HW 06 - saldana(avs387 – homework 06 – Turner –(56705...

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Unformatted text preview: saldana (avs387) – homework 06 – Turner – (56705) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 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 correct 2. Q encl = 0 3. insufficient information 4. Q encl = 2 ǫ E a 2 5. Q encl = 3 ǫ E a 2 6. Q encl = 1 2 ǫ E a 2 Explanation: Electric flux through a surface S is, by con- vention, positive for electric field lines going out of the surface S and negative for lines going in. Here the surface is a cube and no flux goes through the vertical sides. The top receives Φ top =- E a 2 (inward is negative) and the bottom Φ bottom = 2 E a 2 . The total electric flux is Φ E =- E a 2 + 2 E a 2 = E a 2 . Using Gauss’s Law, the charge inside the box is Q encl = ǫ Φ E = ǫ E a 2 . 002 10.0 points A closed surface with dimensions a = b = . 257 m and c = 0 . 257 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 = 6 N / C, and β = 2 N / (C m 2 ). E y x z a c b a What is the magnitude of the net charge enclosed by the surface? Correct answer: 2 . 31757 × 10- 13 C. Explanation: Let : a = b = 0 . 257 m , c = 0 . 257 m , α = 6 N / C , and β = 2 N / (C m 2 ) ....
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This note was uploaded on 10/26/2010 for the course PHYS 56705 taught by Professor Turner during the Spring '10 term at University of Texas.

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HW 06 - saldana(avs387 – homework 06 – Turner –(56705...

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