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Unformatted text preview: Physics 217  Homework 9 Solutions Ruxandra Bondarescu April 29, 2005 Problem 10.1 One way to build the capacitor is to put three layer of tape on top of one another: the first layer of tape would be Aluminum tape, the second one polyethylene tape and the third one Aluminum tape again. The capacitance is C == 4 . 5 × 10 4 cm = 0 . 05 μ F (1) For a parallel plate capacitor C = Aepsilon1 4 πs (2) where s = 0 . 001in is the width of the polyethylene tape. The width of the Aluminum tape is w = 2in. So, the length of the tape is L = A 2in = 4 πsC epsilon1w = 4 π 45 × 10 3 cm × . 001in 2 . 3 × 2in ∼ 123cm (3) When we roll it up the volume remains the same Lw (0 . 001in + 0 . 0005in + 0 . 0005in) = π r 2 w (4) So, the radius is r = radicalbig 123 × . 002 × 2 . 54 /π ∼ . 45cm. Thus the cylinder will have a diameter of about 0 . 9cm and the amount of tape needed is 246cm of Aluminum and about 123cm of polyethy lene. Problem 10.3 The dipole moment vector is p = integraldisplay r prime ρdv prime (5) All charge distributions considered in this problem are electrically neutral. So, the magnitude of the dipole moment is independent of where we choose the origin. (a) I chose the origin at left cornerthe dipole moment is independent of where we choose the origin....
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 Spring '06
 LECLAIR, A
 Work, Electric charge, 1.5 cm, 103 cm, 104 cm, 0.9 cm, 0.45 cm

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