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Unformatted text preview: homework 04 – KIM, JI – 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 0 . 7 mm, calculate the required surface area of each plate. Correct answer: 94 . 8704 km 2 (tolerance ± 1 %). Explanation: Let : C = 1 . 2 F , d = 0 . 7 mm = 0 . 0007 m , and ǫ = 8 . 85419 × 10 − 12 C 2 / N · m 2 . The capacitance is C = ǫ A d A = C d ǫ = (1 . 2 F) (0 . 0007 m) 8 . 85419 × 10 − 12 C 2 / N · m 2 × parenleftbigg 1 km 2 10 6 m 2 parenrightbigg = 94 . 8704 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 θ d Determine the capacitance as a function of the angle of rotation θ (in rad) , where θ = 0 corresponds to the maximum capacitance. 1. C = ǫ N R 2 ( π − θ ) d 2. C = ǫ R 2 ( π − θ ) d 3. C = ǫ N d 2 θ R 4. C = N R 2 ( π − θ ) d 5. C = ǫ N R 2 θ d 6. C = ǫ N R 2 π d 7. C = ǫ (2 N − 1) R 2 ( π − θ ) d correct 8. C = ǫ (2 N ) dθ R 2 9. C = 2 ǫ R 2 (2 π − θ ) d 10. C = ǫ (2 N ) R 2 θ d Explanation: Considering the situation of θ = 0, the two sets of semicircular plates in fact form 2 N − 1 capacitors connected in parallel, with each one having capacitance C = ǫ A d 2 homework 04 – KIM, JI – Due: Feb 10 2008, 4:00 am 2 = ǫ π R 2 2 d 2 = ǫ π R 2 d . So the total capacitance would be C = (2 N − 1) ǫ π 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 θ is C = ǫ (2 N − 1) R 2 ( π − θ ) 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 4 π r 2 2. bardbl vector E bardbl = Q 2 π ǫ r L correct 3. bardbl vector E bardbl = Q 2 2 π ǫ r L 4. bardbl vector E bardbl = Q 2 2 π rL 5. None of these 6. bardbl vector E bardbl = Q 2 4 π ǫ r 2 7. bardbl vector E bardbl = Q 4 π r 2 8. bardbl vector E bardbl = Q 2 π ǫ r 2 9. bardbl vector E bardbl = Q 2 π rL 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 05/05/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Capacitance, Work

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