What is the electric flux through one side of a cube that has a single point charge of -3.80 µC placed at its center? Hint: You do not need to integrate any equations to get the answer.

? N·m^2/C

φ = -7.16 × 10⁴ Nm²/C

Explanation:

q = -3.80 × 10⁻⁶ C

ε₀ = 8.85 10⁻¹² C²/Nm²

By Gauss's law

φ = q/ε₀

as the cube has 6 faces so for one side the flux will

φ =1/6 ( q/ε₀)

φ = 1/6 ( -3.80 × 10⁻⁶ C / 8.85 10⁻¹² C²/Nm²)

φ = -71,563.088 Nm²/C

φ = -7.16 × 10⁴ Nm²/C

-7.00 x 10⁴Nm²/C

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.

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