Phys 181b Problem Set 5 Solution

Phys 181b Problem Set 5 Solution - 45. During charging, the...

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45. During charging, the charge on the positive plate of the capacitor is given by qC e t =− ε τ 1 c h , where C is the capacitance, is applied emf, and = RC is the capacitive time constant. The equilibrium charge is q eq = C . We require q = 0.99 q eq = 0.99 C , so 099 1 .. e t Thus, e t = 001 . . Taking the natural logarithm of both sides, we obtain t / = – ln 0.01 = 4.61 or t = 4.61 .
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51. (a) The initial energy stored in a capacitor is given by 2 0 /2 , C UqC = where C is the capacitance and q 0 is the initial charge on one plate. Thus qC U C 0 63 22 1 0 1 0 0 5 0 1 0 1 0 == × = × −− .. . FJ C . c hb g (b) The charge as a function of time is given by qq e t = 0 τ , where is the capacitive time constant. The current is the derivative of the charge i dq dt q e t =− = 0 , and the initial current is i 0 = q 0 / . The time constant is RC () 66 1.0 10 F 1.0 10 1.0s ×× = . Thus i 0 33 10 10 10 = × . Cs A c h b g . (c) We substitute 0 t e = into V C = q / C to obtain 3 1.0 s 3 1.0 0 6 1.0 10 C 1.0 10 V , 1.0 10 F tt t C q Ve e e C −− − §· × = × ¨¸ × ©¹ where t is measured in seconds.
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This note was uploaded on 07/18/2008 for the course PHYS 181 taught by Professor Stephenirons during the Spring '08 term at Yale.

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Phys 181b Problem Set 5 Solution - 45. During charging, the...

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