Consider a Gaussian cylindrical dotted surface, $$S$$ at a distance $$r$$ from the centre of the cylinder of radius $$r_{0}$$ of infinite length.
The electric field lines are radial and perpendicular to the surface.
Let electric field intensity on Gaussian surface at $$\mathrm{P}$$ is $$\mathrm{E}$$, and total charge $$\mathrm{q}$$ on cylinder will be $$q=\lambda l$$
So, by Gauss's law,
$$\oint
\limits _{S} E_{r} d s=\dfrac{\lambda l}{\varepsilon_{0}} \Rightarrow[E _r \cos
\theta]_{0}^{2\pi i}=\dfrac{\lambda l}{\varepsilon_{0}}$$
$$\mathrm{E}_r
2 \mathrm{2\pi rl} \cos 90^{\circ}=\dfrac{\lambda
l}{\varepsilon_{0}}\left[\angle \theta \text{ } \mathrm{is} \text { between }
\mathrm{E}_{\mathrm{r}} \text { and curved surface of dotted cylinder is }
90^{\circ} \mathrm{J}\right.$$
$$E_{r}=\dfrac{\lambda}{2
\pi r \varepsilon_{0}}$$
We know that electric field $$\mathrm{E}_{\mathrm{r}}$$ at distance $$\mathrm{r}$$ from
centre of cylinder $$E_r=\dfrac{-d V}{d r}$$
So potential difference d at distance $$\mathrm{r}_{0}$$ and $$\mathrm{r}$$ from the centre of cylinder,
$$\mathrm{d}
\mathrm{V}=-\mathrm{E}_{\mathrm{r}} \cdot \mathrm{d} \mathrm{r}\left[\because
E=\dfrac{-d V}{d r}\right]$$
$$V(r)-V\left(r_{0}\right)=-\int
E_{r} \cdot d r$$
$$=\int
\limits _{r_{0}}^{r} \dfrac{\lambda}{2 \pi \varepsilon_{0} r} d r=\dfrac{-\lambda}{2
\pi \varepsilon_{0}} \int \limits _{r_{0}}^{r} \dfrac{d r}{r}=\dfrac{-\lambda}{2
\pi \varepsilon_{0}}\left[\log _{e} r\right]_{0}^{r}$$
$$=\dfrac{-\lambda}{2
\pi \varepsilon_{0}}\left[\log _{e} r-\log _{e} r_{0}\right]=\dfrac{-\lambda}{2
\pi \varepsilon_{0}}\left[\log _{e} r\right]_{0}^{r}$$
$$\log _{e}
\dfrac{r}{r_{0}}=\dfrac{-2 \pi
\varepsilon_{0}}{\lambda}\left[V(r)-V\left(r_{0}\right)\right]$$
$$\dfrac{r}{r_{0}}=e^{\dfrac{-2
\pi \varepsilon _0}{\lambda} [V(r)-V\left(r_{0}\right) ]}$$
$$r=r_{0}
e^{\dfrac{-2 \pi \varepsilon_{0}}{\lambda}\left[V(r)-v\left(r_{0}\right)\right]}$$
So
equipotential surfaces are the coaxial curved surfaces of cylinders with given
cylinder of radius r related as above.