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hw4 - homework 04 LIN KEVIN Due 4:00 am Question 1 chap 26...

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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 air-filled parallel-plate 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 ǫ 0 = 8 . 85419 × 10 12 C 2 / N · m 2 . The capacitance is C = ǫ 0 A d A = C d ǫ 0 = (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 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 0 N d 2 θ R 2. C = 2 ǫ 0 R 2 (2 π θ ) d 3. C = ǫ 0 N R 2 π d 4. C = ǫ 0 R 2 ( π θ ) d 5. C = N R 2 ( π θ ) d 6. C = ǫ 0 N R 2 θ d 7. C = ǫ 0 (2 N ) R 2 θ d 8. C = ǫ 0 (2 N ) d θ R 2 9. C = ǫ 0 (2 N 1) R 2 ( π θ ) d correct 10. C = ǫ 0 N R 2 ( π 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 0 A d 2
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homework 04 – LIN, KEVIN – Due: Feb 10 2008, 4:00 am 2 = ǫ 0 π R 2 2 d 2 = ǫ 0 π R 2 d . So the total capacitance would be C = (2 N 1) ǫ 0 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 0 (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 π rL 2. bardbl vector E bardbl = Q 2 π ǫ 0 r L correct 3. bardbl vector E bardbl = Q 2 2 π ǫ 0 r L 4. bardbl vector E bardbl = Q 4 π r 2 5. bardbl vector E bardbl = Q 2 π ǫ 0 r 2 6. None of these 7. bardbl vector E bardbl = Q 2 4 π ǫ 0 r 2 8. bardbl vector E bardbl = Q 2 2 π rL 9. bardbl vector E bardbl = Q 2 4 π 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 . Thus, Φ = contintegraldisplay vector E · d vector A = Q ǫ 0 E 2 π r L = Q ǫ 0 bardbl vector E bardbl = Q 2 π ǫ 0 r L .
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