Find the equation of the equipotential surface for an infinite cylinder of radius $r_0$,carrying a charge of linear density $\lambda$.

(N/A) Using Gauss's law for a cylindrical Gaussian surface of radius $r$ and length $l$:
$\oint \overrightarrow{E} \cdot d\overrightarrow{S} = \frac{q_{enclosed}}{\epsilon_0} = \frac{\lambda l}{\epsilon_0}$
Since the electric field is radial,$E(2\pi rl) = \frac{\lambda l}{\epsilon_0}$,which gives $E = \frac{\lambda}{2\pi\epsilon_0 r}$.
The potential difference between a point at distance $r$ and the surface at $r_0$ is:
$V(r) - V(r_0) = -\int_{r_0}^{r} E dr = -\int_{r_0}^{r} \frac{\lambda}{2\pi\epsilon_0 r} dr$
$V(r) - V(r_0) = -\frac{\lambda}{2\pi\epsilon_0} \ln\left(\frac{r}{r_0}\right)$
For an equipotential surface,$V(r)$ is constant,which implies $r$ must be constant.
Thus,the equation of the equipotential surface is $r = \text{constant}$,which represents a coaxial cylinder.