thought that there might be people like me got stuck....so here i come.
by the way this post is about Question #3 Down The Wire part 3
it says Find the total number of collisions ( Ne) that all free electrons in this extension cord undergo in one second.
did we talk about collision last time? ...I dont think so - -..
anyways...in order to do this question we must know that in the process of current flow..electrons are having collision EVERY second. with this.. it turns the question more sense than it sounds like initially. (collision in the current? wth.....)
sadly. this question is NOT in 4.1, which is what we were talking about last time in class.
this question is actually a current resistivity problem...(which is what we gonna talk about tomorrow...)
but how do we find T(i.e time) in I = n*q*Vd*A equation?...here we go.
1st we know that F=qE (back in the beginning of the electrict field...)
and also F=ma. then we can get ma=qE
since acceleration is just a simply Velocity per unit time... we know that a = V/T
then we can rewrite the equation to.... V = (qE/m)*T and this is the key equation of part 3..
since I = n*q*Vd*A...and we know that V = (qE/m)*T
we can then subsitude and get (I/A) = n*q*T*(qE/m), which equavilen to J = (n*T*q^2/m)E
does anyone wonder what's the p given in the question for?
here is where we use p, turn your book to Pg. 851. we can find a equation call p = E/J (25.5)
nice....now we can combine the two equation we have to generate our T(time) qeuation....
from J = (n*T*q^2/m)E, we can move J and E around and make (m/n*T*q^2) = E/J
and we know form p = E/J (25.5), we know p = (m/n*T*q^2), furthermore, we can move p and T and generate this T equation, T = (m/n*p*q^2), and now we are good to go.
since the question asked collisions per second, we can think about it is the total electrons collisions in the cord every second... therefore, the answer will be (n*V)/T. just plug number
in and you are good to go....
(end)
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