p33_005

p33_005 - t = 1 2 T + nT = 1 2 (2 n + 1) T = (2 n + 1) 2 f...

This preview shows page 1. Sign up to view the full content.

5. (a) We recall the fact that the period is the reciprocal of the frequency. It is helpful to refer also to Fig. 33-1. The values of t when plate A will again have maximum positive charge are multiples of the period: t A = nT = n f = n 2 . 00 × 10 3 Hz = n (5 . 00 µ s) , where n =1 , 2 , 3 , 4 , ··· . (b )W eno tetha ti ttak e s t = 1 2 T for the charge on the other plate to reach its maximum positive value for the ±rst time (compare steps
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: t = 1 2 T + nT = 1 2 (2 n + 1) T = (2 n + 1) 2 f = (2 n + 1) 2(2 10 3 Hz) = (2 n + 1)(2 . 50 s) , where n = 0 , 1 , 2 , 3 , 4 , . (c) At t = 1 4 T , the current and the magnetic eld in the inductor reach maximum values for the rst time (compare steps a and c in Fig. 33-1). Later this will repeat every half-period (compare steps c and g in Fig. 33-1). Therefore, t L = T 4 + nT 2 = (1 + 2 n ) T 4 = (2 n + 1)(1 . 25 s) , where n = 0 , 1 , 2 , 3 , 4 , ....
View Full Document

Ask a homework question - tutors are online