A long cylindrical volume contains a uniformly distributed charge of density $$\rho \text{ C m}^{-3}$$. The electric field inside the cylindrical volume at a distance $$x = \dfrac{2\epsilon_0}{\rho} \text{ m}$$ from its axis is ______ $$\text{V m}^{-1}$$.

We have a long cylindrical volume with uniform charge density $$\rho \text{ C m}^{-3}$$. We need to find the electric field at a distance $$x = \dfrac{2\epsilon_0}{\rho} \text{ m}$$ from the axis.

Consider a cylindrical Gaussian surface of radius $$x$$ and length $$l$$, coaxial with the charged cylinder.

By Gauss's law:

$$\oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\epsilon_0}$$

The enclosed charge is:

$$q_{enc} = \rho \times \pi x^2 l$$

The flux through the curved surface of the Gaussian cylinder is:

$$E \times 2\pi x l = \frac{\rho \pi x^2 l}{\epsilon_0}$$

$$E = \frac{\rho x}{2\epsilon_0}$$

$$E = \frac{\rho}{2\epsilon_0} \times \frac{2\epsilon_0}{\rho} = 1 \text{ V m}^{-1}$$

The electric field at the given distance is $$1 \text{ V m}^{-1}$$.