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Unformatted text preview: homework 04 LIN, KEVIN Due: Feb 10 2008, 4:00 am 1 Question 1, chap 26, sect 1. part 1 of 1 10 points An airfilled parallelplate capacitor is to have a capacitance of 1 . 2 F. The permittivity of free space is 8 . 85419 10 12 C 2 / N m 2 . If the distance between the plates is 1 . 1 mm, calculate the required surface area of each plate. Correct answer: 149 . 082 km 2 (tolerance 1 %). Explanation: Let : C = 1 . 2 F , d = 1 . 1 mm = 0 . 0011 m , and = 8 . 85419 10 12 C 2 / N m 2 . The capacitance is C = A d A = C d = (1 . 2 F) (0 . 0011 m) 8 . 85419 10 12 C 2 / N m 2 parenleftbigg 1 km 2 10 6 m 2 parenrightbigg = 149 . 082 km 2 . Question 2, chap 26, sect 1. part 1 of 1 10 points A variable air capacitor used in tuning cir cuits is made of N very thin semicircular con ducting plates each of radius R and positioned d from each other. A second identical set of plates that is free to rotate is enmeshed mid way between the gap of the first set of plates. Note: One of the outside moving plates lies outside the first set of plates spaced d 2 . R d Determine the capacitance as a function of the angle of rotation (in rad) , where = 0 corresponds to the maximum capacitance. 1. C = N d 2 R 2. C = 2 R 2 (2 ) d 3. C = N R 2 d 4. C = R 2 ( ) d 5. C = N R 2 ( ) d 6. C = N R 2 d 7. C = (2 N ) R 2 d 8. C = (2 N ) d R 2 9. C = (2 N 1) R 2 ( ) d correct 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 homework 04 LIN, KEVIN Due: Feb 10 2008, 4:00 am 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 . Question 3, chap 26, sect 1. part 1 of 3 10 points A capacitor consists of two thin coax ial cylindrical conducting shells of length L . The inner shell has radius a and the outer shell has radius b . Assume the length is much greater than the radii of the cylin ders, L b . There is charge Q on the inner shell and charge + Q on the outer shell. Q + Q r a b The magnitude of the electric field at radius r , between the two shells, is given by 1. bardbl vector E bardbl = Q 2 rL 2. bardbl vector E bardbl = Q 2 r L correct 3. bardbl vector E bardbl = Q 2 2 r L 4. bardbl vector E bardbl = Q 4 r 2 5. bardbl vector E bardbl = Q 2 r 2 6. None of these 7. bardbl vector E bardbl = Q 2 4 r 2 8. bardbl vector E bardbl = Q 2 2 rL 9. bardbl vector E bardbl = Q 2 4 r 2 Explanation: Construct a Gaussian cylinder at radius r (for a < r < b ). The charge enclosed is the charge on the inner shell, Q ....
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This note was uploaded on 03/03/2011 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Capacitance, Work

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