Potential Difference Between Coaxial Hollow Conducting Cylinders

To determine the correct statement, let us analyze the electric field and the potential difference between two coaxial, hollow conducting cylinders.

Let the inner cylinder have radius $R_1$ and the outer cylinder have radius $R_2$, with $R_1 < R_2$.

The potential difference between the inner and outer cylinders is related to the electric field $\vec{E}(r)$ in the region between them ($R_1 < r < R_2$) by:
$V_1 - V_2 = \int_{R_1}^{R_2} E(r) \, dr$

By Gauss's Law, for a cylindrical Gaussian surface of radius $r$ (where $R_1 < r < R_2$) and length $L$, coaxial with the cylinders:
$\oint \vec{E} \cdot d\vec{A} = E(r)(2\pi r L) = \frac{q_{\text{enclosed}}}{\varepsilon_0}$
$E(r) = \frac{\lambda_{\text{enclosed}}}{2\pi \varepsilon_0 r}$
where $\lambda_{\text{enclosed}}$ is the net charge per unit length enclosed within the radius $r$.

1. Option (A): When a charge density is given to the inner cylinder, the charge per unit length on it is $\lambda_1 \neq 0$. In the region between the cylinders ($R_1 < r < R_2$), the enclosed charge per unit length is $\lambda_{\text{enclosed}} = \lambda_1$.
   Therefore, the electric field is:
   $E(r) = \frac{\lambda_1}{2\pi \varepsilon_0 r} \neq 0$
   The potential difference between the cylinders becomes:
   $V_1 - V_2 = \int_{R_1}^{R_2} \frac{\lambda_1}{2\pi \varepsilon_0 r} \, dr = \frac{\lambda_1}{2\pi \varepsilon_0} \ln\left(\frac{R_2}{R_1}\right) \neq 0$
   Thus, a potential difference appears between the two cylinders. Hence, (A) is correct.

2. Option (B): When a charge density is given only to the outer cylinder, all of this charge resides on the outer surface of the outer cylinder. In the region $R_1 < r < R_2$, the enclosed charge is zero ($\lambda_{\text{enclosed}} = 0$). Hence, $E(r) = 0$, which means $V_1 - V_2 = 0$. Thus, no potential difference appears. Hence, (B) is incorrect.

3. Option (C): When a uniform line charge with linear charge density $\lambda$ is placed along the common axis, it induces a charge $-\lambda$ on the inner surface of the inner cylinder and $+\lambda$ on the outer surface of the inner cylinder (since the inner cylinder is neutral). The net enclosed charge for a Gaussian surface of radius $r$ ($R_1 < r < R_2$) is:
   $\lambda_{\text{enclosed}} = \lambda - \lambda + \lambda = \lambda \neq 0$
   Consequently, an electric field exists between the cylinders, resulting in a non-zero potential difference. Hence, (C) is incorrect.

4. Option (D): If the same charge density is given to both cylinders, the inner cylinder still carries a non-zero charge, meaning $\lambda_{\text{enclosed}} \neq 0$ in the space between the cylinders. Thus, an electric field exists and a potential difference appears. Hence, (D) is incorrect.

Therefore, the correct option is (A).