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Physics 18 Online
OpenStudy (anonymous):

How far must the sea of electrons in the wire move to deliver -23.0 of charge to an electrode?

OpenStudy (anonymous):

that's 23nC

OpenStudy (anonymous):

1mm gold wire, n_e=5.9*10^28m^-3 i got 3.11*10^-12m but it's not right. what i did was... Ne=(23*10^-9C)/(1.6*10^-19C)=1.44*10^11 electrons Ne=n_eAd 1.44*10^11=5.9*10^28m^-3*pi*(.0005m)^2*d solved for d

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