Find out the surface charge density at the intersection of point $$x = 3$$ m plane and $$x$$-axis, in the region of uniform line charge of 8 nC m$$^{-1}$$ lying along the $$z$$-axis in free space.

A uniform line charge of linear charge density $$\lambda = 8$$ nC/m lies along the $$z$$-axis. The electric field at a perpendicular distance $$r$$ from an infinite line charge is $$E = \frac{\lambda}{2\pi\varepsilon_0 r}$$.

At the point where the $$x = 3$$ m plane intersects the $$x$$-axis, the perpendicular distance from the $$z$$-axis is $$r = 3$$ m.

The surface charge density at this point is defined as $$\sigma = \varepsilon_0 E$$. Substituting the expression for $$E$$: $$\sigma = \varepsilon_0 \times \frac{\lambda}{2\pi\varepsilon_0 r} = \frac{\lambda}{2\pi r}$$.

Computing: $$\sigma = \frac{8 \times 10^{-9}}{2\pi \times 3} = \frac{8 \times 10^{-9}}{6\pi} = \frac{8 \times 10^{-9}}{18.85} = 0.4244 \times 10^{-9}$$ C/m$$^2$$ $$= 0.424$$ nC/m$$^2$$.