Question

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)

Answer 1

The electric field vector at the given point (-0.50 cm, 0.00 cm, 0.00 cm) is 8.85 N/C.

Step-by-step explanation:

To find the electric field vector at the given point, we can consider the contributions from both plates. The electric field due to plate A is directed towards the point, since it has a negative surface charge density. The electric field magnitude can be calculated using the formula:

E = ϵ0 * σ

where E is the electric field magnitude, ϵ0 is the permittivity of free space, and σ is the surface charge density. Substituting the given values, we get:

EA = (8.85 x 10-12 C2/N · m2) * (-1.00 x 10-6 C/m2) = -8.85 N/C

The electric field due to plate B is also directed towards the point, since it has a positive surface charge density. Using the same formula, we find:

EB = (8.85 x 10-12 C2/N · m2) * (2.00 x 10-6 C/m2) = 17.7 N/C

Since the electric field vectors from both plates have the same direction, we can add their magnitudes to find the resultant electric field:

E = EA + EB = -8.85 N/C + 17.7 N/C = 8.85 N/C

Answer 2

E ≈ 1.70 10⁵ N/C

Step-by-step explanation:

The electric field is a vector quantity, so we can calculate the field of each plate and then add them. To calculate the field of a plate let's use Gauss's law

Φ = ∫ E. dA =
`q_(int)` / ε₀

To apply this law we must create a Gaussian surface that takes advantage of the symmetry of the problem. The electric field lines on the surface are perpendicular, so the Gaussian surface that will be a cylinder with the base parallel to the plate.

On this surface the normal to the base (A) is parallel to the field lines whereby the scalar product is reduced to the ordinary product. The normal on the sides of the cylinder is perpendicular to the field, therefore, the product scale is zero.

∫I E dA =
`q_(int)` /ε₀

Let's look for the load under the cylinder, let's use the concept of load density

σ =
`q_(int)` / A

`q_(int)` = σ A

Let's write Gauss's law for this case

E A =
`q_(int)` /ε₀

E A = σ A / ε₀

E = σ / ε₀

As the field is emitted for each side of the plate the value to only one side is

E = G / 2ε₀

This expression is the same for each plate, now let's add the electric field at the requested point

R = (0.50, 0.00, 0.00) cm

We see that this point is on the X axis, between the plates that are at the points y = -1.0 cm and y = 1.0 cm, as the plates are very large the test point is between them

The negative plate has an incoming field and the positive plate has an outgoing field, the test load is always positive. The field due to the negative plate goes to the left, the field through the positive plate goes to the left at this point whereby two are added

E = E_ + E +

E = σ1 / 2ε₀ + σ2 / 2ε₀

E = 1 / 2o (σ1 + σ2)

Let's calculate the value

E = 1/2 8.85 10⁻¹² (1.00 10⁻⁶ + 2.00 10⁻⁶)

E = 3 10⁻⁶ / 17.7 10⁻¹²

E = 1,695 10⁵ N / C

E ≈ 1.70 10⁵ N/C