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# hw6 - vo(ytv59 homework 06 Turner(56705 This print-out...

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vo (ytv59) – 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 = 2 ǫ 0 E a 2 2. insufficient information 3. Q encl = 3 ǫ 0 E a 2 4. Q encl = 0 5. Q encl = 1 2 ǫ 0 E a 2 6. Q encl = ǫ 0 E a 2 correct 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 0 Φ E = ǫ 0 E a 2 . 002 10.0 points A closed surface with dimensions a = b = 0 . 554 m and c = 0 . 6648 m is located as in the figure. The electric field throughout the region is nonuniform and given by vector E = ( α + β x 2 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? Correct answer: 1 . 28109 × 10 - 11 C. Explanation: Let : a = b = 0 . 554 m , c = 0 . 6648 m , α = 2 N / C , and β = 4 N / (C m 2 ) . The electric field throughout the region is directed along the x -axis and the direction of d vector A is perpendicular to its surface. Therefore, vector E

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