A

The electric field intensity at the surface of a conductor in electrostatics can be determined using Gauss's Law. According to this law, the electric field $E$ just outside the surface of a charged conductor can be expressed as:
$$E = \frac{q}{\epsilon_0}$$
where $q$ is the charge per unit area and $\epsilon_0$ is the permittivity of free space. This electric field is directed normal to the surface of the conductor. Thus, at the surface, the electric field intensity is indeed normal to the surface. Hence, the correct answer is option A.