Arrive at an expression for drift velocity.

Explanation

`F=-eE and a=-(F)/(m)=-(evec(E))/(m)`
`vecV_(d)=vecV_(0)+vec(a)tau=0+((-evec(E))/(m))tau`
Arriing `vecV_(d)=-(evecE)/(m) tau or |vecV_(d)|=(eEtau)/(m)`
Detailed Answer
If V is the potential difference applied across the ends of a conductor of length 1, the magnitude of electric field set up is `E=(V)/(l)`
The charge of electron is `-e`, each free electron in the conductor experience force `overline(F)=-e.overline(E)`
If m is the mass of an electron, the acceleration of each electron is `veca=(-e.vec(E))/(m)`
At any instant of time, the velocity acquired by electron having velocity`vecu_(1)` will be `vecv_(1)=vecu_(1)+vec(at)_(1),vecv_(2)=vecu_(2)+vec(at)_(2)`
`therfore` The average velocity of all the electrons in the conductor under the effect of external electric field is the drift velocity of the free electrons.
Thus `vecv_(d)=(vecv_(1)+vecv_(2)+vecv_(3) . . +vecv_(n))/(n)`
`vecv_(d)=(vecu_(1)+vecu_(2)+vecu_(3) . . .+vecu_(n))/(n)+(veca(t_(1)+t_(2)+t_(3) . . +t_(n)))/(n)`
`vecv_(d)=0+vec(a)t`
where `t=(T_(1)+T_(2)+T_(3) . . .T_(n))/(n)`
`therefore vecv_(d)=at,vecv_(d)=(-evecE)/(m)T`
`therefore V_(d)=(eET)/(m)`