05 - Solutions

# 05 - Solutions - Solutions for Homework 5 7. (a) Use Gausss...

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Solutions for Homework 5 ( a ) Use Gauss’s law to determine the electric flux. 6 5 2 encl E 12 2 2 1.0 10 C 1.1 10 N m C 8.85 10 C N m o Q ! " " " # \$ = = = " # % # % ( b ) Since there is no charge enclosed by surface A 2 , E 0 ! = . 9. The only contributions to the flux are from the faces perpendicular to the electric field. Over each of these two surfaces, the magnitude of the field is constant, so the flux is just E A r r & on each of these two surfaces. ( ) ( ) 2 2 encl E right left right left 0 Q E E " = + = # = \$ E A E A r r r r & & ll ( ) ( )( ) ( ) 2 2 12 2 2 7 encl right left 0 410N C 560N C 25m 8.85 10 C N m 8.3 10 C Q E E " " = " = " # \$ = " # l 27. ( a ) In the region 1 0 , r r < < a gaussian surface would enclose no charge. Thus, due to the spherical symmetry, we have the following. ( ) 2 encl 0 4 0 0 Q d E r E " = = = # = \$ E A r r & & ( b ) In the region 2 1 , r r r < < only the charge on the inner shell will be enclosed. ( ) 2 2 2 encl 1 1 1 1 2 0 0 0 4 4 Q r r d E r E r # = = = \$ = % E A r r & & ( c ) In the region 2 , r r < the charge on both shells will be enclosed. ( ) 2 2 2 2 2 encl 1 1 2 2 1 1 2 2 2 0 0 0 4 4 4 Q r r r r d E r E r + + = = = \$ = % E A r r & & ( d ) To make 0 E = for 2 , r r < we must have 2 2 1 1 2 2 0 . r r + = This implies that the shells are of opposite charge. ( e ) To make 0 E = for 2 1 , r r r < < we must have 1 0 . = Or, if a charge 2 1 1 4 Q r !" = # were placed at the center of the shells, that would also make 0. E = 46. Because the slab is very large, and we are considering only distances from the slab much less than its height or breadth, the symmetry of the slab results in the field being perpendicular to the slab, with a constant magnitude for a constant distance from the center. We assume that E 0 > and so the electric field points away from the center of the slab. ( a ) To determine the field inside the slab, choose a cylindrical gaussian surface, of length 2 x d < and cross-sectional area A . Place it so that it is centered in the slab. There will be no 1 2 d 1 2 d x x E r E r

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