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Unformatted text preview: Chapter 21 Gauss’s Law (Additional Questions) 1) Find the net flux through the cube given in figure if the electric field is given by (a) ˆ 3.00 j E y = r and (b) ˆ ˆ 4.00i (6.00 3.00 )j E y =  + + r . E is in newtons per coulomb, and y is in meters. (c) In each case, how much charge is enclosed by the cube? Solution: (a) Let A = (1.40 m) 2 . Then (b) The electric field can be rewritten as ˆ 3.00 j E y E = + r r , where ˆ ˆ 4.00i 6.00j E =  + r is a constant field which does not contribute to the net flux through the cube. Thus Φ is still 8 .23 N.m 2 /C. (c) The charge is given by in each case. 2) Consider a closed triangular box resting within a horizontal electric field of magnitude E = 7.80×10 4 N/C, as shown in the figure. Calculate the electric flux through (a) the vertical rectangular surface, (b) the slanted surface, and (c) the entire surface of the box. Solution: (c) The bottom and the two triangular sides all lie parallel to E , so Φ E = 0 for each of these....
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 Spring '09
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 Charge, Flux, Electric charge

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