Problem Setup

Let the inner conducting cylinder have radius $R_1$ and the outer conducting cylinder have radius $R_2$, with $R_2 > R_1$. The potential difference between the two cylinders, $V_{12} = V_{inner} - V_{outer}$, is given by the integral of the electric field $\vec{E}$ in the region between them ($R_1 < r < R_2$): $V_{12} = -\int_{R_2}^{R_1} \vec{E} \cdot d\vec{r} = \int_{R_1}^{R_2} E(r) dr$ A potential difference exists ($V_{12} \neq 0$) if and only if there is a non-zero electric field in the region between the cylinders.

To find the electric field $E(r)$ in the region $R_1 < r < R_2$, we use Gauss's Law for a cylindrical Gaussian surface of radius $r$ and length $L$: $\oint \vec{E} \cdot d\vec{A} = E(2\pi r L) = \frac{q_{enc}}{\epsilon_0}$ Here, $q_{enc}$ is the total charge enclosed within the Gaussian surface. Let $\lambda_{enc}$ be the linear charge density enclosed. Then $q_{enc} = \lambda_{enc} L$. The electric field is: $E = \frac{\lambda_{enc}}{2\pi\epsilon_0 r}$ So, a potential difference exists if and only if the net linear charge density enclosed by a Gaussian surface between the cylinders, $\lambda_{enc}$, is non-zero.

Let's analyze each option:

Step-by-step analysis of options

A: A potential difference appears between the two cylinders when a charge density is given to the inner cylinder

1. Let a linear charge density $\lambda$ be given to the inner cylinder. This charge will reside on its outer surface (at radius $R_1$).
2. For a Gaussian surface with radius $r$ such that $R_1 < r < R_2$, the enclosed charge per unit length is $\lambda_{enc} = \lambda$.
3. Since $\lambda \neq 0$, the electric field between the cylinders is non-zero: $E = \frac{\lambda}{2\pi\epsilon_0 r}$.
4. This non-zero electric field creates a potential difference: $V_{12} = \int_{R_1}^{R_2} \frac{\lambda}{2\pi\epsilon_0 r} dr = \frac{\lambda}{2\pi\epsilon_0} \ln\left(\frac{R_2}{R_1}\right)$ Since $R_2 > R_1$ and $\lambda \neq 0$, $V_{12} \neq 0$.
5. Therefore, this statement is correct.

B: A potential difference appears between the two cylinders when a charge density is given to the outer cylinder

1. Let a linear charge density $\lambda$ be given to the outer cylinder. The inner cylinder is neutral.
2. In electrostatic equilibrium, the electric field inside the material of the outer conductor must be zero. By Gauss's law, the net charge inside any Gaussian surface drawn within the conductor's material must be zero. This implies that any charge given to the outer cylinder resides on its outer surface. There is no charge on the inner surface of the outer cylinder because the inner cylinder is uncharged.
3. For a Gaussian surface with radius $r$ such that $R_1 < r < R_2$, the enclosed charge is the charge on the inner cylinder, which is zero. So, $\lambda_{enc} = 0$.
4. Therefore, the electric field between the cylinders is $E = 0$.
5. With no electric field between them, the potential difference is zero: $V_{12} = \int_{R_1}^{R_2} 0 \cdot dr = 0$.
6. The statement claims a potential difference appears, which is false. This statement is incorrect.

C: No potential difference appears between the two cylinders when a uniform line charge is kept along the axis of the cylinders

1. Let a line charge with linear density $\lambda$ be placed along the axis. Both cylinders are neutral.
2. For a Gaussian surface with radius $r$ such that $R_1 < r < R_2$, the enclosed charge consists of the line charge on the axis ($"\lambda"$) and the total charge of the inner cylinder. Since the inner cylinder is neutral, its total charge is zero.
3. The enclosed linear charge density is $\lambda_{enc} = \lambda + 0 = \lambda$.
4. Since $\lambda \neq 0$, the electric field between the cylinders is non-zero: $E = \frac{\lambda}{2\pi\epsilon_0 r}$.
5. This creates a non-zero potential difference between the cylinders.
6. The statement claims no potential difference appears, which is false. This statement is incorrect.

D: No potential difference appears between the two cylinders when same charge density is given to both the cylinders

1. Let the same linear charge density $\lambda$ be given to both cylinders.
2. We want to find the electric field in the region $R_1 < r < R_2$. For a Gaussian surface in this region, the enclosed charge is simply the charge on the inner cylinder.
3. The enclosed linear charge density is $\lambda_{enc} = \lambda$.
4. Since $\lambda \neq 0$, the electric field between the cylinders is non-zero: $E = \frac{\lambda}{2\pi\epsilon_0 r}$.
5. This creates a non-zero potential difference.
6. The statement claims no potential difference appears, which is false. This statement is incorrect.

Conclusion

Based on the analysis, only option A is a correct statement.