# How is electric field strength related to potential difference?

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?

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