Charge of uniform surface density (0.2 nC/m2) is distributed over the entire xy plane. Determine the magnitude of the electric field at any point having z = 2 m.
To determine the magnitude of the electric field at any point with z = 2 m due to a charge of uniform surface density distributed over the entire xy plane, we can use Gauss's Law. The electric field due to a uniformly charged infinite plane is constant and directed perpendicular to the plane.
Gauss's Law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (ε₀).
Let's consider a Gaussian surface in the form of a cylinder with its symmetry axis along the z-axis and a height (h) extending from z = -∞ to z = +∞. Since the charge is distributed indefinitely in the xy plane, the electric field will have the same magnitude and direction throughout the cylinder.
The charge enclosed by the Gaussian surface is obtained by multiplying the surface charge density (∂) by the area of the Gaussian surface (A), which is the product of the length (L) and width (W) of the cylinder:
Q = ∂ * A
Since the charge is distributed over the entire xy plane, the length (L) and width (W) of the cylinder are arbitrary.
Now, we can calculate the electric flux (Φ) through the Gaussian surface:
Φ = E * A
where E is the magnitude of the electric field and A is the surface area of the cylinder.
By Gauss's Law, Φ = Q / ε₀, so we can rewrite the equation as:
E * A = ∂ * A / ε₀
The surface area of the cylinder is given by:
A = 2πrh
where r is the radius of the cylinder and h is the height.
Substituting the value of A into the previous equation, we get:
E * 2πrh = ∂ * A / ε₀
Canceling out the surface area (A) and rearranging the equation, we obtain:
E = ∂ / (2πε₀r)
Since we are interested in finding the electric field at a point with z = 2 m, we need to calculate the value of r. The distance between the xy plane and the point of interest is the z-coordinate, so r = 2 m.
Finally, substituting the given surface charge density (∂ = 0.2 nC/m²), the permittivity of free space (ε₀ = 8.85 x 10⁻¹² C²/(N·m²)), and the value of r into the equation, we can calculate the magnitude of the electric field (E):
E = (0.2 nC/m²) / (2π(8.85 x 10⁻¹² C²/(N·m²))(2 m))
E ≈ 11.3 N/C
Therefore, the magnitude of the electric field at any point with z = 2 m is approximately 11.3 N/C.