f5ch24 - CHAPTER 24 GAUSS LAW 659 CHAPTER 24 Answer to...

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CHAPTER 24 GAUSS’ LAW 659 C HAPTER 24 Answer to Checkpoint Questions . ( a ) + EA ; ( b ) EA ; ( c ) ; ( d ) . ( a ) ; ( b ) ; ( c ) . ( a ) equal; ( b ) equal; ( c ) equal . + e ; ( b ) e . and tie, then , Answer to Questions . ( a ) N m /C; ( b ) . ( a ) a ; ( b ) r ; ( c ) rh . ( a ) all tie (zero); ( b ) all tie . ( a ) all four; ( b ) neither (they are equal) . + q= . ( a ) less (zero); ( b ) greater; ( c ) equal; ( d ) less (zero) . all tie . ( a ) S , S , S ; ( b ) all tie; ( c ) S , S , S ; ( d ) all tie . all tie . ( a ) impossible for E to be zero there; ( b ) q B = q , q c = any value; ( c ) any values giving q B + q C = q , , ; or , , . ( a ) , , ; ( b ) all tie (+ q ) . ( a ) all tie ( E = ); ( b ) all tie . ( a ) e= r , inward; ( b ) ; ( c ) e= r , inward
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35 o θ E A 660 CHAPTER 24 GAUSS’ LAW . ( a ) same ( E = ); ( b ) decrease to zero; ( c ) decrease (to zero); ( d ) same E ( a ) The mass ux is wd v = ( : m)( : m)( kg/m )( : m/s) = kg/s : ( b ) Since water ows only through an area wd , the ux through a larger area is still kg/s. ( c ) Now the mass ux is ( wd= ) v = ( kg/s) = = kg/s : ( d ) In this case the water ows through an area ( wd= ), so the ux is kg/s. ( e ) Now the ux is ( wd cos ) v = ( kg/s)(cos ) = kg/s : E The vector area A and the electric eld E are shown on the diagram to the right. The angle between them is = so the electric ux through the area is = E A = EA cos = ( N/C)( : m) cos = : N m /C. E Use = E A , where A = A j = ( : m) j . ( a ) = ( : N/C) i ( : m) j = : ( b ) = ( : N/C) j ( : m) j = : N m = C. ( c ) = [( : i + : )( N/C)] k ( : m) j = . ( d ) The total ux of a uniform eld through an enclosed surface is always zero. P Use = R E d A . Note that the side length of the cube is : m : m = : m. ( a ) On the top face of the cube y = : m and d A = ( dA ) j . So E = i ( : + ) j = i j . Thus the ux is = Z top E d A = Z top ( i j ) ( dA ) j = Z top dA = ( )( : ) N m = C = N m = C :
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CHAPTER 24 GAUSS’ LAW 661 ( b ) On the bottom face of the cube y = and d A = ( dA )( j ). So E = i ( + ) j = i j . Thus the ux is = Z bottom E d A = Z bottom ( i j ) ( dA )( j ) = Z bottom dA = ( : ) N m = C = + N m = C : ( c ) On the left face of the cube d A = ( dA )( i ). So = Z left E d A = Z left ( i + E y j ) ( dA )( i ) = Z bottom dA = ( : ) N m = C = N m = C : ( d ) On the back face of the cube d A = ( dA )( k ). But since E has no z component E d A = . Thus = . ( e ) We now have to add the ux through all the six faces. You can easily verify that the ux throught the front face is zero, while that through the right face is the opposite of that through the left one, or + N m = C. Thus the net ux through the cube is = ( + + + + ) N m = C = N m = C. E
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This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.

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f5ch24 - CHAPTER 24 GAUSS LAW 659 CHAPTER 24 Answer to...

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