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

We can use the formula for electric field due to a charged plate:
$E = \frac{\sigma}{2\epsilon_0}$, where $\sigma$ is the surface charge density and $\epsilon_0$ is the permittivity of free space.

Since plate A has a negative charge density, the electric field due to plate A at the given point is directed towards the positive y-axis. Similarly, since plate B has a positive charge density, the electric field due to plate B at the given point is also directed towards the positive y-axis.

Thus, the total electric field at the given point is the vector sum of the electric fields due to plates A and B. Since the two fields are in the same direction, we can simply add their magnitudes.

$E_{total} = \frac{\sigma_A}{2\epsilon_0} + \frac{\sigma_B}{2\epsilon_0}$

Substituting the values given in the problem, we get:

$E_{total} = \frac{-1.00\mu C/m^2}{2\epsilon_0} + \frac{2.00\mu C/m^2}{2\epsilon_0}$

$E_{total} = \frac{1.00\mu C/m^2}{\epsilon_0}$

$E_{total} = 1.13 	imes 10^{15} N/C$ (approximately)

Since the field is directed towards the positive y-axis, we can represent it as:

$\vec{E}_{total} = (0, 1.13 	imes 10^{15}, 0) N/C$

Hence, the correct answer is A.

Two Very Large, Flat Plates Are Parallel to Each Other