How to calculate flux density between perfect parallel-plate capacitor?
The flux density is &quot;created&quot; by the charged particles just as our friend Maxwell tells us: \$\nabla \cdot \vec{D} = \rho \$. The other way to look at this relationship is to think that the net charge contained in any volume can be calculated by taking a surface integral of \$\vec{D}\$ over the surface of this volume. Note that \$\vec{D}\$ is a vector quantity.
See the three steps in the drawing below, red marks positive charge, blue negative, and black is the flux density \$\vec{D}\$:

For a point charge, the \$\vec{D}\$ can be calculated just by drawing a sphere around the charge and saying that at distance \$r\$ the surface density is \$ D = Q / 4\pi r^2\$.

For a single (very large) plate that has a charge density of \$ 7~C/m^2\$, the \$\vec{D}\$ can be calculated by taking a small (square meter) section of the plate. We can see that the charge contained in the section is  7 C and the surface area over which the 2 square meters (one surface on both sides of the plate). So \$D = \frac{7~C/m^2}{2} = 3.5~C/m^2\$.

In phase three, we take a second plate that has a charge of \$-7~C/m^2\$. This second plate creates a \$\vec{D}\$ around itself just like the first one, just flux going in the opposite direction, that is, towards the plate. The vector sum of these two fields is \$\vec{D} = 7~C/m^2\$ towards the negatively charged plate. The sign of the answer is just a matter of how you set your coordinate system. Normally we say that z-axis points upwards and in your question, the top plate is the positively charged one. As a conclusion, the \$\vec{D} = - 7 C/m^2 \vec{\hat{z}}\$, where \$\vec{\hat{z}}\$ is the unit vector.