Q.Find the equation of the equipotentials for an infinite cylinder of radius $r_0$, carrying charge of linear density $\lambda$.

Concept: Electric Potential — For an infinite line charge, the potential varies logarithmically with radial distance.

Reasoning:

1. By Gauss’s law, the electric field outside the cylinder ($r \ge r_0$) is the same as that of an infinite line charge:

   $E(r) = \frac{\lambda}{2\pi\varepsilon_0 r}$, directed radially outward.

2. The potential difference between a reference point at $r = r_0$ and a point at distance $r$ is:

   $V(r) - V(r_0) = -\int_{r_0}^{r} E \, dr = -\frac{\lambda}{2\pi\varepsilon_0} \ln\left(\frac{r}{r_0}\right)$. …

The core idea: symmetry and potential from a line charge

An infinite cylinder with uniform linear charge density $\lambda$ produces the same electric field outside itself as an infinite line charge along its axis. This is a direct consequence of Gauss's law — the cylindrical symmetry means the field is radial and depends only on the distance $r$ from the axis.

For $r \ge r_0$, the field is:

The direction is radially outward (if $\lambda > 0$). Inside the cylinder ($r < r_0$), for a conductor the field is zero; for a uniformly charged insulator the field grows linearly with $r$. The problem likely means a conducting cylinder or a thin cylindrical shell, so we focus on $r \ge r_0$.

Step-by-step derivation

1. Recall the relation between potential and field. Electric potential difference between two points is the negative line integral of the electric field:

For a radial field, the simplest path is along a radial line, so $d\mathbf{l} = dr\,\hat{r}$ and $\mathbf{E} \cdot d\mathbf{l} = E(r)\,dr$.

2. Set up the integral from a reference point. Choose a reference radius $r_{\text{ref}}$ where the potential is $V_{\text{ref}}$. Then at any $r$:

The integral is straightforward:

So:

3. Choose a convenient reference. A natural choice is to set $V = V_0$ at the cylinder's surface $r = r_0$. Then:

Equivalently:

4. What defines an equipotential? …

Method: Finding the Equipotential Surfaces of a Symmetric Charge Distribution

Use this method whenever you're asked for the shape/equation of equipotential surfaces around a charge distribution with clear symmetry (spherical, cylindrical, or planar).

Steps

Step 1: Find the field using the distribution's symmetry (Gauss's law)

Identify the symmetry (here, cylindrical) and use Gauss's law with a Gaussian surface matching that symmetry to get $E(r)$ as a function of the single relevant coordinate. For an infinite charged cylinder viewed from outside, the field matches that of an infinite line charge:

Step 2: Integrate the field to get the potential, relative to a convenient reference

Choose a FINITE reference point (never infinity, if $V$ diverges there — as it does for an infinite line/cylinder). A natural choice is the distribution's own surface, $r=r_0$.

Step 3: Recognise which coordinate(s) $V$ depends on …