Final answer:
The electric field at z < -5 for two infinite uniform sheets of charge, both at z = ±5, is found by using the principle of superposition. Given their symmetrical placement and like charge densities, the fields they produce in that region add together, doubling the magnitude of the electric field compared to a single sheet.
Step-by-step explanation:
The question involves finding the electric field (E) at a position where z < -5, given two infinite uniform sheets of charge with density ρ_s located at z = ±5. To determine this, we'll use the principle of superposition and the fact that an infinite sheet of charge with surface charge density ρ_s produces an electric field with magnitude E = ρ_s / (2ε0) that is directed perpendicularly away from the sheet on both sides.
For z < -5, we are on the side of the sheet located at z = -5 which is opposite to the other sheet located at z = +5. The electric field due to each sheet will be directed away from the respective sheets. Hence, for the sheet at z = -5, the field points in the negative z-direction, and for the sheet at z = +5, it also points in the negative z-direction. Adding these fields, we get:
E (z < -5) = 2 * (ρ_s / (2ε0))
This is because the contributions from both sheets add up as they are in the same direction. So, the total electric field at z < -5 will be equal to the magnitude of the field due to a single sheet, ρ_s / (2ε0), multiplied by 2.
To answer this question fully, one must understand both the concept of an electric field produced by an infinite plane of charge and the principle of superposition. This scenario exemplifies the simplicity that arises in electrostatics when dealing with symmetrical distributions of charge in an idealized context.