Unformatted text preview: I = d q dt or dq = I dt . To fnd the total charge in coulombs that passes through the conductor, one must integrate the current over the time interval. q = Z t 1 dq = Z t 1 I dt = Z t 1 ( a t 2b t + c ) dt = • a 3 t 3 1b 2 t 2 1 + c t ‚ f f f f t 1 = a 3 t 3 1b 2 t 2 1 + c t 1 . 002 (part 2 oF 2) 0 points IF I is in A, and a = 4 C / s 3 , b = 3 C / s 2 , and c = 13 C / s, what quantity oF charge moves across the section oF the conductor From t 1 = 2 s to t 2 = 4 s? Correct answer: 82 . 6667 C. Explanation: Let : a = 4 C / s 3 , b = 3 C / s 2 , c = 13 C / s , t 1 = 2 s , and t 2 = 4 s . Since q = • a 3 t 3b 2 t 2 + c t ‚ f f f f t 2 t 1 , then q 2 = ( 4 C / s 3 ) 3 (4 s) 3( 3 C / s 2 ) 2 (4 s) 2 + (13 C / s)(4 s) = 113 . 333 C , and q 1 = ( 4 C / s 3 ) 3 (2 s) 3( 3 C / s 2 ) 2 (2 s) 2 + (13 C / s)(2 s) = 30 . 6667 C . so that q 21 = q 2q 1 = 82 . 6667 C . keywords:...
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 Spring '08
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