Suppose a charge Q is uniformly distributed over a large plastic plate, producing an electric field of 10 V m ⁻¹ at a point P close to the centre of the plate. If this plastic plate is replaced by a copper plate having identical geometrical dimensions and carrying the identical charge Q , the electric field at point P will be

Step-by-step solution

For a large non-conducting (plastic) plate with total charge Q and surface area A , the electric field at a point close to its center can be found using Gauss's law. Taking a cylindrical Gaussian surface piercing through the plate, the total flux is 2EA = Q ₀ E = Q 2A ₀ . Given E = 10 V m ⁻¹ . When the plastic plate is replaced by a conducting (copper) plate of the same dimensions and carrying the same total charge Q , the charge distributes equally on its two outer surfaces. The charge on each surface is Q 2 , and