ass7 - SOLUTIONS, ASSIGNMENT 7 22.13. rr E (a) IDENTIFY and...

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SOLUTIONS, ASSIGNMENT 7 22.13. (a) I DENTIFY and S ET U P : It is rather difficult to calculate the flux directly from d Φ = E A r r ú since the magnitude of E r and its angle with d A r varies over the surface of the cube. A much easier approach is to use Gauss's law to calculate the total flux through the cube. Let the cube be the Gaussian surface. The charge enclosed is the point charge. E XECUTE : 6 6 2 encl 0 12 2 2 9.60 10 C / 1.084 10 N m /C. 8.854 10 C / N m E Q - - × Φ = = = × × P By symmetry the flux is the same through each of the six faces, so the flux through one face is ( 29 6 2 5 2 1 6 1.084 10 N m /C 1.81 10 N m /C. × = × (b) E VALUATE : In part (a) the size of the cube did not enter into the calculations. The flux through one face depends only on the amount of charge at the center of the cube. So the answer to (a) would not change if the size of the cube were changed. 22.16.
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This note was uploaded on 03/11/2010 for the course PHYSICS 1523 taught by Professor Dfs during the Fall '10 term at Dallas.

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ass7 - SOLUTIONS, ASSIGNMENT 7 22.13. rr E (a) IDENTIFY and...

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