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Unformatted text preview: this, we calculate the derivative of q with respect to time, evaluated for t = 0. We obtain Q sin , which we wish to be positive. Since sin(+46 . 9 ) is positive and sin( 46 . 9 ) is negative, the correct value for increasing charge is = 46 . 9 . (e) Now we want the derivative to be negative and sin to be positive. Thus, we take = +46 . 9 ....
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 Fall '08
 SPRUNGER
 Physics, Charge, Energy

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