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ch26_p71 - towards the negative terminal(low potential then...

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7 1. (a) The current is 4.2 × 10 18 e divided by 1 second. Using e = 1.60 × 10 19 C we obtain 0.67 A for the current. (b) Since the electric field points away from the positive terminal (high potential) and
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Unformatted text preview: towards the negative terminal (low potential), then the current density vector (by Eq. 26-11) must also point towards the negative terminal....
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