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Unformatted text preview: ECE329 Spring 2009 Homework 5 Solution February 17, 2009 1. Verifying vector calculus identities, ( ) = and ( A ) = 0 . a) The gradient of a scalar field is defined as = x x + y y + z z. Taking the curl, we obtain ( ) = x y z x y z x y z = y z z y x + z x x z y + x y y x z = . b) The curl of a vector field A = A x x + A y y + A z z is defined as A = x y z x y z A x A y A z = A z y A y z x + A x z A z x y + A y x A x y z. Taking the divergence, we obtain ( A ) = x A z y A y z + y A x z A z x + z A y x A x y = 2 A z xy 2 A y xz + 2 A x yz 2 A z yx + 2 A y zx 2 A x zy = 0 . 2. Let us verify if A = ( x y ) x + ( x + y ) y satisfies the following vector identity ( A ) = ( A ) 2 A . The lefthand side gives ( A ) = x y z x y z x y x + y = (2 z ) = , 1 ECE329 Spring 2009 and the right hand side gives ( A ) 2 A = ( (( x y ) x + ( x + y ) y )) 2 (( x y ) x + ( x + y ) y ) = (2) = , thus, the identity is verified. 3. The charge distribution associated with a point charge Q = 4 o C can be expressed as ( x,y,z ) = Q ( x ) ( y ) ( z ) where ( ) is the delta function. a) Calculating the closed volume integral V dV. i. If V contains the origin, we have V dV = V Q ( x ) ( y ) ( z ) dxdydz = Q. ii. If V excludes the origin, we simply have V dV = 0 . b) Above, we found that no matter what the shape or size of the volume V are, as long as V contains the charge distribution , the charge inside the volume is Q. Notice, that we would have found the same, if we were considered instead a point charge Q placed at the origin. Thus, we can argue that the charge density function ( x,y,z ) = Q ( x ) ( y ) ( z ) is a valid description of a point charge.point charge....
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 Spring '08
 FRANKE
 Electromagnet

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