Ch27-p069 - 69(a The charge on the positive plate of the capacitor is given by c h q = C 1 e t where is the emf of the battery C is the capacitance

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69. (a) The charge on the positive plate of the capacitor is given by qC e t =− ε τ 1 c h , where is the emf of the battery, C is the capacitance, and is the time constant. The value of is = RC = (3.00 × 10 6 )(1.00 × 10 –6 F) = 3.00 s. At t = 1.00 s, t / = (1.00 s)/(3.00 s) = 0.333 and the rate at which the charge is increasing is () ( ) 6 0.333 7 1.00 10 F 4.00V 9.55 10 C s. 3.00s t dq C ee dt −− × == = × (b) The energy stored in the capacitor is given by 2 , 2 C q U C = and its rate of change is dU dt q C dq dt C = . Now e e t =−=× = × − − 1 100 10 4 00 1 113 10 6 0 333 6 c h c h bg c h .. . . VC , so 6 76 6 1.13 10 C 9.55 10 C s 1.08 10 W. 1.00 10 F C dU qdq dt C dt ⎛⎞ × × = × ⎜⎟ × ⎝⎠ (c) The rate at which energy is being dissipated in the resistor is given by
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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