P28_049 - 49. (a) The charge on the positive plate of the...

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49. (a) The charge on the positive plate of the capacitor is given by q = C E (1 e t/τ ) , where E is the emf of the battery, C is the capacitance, and τ is the time constant. The value of τ is τ = RC =(3 . 00 × 10 6 Ω)(1 . 00 × 10 6 F) = 3 . 00 s. At t =1 . 00 s, t/τ =(1 . 00 s) / (3 . 00 s) = 0 . 333 and the rate at which the charge is increasing is dq dt = C E τ e t/τ = (1 . 00 × 10 6 )(4 . 00 V) 3 . 00 s e 0 . 333 =9 . 55 × 10 7 C / s . (b) The energy stored in the capacitor is given by U C = q 2 2 C . and its rate of change is dU C dt = q C dq dt . Now q = C E (1 e t/τ )=(1 . 00 × 10 6 )(4 . 00 V)(1 e 0 . 333 )=1 . 13 × 10 6 C , so dU C dt = µ 1 . 13 × 10 6 C 1 . 00 × 10 6 F (9 . 55 × 10 7 C / s) = 1 . 08 × 10 6 W . (c) The rate at which energy is being dissipated in the resistor is given by
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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