Question

An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2.4 μC/m^2. A thick, infinite conducting slab, also oriented perpendicular to the x-axis occupiees the region between a = 2.9 cm and b = 4 cm. The conducting slab has a net charge per unit area of σ2 = 66 μC/m^2. What is Ex(P), the value of the x-component of the electric field at point P, located a distance 6.9 cm from the infinite sheet of charge?

Answer 1

The x-component of the electric field at point P is determined by the sheet of charge alone, calculated using Gauss's law, and is constant in magnitude and direction in space.

Step-by-step explanation:

To calculate the x-component of the electric field at point P, located 6.9 cm from the infinite sheet of charge, we must consider the contributions of both the sheet of charge and the conducting slab. Using Gauss's law, the electric field due to the infinite sheet of charge at any point in space is given by E = σ_1 / (2ε_0), where σ_1 is the surface charge density of the sheet and ε_0 is the permittivity of free space. Since the conducting slab is neutral overall, it does not contribute to the electric field outside the slab. Therefore, only the sheet of charge affects the electric field at point P.

The electric field due to the sheet of charge is constant in magnitude and direction at all points in space. Thus, the value of the electric field at point P, Ex(P), is given by E = σ_1 / (2ε_0) = -2.4 μC/m^2 / (2ε_0), where the negative sign indicates the direction of the field is towards the negative x-axis (since the sheet has a negative charge).