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

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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
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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 , ....
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