if a charged plane has a uniform surface charge density of −1.5×10−7C/m2 , what is the magnitude of the electric field just outside the plane’s surface?

Express your answer to two significant figures and include the appropriate units.

To find the magnitude of the electric field just outside the plane's surface, we can use Gauss's law. Gauss's law states that the electric flux through a closed surface is proportional to the charge enclosed by that surface.

In this case, let's consider a Gaussian surface just outside the plane's surface. Since the plane has a uniform surface charge density, the enclosed charge by this Gaussian surface will be the product of the surface charge density and the area of the Gaussian surface.

The electric flux through the Gaussian surface is given by the equation:

Φ = E * A

where Φ is the electric flux, E is the electric field, and A is the area of the Gaussian surface.

Using Gauss's law, we can relate the electric flux to the charge enclosed:

Φ = q / ε₀

where q is the charge enclosed and ε₀ is the vacuum permittivity.

Now, let's find the area of the Gaussian surface. Assume the Gaussian surface has an area of A. The enclosed charge q is then given by:

q = σ * A

where σ is the surface charge density.

Substituting this value of q into the equation Φ = q / ε₀, we get:

E * A = (σ * A) / ε₀

Simplifying further, we find:

E = σ / ε₀

Now, we can calculate the magnitude of the electric field just outside the plane's surface using the given surface charge density.

σ = -1.5×10⁻⁷ C/m² (given)

ε₀ = 8.85×10⁻¹² C²/N·m² (vacuum permittivity)

Substituting the values, we get:

E = (-1.5×10⁻⁷ C/m²) / (8.85×10⁻¹² C²/N·m²)

Calculating this, we find:

E ≈ 1.7×10⁵ N/C

Therefore, the magnitude of the electric field just outside the plane's surface is approximately 1.7×10⁵ N/C.