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quiz3-solutions - ECE 220 — Quiz 3 Please write your name...

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Unformatted text preview: ECE 220 — Quiz 3 (03/01/04) Please write your name at the top right corner before you start the exam Remember to write the units of the various quantifies that you calculate Use the back side of each page if you need extra space. All formulas and units that you need are given on the last page 1. (20 points) A microelectronic capacitor consists of a 0.1 pm thick, square film of SiO2 (er = 4) dielectric sandwiched between two conducting layers. The conducting layers are square strips with each side measuring 50 um. Calculate the capacitance of the capacitor. (assume 60 = 10''1 F/ m) a) Another capacitor of the same dimensions has a layer of Gallium Arsenide (er = 13) in between. For a voltage of SOmV (50 x 10'3V), how much additional charge does this capacitor store? (relative to the SiO2 capacitor) 1 b) What is the maximum charge we can store in the SiO2 capacitor, if the dielectric strength of SiO2 is 3 x 107 V/m (dielectric strength is the maximum electric field that is admissible in SiOz). (Hint: V = Ed) 34:04 um : axuo‘7m. Er=4 A: 50 mesoum ED —. :0‘” F/m I 12500 KID—l2 m1 - -1! 42, (2:0er Cl . £4 1 4Ler XZBOOXKD “1‘04sz 8'01) Ol IX 10‘ 7 a), Gain/mo Arseniole ~> 2,:I3. (7ale 6; : \3xro’llx2‘500xuflz 1 3 "2 Amen/Cola x (o F -1 H \x1o' For V: 50 X10"? (M) c:Gl/V ; Q:Cv, $352.:E” C‘ (Swag. ; C.V 1 lO'Qé50xzo'3). 01; :. C ‘V :Lgnxlorn (50"(0’3). 1+ Additional charge = @L'Glf" (L314) Io’Viéomo"3 : 2-25 X (6% some“? :6) u {V1006 Mmm VO\ koae’ me ’ QI/Max CI VMOkX x3 IOvI'Z ' m cln aurae’ b5; $WYe—Cq SiO; Maximu M W in WE, capo/mi ("CV 3x107 V/M. :7 zgxtd’xfl X10 :BV. 2. (10 points) a) In a dielectric slab shown below, E = kp‘c + k2 )7 . If the voltage difference between A and B is 5d (V) and the voltage is the same at B and C, find the values of k1 and k2. d b) Can we have a non-zero E field as well as a constant voltage as we move in any direction? (in - any general case, not restricted to the problem C Y above). Give your reasoning. I X. . A B 0‘ “#3 a : not: U36 “(8’ a on > VAB 50‘ 3 (5' m m x olM€J5t VAB : “S E AK A B t. - S K. O“ 3 y K IA A O 0 o k K I :— '5— c C o a k 9 [ k 1 I O ' VB = 2 -g E ;— L63 0 ° c {S B B When we, move, ‘m U69, cat/{vean 0L We, Mn'zero E) Elm; E «etc/lei Mes wovk amok voltage, otearwses, Lance max/6 againsk Wue E) l”ch we, hon/e, 1-0 0&0 u)ka & Voltage, increases, 9:3 in We akove PVobIe/m} be,wa B 4 c We: 5 “dd is zero. Olga/Q 3. (10 points) Finite differences a) In cylindrical coordinates, the equation given below can be used to find the flux density, D from the volume charge density. While simulating the fields, we’ve stored the p values at subsequent discrete radii in an array rho i.e. rho[i] is the charge density at the radius r[i]. Find an expression for D[i]. _1_a(rD) = r 3r p b) Rearrange the following finite difference equation and obtain the corresponding partial differential equation (similar to the one above). (D — flux density , A - Area , r - radius , rho - charge density, dr —- spacing between consecutive r[i]) A[i—1]D[i—l]+ rho [i]A[i]dr A[i] D[i]= 4. (10 points) A spherical charge distribution exists in free space in the region 0 < r < a and is given by p(r) = po(1 — rz/az) (C/m3). The electric field in the region outside the spherical charge is given by E : 2/00a3 (V/m) forr>a 15801‘2 Find the voltage difference between two points A (at radius r3) and B (at radius rb). (Assume A is further away from the origin than B'and ra , rb > a, i.e. both points are outside the charged region) List of units 1) Charge - Coulombs (C) 2) Electric Flux Density(D) & Surface Charge Density(ps) — Coulombs/meter2(C/mz) 3) Volume Charge Density (pv) — Coulombs/meter3 (C/m3) 4) Electric Field Intensity (E) - Newtons/Coulomb (N/C) or Volts/meter (V/m) 5) Voltage — Volts(V) 6) Polarization — Coulombs/meter2(C/m2) 7) Permittivity (8) — Farads/meter (F/m) 8) Capacitance (C) — Farads List of formulas and quantities 1) Surface area of a sphere of radius ‘R’(m) — 47: R2 (m2) 2) Perrnittivity of free space (air or vacuum, where a, = 1) 80 = 8.854 * 10'12 (F/m) 3) Capacitance C = Q / V 4) Capacitance of a parallel plate capacitor = eA/d 5) Gauss’s law 535 = Qm 6) Relation between E and D D = ersoE 7) Boundary conditions for electric fields at any interface Dnz — Dnl = ps (normal components are related by the surface charge density) Ea — E1 = 0 (tangential components are continuous) 8) Relationship between Electric field and Voltage Va =41?ch ...
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