## College Physics (4th Edition)

$I = 8.0\times 10^{-3}~A$
Since the positive charges and negative charges move in opposite directions, we can add the magnitudes of the flow of charges to find the total current. $I = (1.6\times 10^{-19}~C)(3.8\times 10^{16}+1.2\times 10^{16})~s^{-1}$ $I = 8.0\times 10^{-3}~A$