Given:

Radius of infinite cylinder = r_0.

Linear charge density of infinite cylinder = λ.

An equipotential surface is a surface on which all points have the same electric potential. For an infinite cylinder of radius r₀, consider an imaginary cylinder of radius r concentric with the cylinder of radius r₀ as shown in the figure. Let the cylinder be charged with a positive charge of given linear density λ such that the electric field lines are radially outward.

We know that electric field and potential difference between two points is given by the integral relation:

The electric field due to the infinite cylinder of radius r₀ at a distance r from its center can be found by applying Gauss’s law with the imaginary cylinder of radius r. Gauss’s law states that the net electric flux enclosed through a closed surface is equal to the charge enclosed within the surface divided by the permittivity i.e.

Therefore, for this case the closed surface is the cylinder r₀ and the enlosed charge is the charge on the smaller cylinder of radius r , thus,

Where is the permittivity of free space, is the length of the imaginary cylinder, is the charge enclosed within the imaginary surface and is the area vector normal to the surface.

Now,

(dot product of vectors)

Since the electric field is radially outward from the cylinder and therefore normal to the surface of the cylinder, therefore the angle between the surface normal and electric field is for the curved surface, and at the ends is .

Now from equations (1) and (2)

Now for a given V

Now for a given r the potential difference is constant and does not depend on the position of the object on the surface of the cylinder r. Therefore, the equation of equipotential surface for the infinite cylinder of radius r₀is given by equation (4) and thus the equipotential surface is a cylinder of radius r.