04homework04

04homework04 - Husain Zeena – Homework 4 – Due 4:00 am – Inst Sonia Paban 1 This print-out should have 23 questions Multiple-choice questions

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Unformatted text preview: Husain, Zeena – Homework 4 – Due: Feb 16 2004, 4:00 am – Inst: Sonia Paban 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points Given: V = αx 2 y 2 + β z 2 ( x 2- γ ) + δ y 3 z , where α = 5 V / m 4 , β = 5 . 3 V / m 4 , γ = 7 . 3 m 2 , and δ = 5 . 5 V / m 4 . What is the y component of the electric field E y at (7 . 8 m,- 2 m, 2 . 5 m)? Correct answer: 1051 . 8 V / m. Explanation: Given : α = 5 V / m 4 , β = 5 . 3 V / m 4 , γ = 7 . 3 m 2 , δ = 5 . 5 V / m 4 , and ( x,y,z ) = (7 . 8 m ,- 2 m , 2 . 5 m) . E y =- ∂V ∂y =- £ αx 2 (2 y ) + δ (3 y 2 ) z / =- (5 V / m 4 )(7 . 8 m) 2 [2(- 2 m)]- (5 . 5 V / m 4 ) £ 3(- 2 m) 2 / 2 . 5 m = 1051 . 8 V / m . 002 (part 1 of 3) 10 points Given a parallel plate capacitor system with plate charge Q ( Q > 0) and cross section A of each plate. + Q d- Q q upper q lower Top plate Bottom plate Denote the charges on the upper and lower surfaces of the top plate by q upper and q lower , and the magnitude of the potential difference between the two plates by V . The three quantities q upper , q lower and V are 1. q u = Q 2 , q l = Q 2 , V = Q 2 ² dA . 2. q u = Q 2 , q l = Q 2 , V = Qd 2 ² A . 3. q u = Q 2 , q l = Q 2 , V = Q ² dA . 4. q u = 0 , q l = Q, V = Qd 2 ² A . 5. q u = Q,q l = 0 , V = Qd 2 ² A . 6. q u = Q, q l = 0 , V = Q ² dA . 7. q u = Q, q l = 0 , V = Qd ² A . 8. q u = Q 2 , q l = Q 2 , V = Qd ² A . 9. q u = 0 , q l = Q, V = Qd ² A . correct 10. q u = 0 , q l = Q, V = Q ² dA . Explanation: Above the first plate, we can use Gauss’s law and enclose both plates in a Gaussian surface. We find that the enclosed charge is + Q- Q = 0 so the electric field has to be zero. The electric field outside of a conductor is given by E ⊥ = Δ Q encl Δ A² (1) Since the field is zero above the top plate, the charge on the upper surface must be zero: q upper = 0 Since the net charge on the top plate is + Q , this charge must then all reside on the lower surface of the top plate: q lower = Q This means, from equation (1), that the elec- tric field inside the gap E gap is given by E gap = Q ² A Husain, Zeena – Homework 4 – Due: Feb 16 2004, 4:00 am – Inst: Sonia Paban 2 Since the E gap is constant, the potential dif- ference across the gap is V = E gap d = Qd ² A . 003 (part 2 of 3) 10 points Now insert a dielectric material with dielectric constant 3 . 87 which fills the gap. The plate charge is 4 . 3 μ C, the gap width 2 . 83 mm, and the plate area 0 . 25 m 2 . Find the electric field across the capacitor plates. Correct answer: 501960 N / C....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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04homework04 - Husain Zeena – Homework 4 – Due 4:00 am – Inst Sonia Paban 1 This print-out should have 23 questions Multiple-choice questions

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