Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Derive the expression for drift velocity

Derive the expression for drift velocity

Answers (1)

When a potential difference is applied across a conductor an electric field \vec E is created.
Due to this electric field, the electrons accelerate colliding with each other and acquires an average constant velocity known as drift velocity.
The acceleration of electron
a= \frac{-e\vec{E}}{m}
where e is the charge of electron and m is the mass of the electron.
for an electron accelerating the final velocity just before the collision
V_{1}= u_{1}+a\, t_{1}
u1 is the initial thermal velocity of electron where t1 is the time interval between successive collision
Similarly for n electrons
V_{2}= u_{1}+a\, t_{2}
\vdots
V_{n}= u_{n}+a\, t_{n}
\frac{V_{1}+V_{2}+\cdots V_{n}}{n}= \frac{u_{1}+u_{2}+\cdots u_{n}}{n}+a\left ( \frac{t_{1}+t_{2}+\cdots t_{n}}{n} \right )
\frac{V_{1}+V_{2}+\cdots V_{n}}{n}= V_{d}\left ( driff\; velocity \right )
\frac{u_{1}+u_{2}+\cdots u_{n}}{n}= 0\left ( average\; thermal\; velocity \right )
\frac{t_{1}+t_{2}+\cdots t_{n}}{n}= \tau \left ( relaxation\; \; time \right )
\therefore V_{d}= 0+a\tau
  V_{d}=\frac{-Ee}{m}\tau

Posted by

Safeer PP

View full answer

Similar Questions

Latest Question