrchal...

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: taherisefat (mt23852) – hw 3 – Opyrchal – (121012) 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 10.0 points An electric field of magnitude 25000 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 3 . 5 m by 2 . 5 m is traveling along a road that is inclined 6 ◦ relative to the ground. Determine the electric flux through the bot- tom of the truck. Correct answer: 2 . 17552 × 10 5 N · m 2 / C. Explanation: Let : E = 25000 N / C , ℓ = 3 . 5 m , w = 2 . 5 m , and θ = 6 ◦ . By Gauss’ law, Φ = vector E · vector A The flux through the bottom of the car is Φ = E A cos θ = E ℓ w cos θ = (25000 N / C) (3 . 5 m) × (2 . 5 m) cos(6 ◦ ) = 2 . 17552 × 10 5 N · m 2 / C . 002 10.0 points A nonconducting wall carries a uniform charge density of 12 . 82 μ C / cm 2 . What is the electric field 11 cm in front of the wall? Correct answer: 7 . 23951 × 10 9 N / C. Explanation: Let : σ = 12 . 82 μ C / cm 2 and r = 11 cm . The electric field of an infinite plane of surface charge density σ is E = σ 2 ǫ = 12 . 82 μ C / cm 2 2 (8 . 85419 × 10 − 12 C 2 / N · m 2 ) × parenleftbigg 1 C 10 6 μ C parenrightbigg · parenleftbigg 100 cm 1 m parenrightbigg 2 = 7 . 23951 × 10 9 N / C . 003 (part 1 of 2) 10.0 points A thin spherical shell of radius 4 . 95 m has a total charge of 4 . 84 C distributed uniformly over its surface. Let : k e = 8 . 988 × 10 9 N · m 2 / C 2 . 4 . 9 5 m vector E + + + + + + + + + + + + + + + + + + + + Find the electric field vector E 11 m from the center of the shell (outside the shell). Correct answer: 3 . 5952 × 10 8 N / C. Explanation: Let : a = 4 . 95 m , Q = 4 . 84 C , and r = 11 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 field at a point outside the shell is equiva- lent to that of a point charge Q at the center....
View Full Document

This note was uploaded on 01/31/2011 for the course PHYS 111 taught by Professor Moro during the Spring '08 term at NJIT.

Page1 / 4

solution3_pdf - taherisefat (mt23852) – hw 3 – Opyrchal...

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