This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: huang (jh45932) – Gauss’s Law – hester – (APCE&M) 1 This printout should have 35 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. Assignment is due Sunday, 1/17, 9pm Cali fornia time. 001 10.0 points A cubic box of side a , oriented as shown, con tains an unknown charge. The vertically di rected electric field has a uniform magnitude E at the top surface and 2 E at the bottom surface. a E 2 E How much charge Q is inside the box? 1. Q encl = ǫ E a 2 2. Q encl = 0 3. Q encl = 2 ǫ E a 2 4. Q encl = 3 ǫ E a 2 5. Q encl = 1 2 ǫ E a 2 6. insufficient information 002 (part 1 of 2) 10.0 points A closed surface with dimensions a = b = . 562 m and c = 1 . 0678 m is located as in the figure. The electric field throughout the region is nonuniform and given by vector E = ( α + β x 2 )ˆ ı where x is in meters, α = 2 N / C, and β = 4 N / (C m 2 ). E y x z a c b a What is the magnitude of the net charge enclosed by the surface? Answer in units of C. 003 (part 2 of 2) 10.0 points What is the sign of the charge enclosed in the surface? 1. Cannot be determined 2. positive 3. negative 004 (part 1 of 2) 10.0 points Given : V sphere = 4 π R 3 3 , and A sphere = 4 π R 2 . Consider a sphere, which is an insula tor, where charge is uniformly distributed throughout. Consider a spherical Gaussian surface with radius R 2 , which is concentric to the sphere with a radius R . R R 2 p Q is the total charge inside the sphere. The total amount of flux flowing through the Gaussian surface is given by huang (jh45932) – Gauss’s Law – hester – (APCE&M) 2 1. Φ = 4 Q ǫ . 2. Φ = Q 2 ǫ . 3. Φ = 2 Q ǫ . 4. Φ = Q ǫ . 5. Φ = Q 8 ǫ . 6. Φ = Q 4 ǫ . 005 (part 2 of 2) 10.0 points The magnitude of the electric field bardbl vector E bardbl at R 2 is given by 1. bardbl vector E bardbl = k Q R 2 . 2. bardbl vector E bardbl = k Q 2 R 2 . 3. bardbl vector E bardbl = 2 k Q R 2 . 4. bardbl vector E bardbl = k Q 2 R 2 . 5. bardbl vector E bardbl = 2 k Q 2 R 2 . 6. bardbl vector E bardbl = k Q 2 2 R 2 . 006 10.0 points Consider an infinitely long line of charge hav ing uniform charge per unit length 2 μ C / m. Determine the total electric flux through a closed right circular cylinder of length 6 . 9 m and radius 97 m that is parallel to the line charge, if the distance between the axis of the cylinder and the line of charge is 1 . 7 m. Answer in units of N · m 2 / C. 007 10.0 points Two large, parallel, insulating plates are charged uniformly throughout the plates. Each plate has the same amount of charge per unit area, + σ . The permittivity of free space ǫ = 1 4 π k e ....
View
Full
Document
This note was uploaded on 01/31/2010 for the course PHYSICS 103 taught by Professor Mander during the Spring '10 term at Westminster UT.
 Spring '10
 mander
 Magnetism

Click to edit the document details