Two very large,flat plates are parallel to each other. Plate A,located at y = 1.0 cm,is along the xz-plane and carries a uniform surface charge density -1.00 μC/m². Plate B is located at y = -1.0 cm and carries a uniform surface charge density +2.00 μC/m².What is the electric field vector at the point having x,y,z coordinates (-0.50 cm,0.00 cm,0.00 cm)? (ε₀ = 8.85 × 10^-12 C²/N • m²) A) (+1.13 × 10⁵ N/ B) (-2.83 × 10⁵ N/ C) (+1.19 × 10⁵ N/ D) (+1.69 × 10⁵ N/ E) (-1.19 × 10⁵ N/

Question

Quizplus · Accepted Answer

Using Gauss's law, we know that the electric field between two large parallel plates is given by: 
E = σ/ε0 
where σ is the surface charge density and ε0 is the vacuum permittivity. 
The electric field due to plate A is: 
EA = -σ/ε0 = (1.00 μC/m^2)/(8.85 × 10^-12 C^2/N·m^2) = -1.13 × 10^5 N/C 
The electric field due to plate B is: 
EB = σ/ε0 = (2.00 μC/m^2)/(8.85 × 10^-12 C^2/N·m^2) = 2.26 × 10^5 N/C 
The net electric field at the point (-0.50 cm,0.00 cm,0.00 cm) is the sum of EA and EB in the x-direction since the two plates are parallel to the xz-plane. 
ENet = EA + EB = (2.26 - 1.13) × 10^5 N/C = 1.13 × 10^5 N/C 
Therefore, the correct answer is A.

Two Very Large,flat Plates Are Parallel to Each Other