Sheets of charge: 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

To find the electric field at a point within the sheet, we need to use Gauss's law. We can imagine a Gaussian cylinder with a radius of 2.00 cm and a height of 10.0 cm (the thickness of the sheet) centered at the point of interest. The electric field will be perpendicular to the faces of the cylinder and have the same magnitude at all points on the curved surface of the cylinder, since the charge density is uniform.

By Gauss's law, the flux through the curved surface of the cylinder is equal to the charge enclosed divided by the permittivity of free space. The charge enclosed is the charge density times the volume of the cylinder:

enclosed charge = (charge density)(volume) = (95.0 × 10S1U1P-9S1S1P0 C/mS1U1P2S1S1P0)(π(0.02 m)S1U1P2S1S1P0)(0.10 m) = 3.77 × 10S1U1P-7S1S1P0 C

The flux through the curved surface is E(2π(0.02 m)(0.10 m)), where E is the magnitude of the electric field. Thus,

E(2π(0.02 m)(0.10 m)) = enclosed charge / εS1U1B10S1U1B0
E = enclosed charge / (2πεS1U1B10S1U1B0(0.02 m)(0.10 m)) = 6.78 × 10S1U1P13S1S1P0 N/C

Therefore, the answer is C.

Sheets of Charge: a Huge (Essentially Infinite) Horizontal Nonconducting Sheet