329sp09hw5sol

329sp09hw5sol - ECE-329 Spring 2009 Homework 5 Solution...

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Unformatted text preview: ECE-329 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 left-hand side gives ( A ) = x y z x y z x- y x + y = (2 z ) = , 1 ECE-329 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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329sp09hw5sol - ECE-329 Spring 2009 Homework 5 Solution...

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