Answer:
Electric field lines from a positive point charge would radially emanate and end on the nearby charged plate. The electric field intensity at the plate's surface can be derived using E = \(\sigma/\epsilon_0\) from Gauss's law, where \(\sigma\) is the surface charge density and \(\epsilon_0\) the permittivity of free space.
Step-by-step explanation:
The task involves the electric fields generated by a positive point charge near a charged plate. To address part (i), we would draw a series of electric field lines emanating radially outwards from the point charge and terminating perpendicularly on the surface of the charged plate.
The density of the lines would represent the field's strength, with denser lines indicating stronger fields closer to the charge.
For part (ii), the electric field intensity (E) at the surface of a charged plate can be derived using Gauss's law, which relates the electric flux through a closed surface to the charge enclosed by that surface.
The electric field at the surface of a charged plate is given by E = \(\sigma/\epsilon_0\), where \(\sigma\) is the surface charge density of the plate and \(\epsilon_0\) is the permittivity of free space.
This formula assumes that the plate is large enough to be considered infinite, and the field is uniform and perpendicular to the surface of the plate.