20092ee1_1_HW4_Solutions

20092ee1_1_HW4_Solutions - EE 1 Homework#4 Solutions(1(15...

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Unformatted text preview: EE 1 - Homework #4 Solutions (1) (15 Points) Given the potential field, V = 1000 xz x 2 +4 . a. Find ~ D at the surface z = 0 b. Show that the z = 0 surface is an equipotential surface. c. Assume that the z = 0 surface is a conductor and find the total charge on that portion of the conductor defined by 0 < x < 2,- 3 < y < 0. a) We must find the electric field intensity, ~ E , first and then relate it to the electric flux density, ~ D . The electric field intensity can be found by: ~ E =-∇ V ~ E =- ˆ x ∂V ∂x- ˆ y ∂V ∂y- ˆ z ∂V ∂z =- ˆ x 1000 z ( x 2 + 4)- 2000 zx 2 ( x 2 + 4) 2- ˆ z 1000 x ( x 2 + 4) Then, knowing that ~ D = ~ E , ~ D = (- ˆ x 1000 z ( x 2 + 4)- 2000 zx 2 ( x 2 + 4) 2- ˆ z 1000 x ( x 2 + 4) ) ~ D ( x, 0) =- ˆ z 1000 x x 2 + 4 b) Looking at the given equation for V, the potential on the plane z = 0 is always equal to zero. In addition, the electric field from part (a) is normal to the plane z=0. The surface is equipotential. c) The total charge for the described region can be found by Gauss’ Law: R S ~ D · d ~ S = Q . Q = Z S ~ D · d ~ S = Z- 3 Z 2- ˆ z 1000 x x 2 + 4 · ˆ zdxdy =- 500 Z- 3 Z 2 2 x x 2 + 4 dxdy =- 500 || y ln( x 2 + 4) | 2 |- 3 =- 500...
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20092ee1_1_HW4_Solutions - EE 1 Homework#4 Solutions(1(15...

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