Question

An object with a charge of \(0.890 \mu \mathrm{C}\) is placed at the center of a cube. What is the electric flux through one surface of the cube?

Short answer

Answer: The electric flux through one surface of the cube is 1.662 x 10³ N.m²/C.

Step-by-step solution

Understanding Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface. It is given by the formula: ∮E ⋅ dA = Q/ε₀ Where E is the electric field, dA is a surface area element, Q is the enclosed charge, and ε₀ is the vacuum permittivity (ε₀ = 8.85 x 10⁻¹² C²/N.m²).

Electric field and the cube

In our problem, there is a charge placed at the center of the cube. The electric field produced by the charge is radially outward and symmetric in all directions. So, the electric field would be same through all faces of the cube.

Use Gauss's law to find flux through the entire cube

For the entire cube, we can write Gauss's Law as: Φ = ∮E ⋅ dA = Q/ε₀ Here, the electric field E has the same value on each face of the cube, so we can rewrite this as: Φ = E * (area of all faces) = Q/ε₀

Calculate the area of all faces

We know that there are 6 faces in the cube and the electric field on each face is the same. Let's denote each face's area as A. So, the area of all faces will be: (area of all faces) = 6A Φ = E * 6A = Q/ε₀

Find the electric flux through one surface of the cube

We want to find the electric flux through only one face of the cube (A). We can divide both sides of the equation by 6: (E * 6A)/6 = (Q/ε₀)/6 E * A = (Q/6ε₀) Now, we have the electric flux through one face of the cube: Φ₁ = (Q/6ε₀)

Plug in values

We know the given charge Q = 0.890 μC = 0.890 x 10⁻⁶ C, and ε₀ = 8.85 x 10⁻¹² C²/N.m². Plug in these values: Φ₁ = ((0.890 x 10⁻⁶)/(6 * 8.85 x 10⁻¹²)) Φ₁ = 1.662 x 10³ N.m²/C So, the electric flux through one surface of the cube is 1.662 x 10³ N.m²/C.