Charge of uniform surface density (0.20 nC/m²) is distributed over the entire xy plane. Determine the magnitude of the electric field at any point having z = 2.0 m. A) 17 N/C B) 11 N/C C) 23 N/C D) 28 N/C E) 40 N/C

Question

Quizplus · Accepted Answer

Since the charge is uniformly distributed over the entire xy plane, we can assume that the sheet of charge is infinite in extent. Using Gauss's law and applying symmetry, we can find that the electric field points upwards and has constant magnitude at all points above and below the plane.

To find the electric field at a point above the plane with z=2.0 m, we can use a Gaussian cylinder of radius r and height h, with its axis perpendicular to the xy plane and its top surface at the point of interest. The electric flux through the top surface of the cylinder is given by:

Φ = E x A

where A is the area of the top surface of the cylinder, which is pi*r^2, and E is the magnitude of the electric field. The electric field is perpendicular to the top surface of the cylinder and has the same magnitude all along that surface. Therefore, the flux through the top surface is:

Φ = E x pi*r^2

We can now apply Gauss's law, which states that the total electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space (ε0). Since the charge is uniformly distributed over the xy plane, the total charge enclosed by the cylinder is:

Q = σ x A x h

where σ is the surface charge density, which is 0.20 nC/m^2 in this case. Therefore, the flux through the top surface of the cylinder is:

Φ = Q / ε0 = σ x A x h / ε0 = σ x pi x r^2 x h / ε0

Setting Φ = E x pi x r^2 equal to this expression, we have:

E = σ x h / ε0

Plugging in the given values, we get:

E = (0.20 nC/m^2) x (2.0 m) / (8.85 x 10^-12 C^2/N m^2) = 11 N/C

Therefore, the answer is B) 11 N/C.

Charge of Uniform Surface Density (0