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I was looking at the derivation of drift velocity of free electrons in a conductor in terms of relaxation time of electrons.

Here a metallic conductor XY of length $\ell$ is considered and having cross sectional area A. A potential difference $V$ is applied across the conductor XY. Due to this potential difference an electric field $E$ is produced. The magnitude of electric field strength is $E = V/\ell$.

I dont understand this part. How is electric field strength potential difference divided by length. Where am I lagging in my concepts?

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The assumption is that the electric flux lines are going to go through the conductor parallel to the conductor. Under those conditions, the electric field within the conductor is going to have a constant magnitude, and point parallel to the conductor. I assume you're familiar with

$$E=-\nabla \phi\ \ ,$$

where $\phi$ is the electric potential. We then integrate along a line through the conductor

$$V=\int_{0}^{\ell} \nabla \phi \ dx = \int_{0}^{\ell} – E \ dx = - E \ell\ \ ,$$

which is the same as $E=V/\ell$ if you consider a positive $E$ to mean a vector pointing in the $-x$ direction instead of the $+x$ direction.

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