Two charges of 15 pC and -40 pC are inside a cube with sides that are of 0.40-m length.Determine the net electric flux through the surface of the cube. A)+2.8 N. m²/C B)-1.1 N .m²/C C)+1.1 N. m²/C D)-2.8 N . m²/C E)-0.47 N . m²/C

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By Gauss's Law, the net electric flux through a closed surface enclosing charges is equal to the algebraic sum of the charges inside the surface divided by the permittivity of free space (ε₀). Using Gauss's Law, we can find the net electric flux through the surface of the cube: Φ = Q_in / ε₀ where Q_in is the algebraic sum of the charges inside the cube. Q_in = 15 pC + 40 pC = 55 pC Φ = (55 pC) / (8.85×10−12 C²/N*m²) = 6.21×10⁹ N⋅m²/C The surface area of the cube is A = 6(0.40 m)² = 0.384 m² The electric flux density (D) is the electric flux per unit area: D = Φ / A = (6.21×10⁹ N⋅m²/C) / (0.384 m²) = 1.62×10¹¹ N/C The net electric flux through the cube is directed outward if the net charge inside the cube is positive and inward if the net charge inside the cube is negative. By the given charges, the net charge inside the cube is positive, so the net electric flux through the surface of the cube is outward. The answer is therefore D, F1F1F1S1F1F1F102.8 N F1F1F1S1F1F1F10 m²/C.

Two Charges of 15 PC and -40 PC Are Inside