329sum11hw4sol

329sum11hw4sol - ECE 329 Homework 4 — Solution Due: June...

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Unformatted text preview: ECE 329 Homework 4 — Solution Due: June 28, 2011, 5PM 1. Two infinite perfectly conducting plates at z = 0 and z = z o are kept at potentials V = 0 and V = V p , respectively. The region between the plates is filled with two slabs of perfect dielectric materials having permittivities 1 for < z < d (region 1) and 2 for d < z < z o (region 2), as shown in the following figure. z z = z o z = d z = 0 2 1 V = V p V = 0 a) Using Laplace’s equation, we can find that the potential in the dielectrics can take the following form V ( z ) = a 1 z + b 1 for < z < d a 2 z + b 2 for d < z < z o . This implies that the associated electric field is simply \$ E ( z ) =-∇ V =- a 1 ˆ z for < z < d- a 2 ˆ z for d < z < z o . Since V = 0 at z = 0 , we get b 1 = 0 . In addition, given that V = V p at z = z o , we have a 2 z o + b 2 = V p . We also know that V must be continuous at the interface x = d , thus, we have a 2 d + b 2- a 1 d = 0 . Maxwell’s boundary conditions tell us that the normal component of the elec- tric flux density \$ D ( z ) must be continuous across the interface z = d. Because of this we can write that ˆ n · ( \$ D 1- \$ D 2 ) = 1 a 1- 2 a 2 = 0 . 1 Combining these three equations, we find a 1 = 2 V p z o 1 + d ( 2- 1 ) , a 2 = 1 V p z o 1 + d ( 2- 1 ) , b 2 = d ( 2- 1 ) V p z o 1 + d ( 2- 1 ) , and, in consequence, the electric potential in the two dielectrics is V ( z ) = 2 V p z o 1 + d ( 2- 1 ) z < z < d, 1 V p z o 1 + d ( 2- 1 ) z + d ( 2- 1 ) V p z o 1 + d ( 2- 1 ) d < z < z o . b) Applying \$ E =-∇ V , the electric field inside the dielectrics is found to be \$ E ( z ) =- 2 V p z o 1 + d ( 2- 1 ) ˆ z < z < d,- 1 V p z o 1 + d ( 2- 1 ) ˆ z d < z < z o . Given that z o = 2 d = 1 m , V p = 5 V , 1 = o , and 2 = 2 o , we can find that \$ E ( z ) =- 20 3 ˆ z V m < z < d,- 10 3 ˆ z V m d < z < z o . The surface charge density at z = z o is given by ρ S | z = z o = \$ D · ˆ n z = z o =- D z ( z o ) =- 2 E z ( z o ) = 20 3 o C m 2 . 2. A vacuum diode consists of a cathode (source of the electrons injected into the vacuum tube) in the x = 0 plane and an anode in the x = d plane, where the anode is held to a constant potential V a = 1 V relative to the cathode. The potential distribution between the plates is given by V ( x ) = V a (...
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This note was uploaded on 11/06/2011 for the course ECE 329 taught by Professor Kim during the Summer '08 term at University of Illinois, Urbana Champaign.

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329sum11hw4sol - ECE 329 Homework 4 — Solution Due: June...

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