A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The upper face carries +95.0 nC/m² while the lower face carries -25.0 nC/ m². What is the magnitude of the electric field at a point within the sheet 2.00 cm below the upper face? (ε0 = 8.85 × 10^-12 C²/N ∙ m²) A) 0.00 N/C B) 3.95 × 10³ N/C C) 6.78 × 10³ N/C D) 7.91 × 10³ N/C E) 1.36 × 10⁴ N/C

Question

Quizplus · Accepted Answer

Since we are within the sheet, we must use Gauss's Law to find the electric field. We choose a Gaussian surface in the shape of a cylinder with a height of 10.0 cm and radius of 2.00 cm, centered on the point of interest within the sheet. The electric field is perpendicular to the top and bottom surfaces of the cylinder and has the same magnitude at every point on the curved surface of the cylinder (by symmetry).

The net charge enclosed by the Gaussian surface is the difference in charge densities between the upper and lower faces multiplied by the surface area of the cylinder:

Q enclosed = (ρ_upper - ρ_lower) × A_cylinder
= (9.5 × 10^-8 C/m^2 - (-2.5 × 10^-8 C/m^2)) × (2π × 0.02 m × 0.1 m)
= 6.78 × 10^-11 C

By Gauss's Law, the electric flux through the Gaussian surface is equal to the total enclosed charge divided by the permittivity of free space:
Φ = Q enclosed / ε0

The electric field is then given by the flux divided by the surface area of the cylinder:
E = Φ / A_cylinder

Plugging in the values, we obtain:
E = (6.78 × 10^-11 C) / (8.85 × 10^-12 C^2/N·m^2) / (2π × 0.02 m × 0.1 m)
E = 6.78 × 10^3 N/C
Therefore, the magnitude of the electric field at a point within the sheet 2.00 cm below the upper face is 6.78 × 10^3 N/C.

A Huge (Essentially Infinite) Horizontal Nonconducting Sheet 10