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 find the magnitude of the electric field at any point with z = 2m due to a uniformly charged surface, we can use Gauss's Law. Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface.
In this case, we want to find the electric field at a point above the xy plane. Since the electric field due to a uniformly charged infinite plane is perpendicular to the plane, the electric field will only have a z-component.
1. Start by considering a Gaussian surface in the shape of a cylinder that is coaxial with the xy plane and has a circular base centered at the point of interest. The cylinder should be closed on the top and bottom with equal areas for simplicity.
2. Apply Gauss's Law, which states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (ε0).
Φ = Q_enclosed / ε0
Here, the electric flux Φ is related to the electric field E through the equation:
Φ = E * A
Where A is the area of the circular base of the Gaussian surface.
3. Since the electric field due to the infinite plane is uniform, the electric field passing through the top and bottom surfaces of the Gaussian surface is the same and cancels out. Therefore, only the curved surface of the cylinder contributes to the electric flux.
Φ = E * A_curved_surface
4. The area of the curved surface can be calculated by multiplying the circumference of the base (2πr) with the height of the cylinder (h). In this case, the height is the distance of the point from the xy plane, which is 2m.
A_curved_surface = 2πr * h
5. Now, substitute this expression for the area into the equation for the electric flux:
Φ = E * 2πr * h
6. The charge enclosed within the Gaussian surface is the product of the charge density (σ) and the area of the base (A_base).
Q_enclosed = σ * A_base
Since the Gaussian surface encloses no net charge, the charge enclosed equals the surface charge density (σ) multiplied by the area of the base.
Q_enclosed = σ * A_base
= σ * πr^2
7. Substitute this expression for the charge enclosed into the equation for the electric flux:
Φ = σ * πr^2 / ε0
8. Finally, equate the two expressions for the electric flux:
σ * πr^2 / ε0 = E * 2πr * h
9. Simplify the expression and solve for E:
E = σ / (2ε0)
10. Now, substitute the surface charge density (σ) into the equation. The given surface charge density is 0.2 nC/m^2, which is equivalent to 2e-7 C/m^2:
E = (2e-7 C/m^2) / (2 * 8.85e-12 C^2/(N * m^2))
11. Calculate the magnitude of the electric field E:
E ≈ 1.13e5 N/C
Therefore, the magnitude of the electric field at any point with z = 2m above the xy plane due to a uniformly charged surface with a surface density of 0.2 nC/m^2 is approximately 1.13e5 N/C.