A cube, whose edges are aligned with the 𝑥, 𝑦 and 𝑧 axes, has a side length 𝑠=1.82 m. The field is immersed in an electric field aligned with the 𝑧 axis. On the left and right faces, the field has a strength 714.76 N/C and 737.55 N/C, respectively. The field along the front and back faces has strengths 750.94 N/C and 756.65 N/C. The field at the bottom and top faces has strengths 951.48 N/C and 1080.39 N/C, respectively. What is the total charge enclosed by the cube?
Question
A cube, whose edges are aligned with the 𝑥, 𝑦 and 𝑧 axes, has a side length 𝑠=1.82 m. The field is immersed in an electric field aligned with the 𝑧 axis. On the left and right faces, the field has a strength 714.76 N/C and 737.55 N/C, respectively. The field along the front and back faces has strengths 750.94 N/C and 756.65 N/C. The field at the bottom and top faces has strengths 951.48 N/C and 1080.39 N/C, respectively. What is the total charge enclosed by the cube?
Solution
To find the total charge enclosed by the cube, we can use Gauss's Law. Gauss's Law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity of free space (ε₀). The electric flux (Φ) through a surface is given by the integral of the electric field (E) over the area (A) of the surface, or Φ = ∫E·dA.
The electric field is given for each face of the cube, and since the field is uniform over each face, we can simplify the integral to a multiplication: Φ = E*A. The area of each face of the cube is s², where s is the side length of the cube.
The total electric flux out of the cube is the sum of the fluxes through each face. Since the electric field is aligned with the z axis, only the top and bottom faces contribute to the flux. The left, right, front, and back faces have electric field lines parallel to them, so the electric field does not pass through these faces.
The total flux is then Φ_total = E_bottomA_bottom + E_topA_top = E_bottoms² + E_tops².
Substituting the given values, we get Φ_total = (951.48 N/C)(1.82 m)² + (1080.39 N/C)(1.82 m)².
Finally, we can find the total charge enclosed by the cube using Gauss's Law: Q = ε₀Φ_total. The permittivity of free space is ε₀ = 8.8510^-12 C²/N·m².
Substitute the calculated total flux and the value of ε₀ into the equation to find the total charge.
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