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
Unformatted text preview: Physics 2Ba quiz 2 solutions (Dated: January 21, 2009 12:59) Problem 1 The figure illustrates two surfaces placed in a uniform external electric field (indicated by the arrows). Surface 1 is a disk and surface 2 a hemisphere. Let 1 and 2 denote the electric flux through surfaces 1 and 2, respectively. By convention the unit normal to these surfaces points out of the region bounded by the surfaces. Choose the best answer: A) 1 =- 2 B) 1 > 2 C) 1 = 2 D) None of the above are correct The correct answer is A . There is no charge inside the surface, and so by Gausss law, the total flux between the two surfaces must be zero. So if tot = 1 + 2 = 0, must conclude A is true. Response C is false since the electric field is directed into surface 1 (negative flux), and the is directed out of surface 2 (positive flux). B is wrong for the same reason since 1 < 0 and 2 > 0, so 1 < 2 . Problem 2 A solid spherical conductor has a spherical, completely empty cavity in its interior. The cavity is not centered on the center of the conductor. If a positive charge is placed on the conductor, the electric field in the cavity A) is zero B) points generally toward the outer surface of the conductor C) points generally away from the outer surface of the conductor D) not enough information given to decide 2 The correct answer is A . When a charge is placed on a conductor it immediately goes to its surface. Now there are two surfaces, so which surfaces does it go to? The electric field inside the conductor is zero, and so if we take a gaussian surface just inside the conductor at the cavities interface, the total flux is zero since E = 0 everywhere inside the conductor. Gausss law tells us that the enclosed charge must be zero, and since there is no charge inside the cavity, there is no charge on the inside surface of the cavity either. As shown in Section 28.6 of your textbook, since there is no charge density on the cavities surface, the electric field must vanish inside the cavity.since there is no charge density on the cavities surface, the electric field must vanish inside the cavity....
View Full Document
This note was uploaded on 03/28/2009 for the course PHYS 2b taught by Professor Schuller during the Winter '08 term at UCSD.
- Winter '08