number of electrons = coulombs * electrons/coulomb
but we know
electrons/coulomb = 1/1.6*10^-19coulombs
= 10/1.6*10^-18
= 6.25*10^18 electrons/coulomb
we know that 2 coulombs/second pass a point which is 12.5*10^18 electrons per second
how many seconds to go a meter?
1/v = 1/1*10^-3 = 1000 seconds/meter
so
12.5 *10^15 electrons in that meter.
Please help and do it step by step all parts of questions and use:
T = time(s)
I=current (amps)
Q=net charge on object (Coulombs)
N=no electrons (and write if added or removed from object and how)
E=elementary charge
Calculate the number of free electrons per metre length of wire if a current of 2.0A produces a drift velocity of 1.0 x 10^-3 ms-1.
As physics - bobpursley Tuesday, December 12, 2017 at 10:27am
current=driftevelocity*area*electrondensity
but number electrons per lenth=electron dinsity)*area
the equation i=anev
n=work out
i=2
a=???
v= 1.0 X 10^-3 ms-1
was what i thought at first but then 2 values were missing when only one should be then you said number "NUMBER electrons per lenth=electron dinsity)*area"
but i dont know the density of the electrons or the area pelase help
As Physics - cylinder today at 10:34am
is there a way to work this out?
As Physics - cylinder today at 10:45am
i cant find the numbers for the equation yo gave me
"NUMBER electrons per lenth=electron dinsity)*area"
As Physics - cylinder today at 11:10am
please help
Thank you Bob i get the 4x thing now, this is the only question i cant get now please help someone
5 answers
Calculate the number of free electrons per metre length of wire if a current of 2.0A produces a drift velocity of 1.0 x 10^-3 ms-1. <<
You are given the drift velocity = 1.0 x 10^-3 m/s.
So it would take 1/(1.0 x 10^-3 m/s) = 1000 seconds to "flush out" all the electrons currently in a one metre length of wire.
2.0 A = 2.0 coulombs/second
So a 1m length of wire contains (2.0 C/s * 1000 s) = 2000 C worth of electrons
Each electron contributes 1.602 x 10^-19 C of that, so your 1 m length of wire represents
2000 / (1.602 x 10^-19) ≈ 1.25 x 10^22 electrons/metre length
You are given the drift velocity = 1.0 x 10^-3 m/s.
So it would take 1/(1.0 x 10^-3 m/s) = 1000 seconds to "flush out" all the electrons currently in a one metre length of wire.
2.0 A = 2.0 coulombs/second
So a 1m length of wire contains (2.0 C/s * 1000 s) = 2000 C worth of electrons
Each electron contributes 1.602 x 10^-19 C of that, so your 1 m length of wire represents
2000 / (1.602 x 10^-19) ≈ 1.25 x 10^22 electrons/metre length
I think Prof Damon erred in his last step,
delectrons/meter=1000sec/m*12.5*10^18elet/sec=12.5e21 electrons/m, which is the same as my number 1.25e22 electrons/meter
delectrons/meter=1000sec/m*12.5*10^18elet/sec=12.5e21 electrons/m, which is the same as my number 1.25e22 electrons/meter
Yes, you are right. I should have multiplied electrons / second times seconds / meter
Thank you so much both of you really helpful!! :)
1 answer