Positive SSL
    14.素養(4)
1 有一導線每分鐘通過3.75×1016個電子,則通過導線之電流為
A 6mA
B 5A
C 0.5A
D 0.1mA

3.75×1016/6.25×1018=0.6×10-2=6mC

I=Q/t=6m/60=0.1mA