hw4sol - Flores, Michael Homework 4 Due: Feb 16 2004, 4:00...

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Unformatted text preview: Flores, Michael 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 = x 2 y 2 + z 2 ( x 2- ) + y 3 z , where = 2 V / m 4 , = 4 . 2 V / m 4 , = 3 . 9 m 2 , and = 4 V / m 4 . What is the y component of the electric field E y at (6 . 4 m,- 3 m, 7 . 1 m)? Correct answer:- 275 . 28 V / m. Explanation: Given : = 2 V / m 4 , = 4 . 2 V / m 4 , = 3 . 9 m 2 , = 4 V / m 4 , and ( x,y,z ) = (6 . 4 m ,- 3 m , 7 . 1 m) . E y =- V y =- x 2 (2 y ) + (3 y 2 ) z / =- (2 V / m 4 )(6 . 4 m) 2 [2(- 3 m)]- (4 V / m 4 ) 3(- 3 m) 2 / 7 . 1 m =- 275 . 28 V / m . 002 (part 1 of 3) 10 points Given a parallel plate capacitor system with plate charge Q ( Q &gt; 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 = 0 , q l = Q, V = Qd A . correct 2. q u = Q 2 , q l = Q 2 , V = Q dA . 3. q u = Q, q l = 0 , V = Q dA . 4. q u = Q,q l = 0 , V = Qd 2 A . 5. q u = Q 2 , q l = Q 2 , V = Qd A . 6. q u = 0 , q l = Q, V = Qd 2 A . 7. q u = Q 2 , q l = Q 2 , V = Qd 2 A . 8. q u = Q, q l = 0 , V = Qd A . 9. q u = Q 2 , q l = Q 2 , V = Q 2 dA . 10. q u = 0 , q l = Q, V = Q dA . Explanation: Above the first plate, we can use Gausss 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 = Q encl A (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 A Flores, Michael 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 A . 003 (part 2 of 3) 10 points Now insert a dielectric material with dielectric constant 3 . 2 which fills the gap. The plate charge is 5 . 47 C, the gap width 1 . 83 mm, and the plate area 0 . 44 m 2 . Find the electric field across the capacitor plates. Correct answer: 438769 N / C....
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This note was uploaded on 02/18/2010 for the course PHYS 303l taught by Professor Panab during the Spring '04 term at North Texas.

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hw4sol - Flores, Michael Homework 4 Due: Feb 16 2004, 4:00...

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