{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hkLQ2u1cH3bN_1210016556_jwk572

# hkLQ2u1cH3bN_1210016556_jwk572 - 2 outside 4 m 9 μ C 3 μ...

This preview shows page 1. Sign up to view the full content.

inclasstest 02 – KIM, JI – Due: Jan 28 2008, 1:00 pm 1 Question 1, chap -1, sect -1. part 1 of 1 1 points This is a question for showing your atten- dance. Please choose the correct answer be- low, or you will lose 25% of this weekly test. 1. This is the wrong answer. 2. This is the correct answer. correct Explanation: Question 2, chap 24, sect 5. part 1 of 1 3 points Consider a conducting spherical shell with inner radius 0 . 6 m and outer radius 1 . 2 m. There is a net charge 3 μ C on the shell. At its center, within the hollow cavity, there is a point charge 9 μ C. 0 . 6 m , Q 2 inside 1 . 2 m Q ′′ 2 outside , 0 . 4 m 9 μ
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2 outside , . 4 m 9 μ C 3 μ C S Determine the Fux through the spherical Gaussian surface S , which has a radius of . 4 m. The permitivity of free space is 8 . 8542 × 10 − 12 C 2 / N m 2 . Correct answer: 1 . 01647 × 10 6 . Explanation: Let : q 1 = 9 μ C = 9 × 10 − 6 C and ǫ = 8 . 8542 × 10 − 12 C 2 / N m 2 . The surface S encloses only the point charge q 1 , so by Gauss’ Law Φ S = c S v E · d v A = q encl ǫ = q 1 ǫ = 9 × 10 − 6 C 8 . 8542 × 10 − 12 C 2 / N m 2 = 1 . 01647 × 10 6 N m 2 / C ....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online