This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: homework 04 – SMITH, TAYLOR – 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 0 . 4 F. The permittivity of free space is 8 . 85419 × 10 − 12 C 2 / N · m 2 . If the distance between the plates is 1 . 3 mm, calculate the required surface area of each plate. Correct answer: 58 . 7293 km 2 (tolerance ± 1 %). Explanation: Let : C = 0 . 4 F , d = 1 . 3 mm = 0 . 0013 m , and ǫ = 8 . 85419 × 10 − 12 C 2 / N · m 2 . The capacitance is C = ǫ A d A = C d ǫ = (0 . 4 F) (0 . 0013 m) 8 . 85419 × 10 − 12 C 2 / N · m 2 × parenleftbigg 1 km 2 10 6 m 2 parenrightbigg = 58 . 7293 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 d 2. C = ǫ N R 2 θ d 3. C = ǫ (2 N ) R 2 θ d 4. C = ǫ N R 2 ( π − θ ) d 5. C = N R 2 ( π − θ ) d 6. C = ǫ (2 N − 1) R 2 ( π − θ ) d correct 7. C = ǫ R 2 ( π − θ ) d 8. C = ǫ (2 N ) dθ R 2 9. C = ǫ N d 2 θ R 10. C = 2 ǫ R 2 (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 A d 2 homework 04 – SMITH, TAYLOR – 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 π ǫ r 2 2. bardbl vector E bardbl = Q 2 π ǫ r L correct 3. bardbl vector E bardbl = Q 2 4 π r 2 4. bardbl vector E bardbl = Q 2 2 π rL 5. None of these 6. bardbl vector E bardbl = Q 2 2 π ǫ r L 7. bardbl vector E bardbl = Q 2 4 π ǫ r 2 8. bardbl vector E bardbl = Q 4 π 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 .....
View
Full Document
 Spring '08
 Turner
 Physics, Capacitance, Work, Correct Answer, μF, Ub Ut Ub Ut Ub Ut Ub Ut Ub Ut Ub Ut Ub Ut Ub Ut Ub Ut Ub

Click to edit the document details