PHY303LHW4S - rosen (arr956) Homework 04 Chiu (58295) 1...

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Unformatted text preview: rosen (arr956) Homework 04 Chiu (58295) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A variable air capacitor used in tuning circuits is made of N semicircular plates each of radius R and positioned d from each other. A second identical set of plates that is free to rotate is enmeshed with the first set. R d Determine the capacitance as a function of the angle of rotation , where = 0 corre- sponds to the maximum capacitance. 1. C = (2 N 1) R 2 ( ) d correct 2. C = N R 2 d 3. C = N R 2 ( ) d 4. C = N d 2 R 5. C = (2 N ) R 2 d 6. C = R 2 ( ) d 7. C = (2 N ) d R 2 8. C = 2 R 2 (2 ) d 9. C = N R 2 d 10. C = N R 2 ( ) d Explanation: Considering the situation of = 0, the two sets of semicircular plates in fact form 2 N 1 capacitors connected in parallel, with each one having capacitance C = A d 2 = R 2 2 d 2 = R 2 d . So the total capacitance would be C = (2 N 1) R 2 d . Note: The common area of the two sets of plates varies linearly when one set is rotating, so the capacitance at angle is C = (2 N 1) R 2 ( ) d . 002 10.0 points Consider a parallel plate capacitor system with plate charge Q ( Q > 0) and cross sec- tion A of each plate. Top plate Bottom plate q u q + Q Q Plate area is A d E gap Denote the charges on the upper and lower surfaces of the top plate by q u and q , and the magnitude of the electric field within the gap by E gap . Find the quantities q u , q and E gap . 1. q u = Q, q = 2 Q, E gap = Q A 2. q u = 0 , q = Q, E gap = Q A correct rosen (arr956) Homework 04 Chiu (58295) 2 3. q u = Q 2 , q = Q 2 , E gap = Q A 4. q u = 0 , q = Q, E gap = Q 2 A 5. q u = Q, q = 0 , E gap = Q A 6. q u = Q 2 , q = 3 Q 2 , E gap = Q A 7. q u = Q 2 , q = 3 Q 2 , E gap = Q A 8. q u = Q 2 , q = Q 2 , E gap = Q 2 A 9. q u = Q, q = 2 Q, E gap = Q 2 A 10. q u = Q, q = 0 , E gap = Q 2 A 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 u = 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 = Q ....
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This note was uploaded on 04/13/2009 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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PHY303LHW4S - rosen (arr956) Homework 04 Chiu (58295) 1...

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