P24_054

# P24_054 - 54 We use = E dA We note that the side length of...

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54. We use Φ = R ~ E · d ~ A . We note that the side length of the cube is 3 . 0m 1 . 0m=2 . 0m. (a) On the top face of the cube y =2 . 0mand d ~ A =( dA ) ˆ j .So ~ E =4 ˆ i 3((2 . 0) 2 +2) ˆ j=4 ˆ i 18 ˆ j .Thu s the ﬂux is Φ= Z top ~ E · d ~ A = Z top (4 ˆ i 18 ˆ j) · ( dA ) ˆ j = 18 Z top dA =( 18)(2 . 0) 2 N · m 2 / C= 72 N · m 2 / C . (b) On the bottom face of the cube y =0and d ~ A =( dA )( ˆ j). So ~ E =4 ˆ i 3(0 2 +2) ˆ j=4 ˆ i 6 ˆ j .Thu s , the ﬂux is Φ= Z bottom ~ E · d ~ A = Z bottom (4 ˆ i 6 ˆ j) · ( dA )( ˆ j) =6 Z bottom dA =6(2 . 0) 2 N · m 2 / C=+24N · m 2 / C . (c) On the left face of the cube d ~ A =( dA )( ˆ i). So Φ= Z left ~ E · d ~ A = Z left (4 ˆ i+ E y ˆ j) · ( dA )( ˆ i) = 4 Z bottom dA = 4(2 . 0) 2 N · m 2 / C= 16 N · m 2 / C
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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