20092ee1_1_HW2_Solutions

# 20092ee1_1_HW2_Solutions - EE 1 - Homework #2 Solutions...

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EE 1 - Homework #2 Solutions Spring 2009 (1) (1pt each) 1

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(2) (10 points) The gradient of a scalar function is given by: V = ˆ ze - 2 z If V = 20 V at z = 0, ﬁnd V ( z )? The gradient of V is essentially the derivative of V . Integrating the gradient will result in the general form of V ( z ): V 2 - V 1 = Z P 2 P 1 V · dl = Z P 2 P 1 ˆ ze - 2 z · xdx + ˆ ydy + ˆ zdz ) Integrating, and choosing P 1 = 0 and P 2 = z the general result is: V ( z ) - V (0) = Z z 0 e - 2 z dz = - 1 2 e - 2 z | z 0 V ( z ) = 1 - e - 2 z 2 + V (0) = 41 - e - 2 z 2 ( V ) 2
(3) (15 Points) For the vector ﬁeld: ~ E = ˆ r 5 e - r - ˆ z 6 z verify the divergence theorem for the cylindrical region enclosed by r = 2, z = 0, and z = 4. To verify the divergence theorem, we merely have to show that the surface integral of the vector ﬁeld is equal to the volume integral of the divergence of the same vector ﬁeld: Z S ~ E · d ~ S = Z V ∇ · ~ EdV First, we set-up the surface integral. For a cylinder, there are three surfaces to

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## This note was uploaded on 06/06/2009 for the course EE EE 1 taught by Professor Ozcan during the Spring '09 term at UCLA.

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20092ee1_1_HW2_Solutions - EE 1 - Homework #2 Solutions...

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