Answer:
-7.00 x 10⁴Nm²/C
Explanation:
Let the single point charge of -3.80µC be Q
i.e
Q = -3.80µC = -3.80 x 10⁻⁶C
From Gauss's law, the total electric flux,φ, through an enclosed surface is equal to the quotient of the enclosed charge Q, and the permittivity of free space, ε₀. i.e;
φ = Q / ε₀ ----------------------(i)
Where;
ε₀ = known constant = 8.85 x 10⁻¹²F/m
Now, If the total flux through the enclosed cube is given by equation (i), then the flux, φ₁, through one of the six sides of the cube is found by dividing the equation by 6 and is given as follows;
φ₁ = Q / 6ε₀ --------------------(ii)
Substitute the values of Q and ε₀ into equation (ii) as follows;
φ₁ = -3.80 x 10⁻⁶ / (6 x 8.85 x 10⁻¹²)
φ₁ = -0.07 x 10⁶
φ₁ = -7.00 x 10⁴
Therefore, the electric flux through one side of the cube is -7.00 x 10⁴ Nm²/C
Note:
The result (flux) above is negative to show that the electric field lines from the point charge points radially inwards in all directions.
