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/C) î B) (-2.83 × 10⁵ N/C) ĵ C) (+1.19 × 10⁵ N/C) ĵ D) (+1.69 × 10⁵N/C) ĵ E) (-1.19 × 10⁵ N/C) ĵ

Question

Quizplus · Accepted Answer

The electric field due to plate A is:$$E_A = \frac{\sigma}{2\varepsilon_0} = \frac{-1.00 	imes 10^{-6} 	ext{ C/m}^2}{2(8.85 	imes 10^{-12} 	ext{ C}^2/	ext{N} \cdot 	ext{m}^2)} = -5.65 	imes 10^5 	ext{ N/C} \hat{j}$$The electric field due to plate B is:$$E_B = \frac{\sigma}{2\varepsilon_0} = \frac{2.00 	imes 10^{-6} 	ext{ C/m}^2}{2(8.85 	imes 10^{-12} 	ext{ C}^2/	ext{N} \cdot 	ext{m}^2)} = 1.13 	imes 10^5 	ext{ N/C} \hat{j}$$The net electric field is:$$E = E_A + E_B = (-5.65 	imes 10^5 	ext{ N/C} + 1.13 	imes 10^5 	ext{ N/C}) \hat{j} = -4.52 	imes 10^5 	ext{ N/C} \hat{j}$$The electric field vector at the point having x, y, z coordinates (-0.50 cm, 0.00 cm, 0.00 cm) is:$$E = -4.52 	imes 10^5 	ext{ N/C} \hat{j} = (-4.52 	imes 10^5 	ext{ N/C}) \hat{j} = -4.52 	imes 10^5 	ext{ N/C} \hat{j}$$

Two Very Large, Flat Plates Are Parallel to Each Other