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Drift occurs when electric field is applied to a semiconductor and free electrons move or drift towards the positive terminal of the voltage source. We know from equation $$Drift Velocity=Mobility*Electric Field$$ that with increase in electric field the drift velocity increases.But why?With increase in electric field only the number of electrons from the negative terminal of the voltage source increases.Why does drift velocity increases with increase in number of electrons from the negative terminal? Thanks for your time.

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Statistically, electrons undergo random collisions with the crystal lattice of the conductor. This makes their velocity vectors just after one of these collisions purely random($v_i$). Now suppose we apply an electric field $\vec E$ across the conductor. Think of this field as an agent who introduces slight order in this chaos of randomly moving electrons. Due to this order, electrons have a tendency to move in a particular direction. Assume the mass of electron to be $m$ and charge to be $e$. Then the velocity $V_i$ of the ith electron between 2 consecutive collisions is given by :

$\vec V_i = \vec v_i + \frac{e.\vec E.t}{m}$ where t is the time elapsed after the first collision.

Averaging both sides over the average time beetween 2 successive collisions (relaxation time $\tau$)

$\lt \vec V_i \gt = \lt \vec v_i \gt + \frac{e\tau.\vec E}{m}$

The lhs is the drift velocity and the 2nd term is zero because just after collision, the direction of motion is random. Thus,

$\vec v_d = \frac{e.\vec E. \tau}{m}$

As the electric field increases, more order is introduced in the system and hence the drift velocity increases. The mathematical variation is given by the last eqn above.

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