RECITATION-CH 24

RECITATION-CH 24 - RECITATION 4 5 The area of each face is A =(1.40 m)2(a Since the field has only a y-component there will be no flux through the

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RECITATION 4 5. The area of each face is A = (1.40 m) 2 . (a) Since the field has only a y -component, there will be no flux through the faces pointing in the x and z directions. Then ( 29 ( 29 ( 29 ( 29 ( 29 net =0 1.40 2 2 ˆ ˆ ˆ ˆ ( )j j ( )j j 0.00 N C 3.00 N C 1.40m 1.40m 8.23 N m C. j j y y E A E A = Φ = ⋅ - + = + = (b) The electric field can be re-written as ( 29 0 ˆ [ 3.00 N C m ] j E y E = + r r , where ( 29 ( 29 0 ˆ ˆ 4.00 N C i 6.00 N C j E = - + r is a constant field which does not contribute to the net flux through the cube (since any element of positive flux will be balanced by an element of negative flux on the opposite face). Thus net Φ is 2 8.23 N m C just as before. (c) The charge is given by ( 29 2 enc net 12 2 11 0 2 C 8.85 10 8.23 N m C 7.29 10 C N m q ε - - = Φ = × = × in each case. 19. For spherically symmetric charge distributions, Eq. 24-11 is ( 29 2 0 1 , 4 q r E r πε = r where ( 29 q r is the function describing how the enclosed charge varies with distance from
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This note was uploaded on 06/01/2008 for the course PHYS 232 taught by Professor Goksu during the Spring '08 term at Millersville.

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RECITATION-CH 24 - RECITATION 4 5 The area of each face is A =(1.40 m)2(a Since the field has only a y-component there will be no flux through the

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