Charge of a uniform density (8.0 nC/m2) is distributed over the entire xy plane.A charge of uniform density (3.0 nC/m2) is distributed over the parallel plane defined by z = 2.0 m.Determine the magnitude of the electric field for any point with z = 3.0 m.
A)0.79 kN/C
B)0.17 kN/C
C)0.62 kN/C
D)0.34 kN/C
E)0.28 kN/C

Question

Quizplus · Accepted Answer

We can use the principle of superposition to find the electric field at a point with z = 3.0 m.

Let's first consider the contribution from the charge density on the xy plane. We can use the formula for the electric field from an infinite plane:

E_xy = σ / (2ε₀)

where σ is the charge density (in C/m²) and ε₀ is the permittivity of free space. Plugging in the given values, we get:

E_xy = (8.0 nC/m²) / (2ε₀) = 4.0 ε₀ nC/C

Notice that the electric field from the xy plane is independent of z.

Now let's consider the contribution from the charge density on the plane defined by z = 2.0 m. We can again use the formula for the electric field from an infinite plane, but we need to adjust for the distance between the two planes. The distance between the planes is |3.0 m - 2.0 m| = 1.0 m. So the electric field from the z = 2.0 m plane is:

E_z2 = σ_z2 / (2ε₀) = (3.0 nC/m²) / (2ε₀) = 1.5 ε₀ nC/C

Note that the direction of E_z2 is opposite to E_xy, because the charges on the z = 2.0 m plane attract charges on the xy plane.

The total electric field at the point with z = 3.0 m is the vector sum of E_xy and E_z2:

E = sqrt(E_xy² + E_z2²) = sqrt((4.0 ε₀ nC/C)² + (1.5 ε₀ nC/C)²)

E ≈ 0.62 kN/C

Therefore, the best choice is C, with the magnitude of the electric field at the point with z = 3.0 m being 0.62 kN/C.

Charge of a Uniform Density (8