A point charge of 17.7 pC is at the center of a cube. What is the electric flux through one face of the cube?

The charge $q = 17.7 \text{ pC}$. Convert picocoulombs to coulombs:

$$q = 17.7 \times 10^{-12} \text{ C}$$

The permittivity of free space is $\epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}$.

Gauss's Law states that the total electric flux ($\Phi_{\text{total}}$) through a closed surface is equal to the total enclosed charge ($q$) divided by the permittivity of free space ($\epsilon_0$).

$$\Phi_{\text{total}} = \frac{q}{\epsilon_0}$$

Substitute the values:

$$\Phi_{\text{total}} = \frac{17.7 \times 10^{-12} \text{ C}}{8.85 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}}$$ $$\Phi_{\text{total}} = \frac{17.7}{8.85} \text{ N m}^2 \text{ C}^{-1}$$ $$\Phi_{\text{total}} = 2 \text{ N m}^2 \text{ C}^{-1}$$

A cube has 6 identical faces. Since the charge is placed at the center of the cube, the total electric flux is distributed equally among all 6 faces.
The flux through one face ($\Phi_{\text{face}}$) is:

$$\Phi_{\text{face}} = \frac{\Phi_{\text{total}}}{6}$$ $$\Phi_{\text{face}} = \frac{2 \text{ N m}^2 \text{ C}^{-1}}{6}$$ $$\Phi_{\text{face}} = \frac{1}{3} \text{ N m}^2 \text{ C}^{-1}$$

The electric flux emerging from each face of the cube is $\frac{1}{3} \text{ N m}^2 \text{ C}^{-1}$.