The two equal charges ($q$ at $x_3 = -d$ and $q$ at $x_3 = d$) create a total electric field ($\mathbf{E}$) on the equidistant plane ($x_3 = 0$) that is purely radial (in the $\mathbf{e}\rho$ direction): $\mathbf{E} = \mathbf{E}{q, \text{lower}} + \mathbf{E}{q, \text{upper}} = \frac{q \rho}{2\pi\epsilon_0 r^3} \mathbf{e}\rho$ , where $r^2 = \rho^2 + d^2$. Since the electric field has no component in the $x_3$ direction ($\mathbf{E}_3 = 0$), the field lines are parallel to the surface at this mid-plane.
The key to understanding the total electric field $\mathbf{E}$ on the equidistant plane ($x_3=0$) is the principle of superposition and symmetry.
When two identical point charges ($q$) are placed symmetrically on the $x_3$-axis (at $x_3 = -d$ and $x_3 = d$), the total electric field $\mathbf{E}$ at any point on the plane $x_3 = 0$ is the vector sum of the fields from the individual charges: $\mathbf{E} = \mathbf{E}{\text{lower}} + \mathbf{E}{\text{upper}}$.
Consider a point $P$ on the $x_3=0$ plane, located a radial distance $\rho$ from the $x_3$-axis. The distance from each charge to $P$ is $r = \sqrt{\rho^2 + d^2}$.
Each field vector can be broken into two components: a radial component ($\mathbf{E}_{\rho}$) and an axial component ($\mathbf{E}_3$ or $\mathbf{E}_z$).
Because the charges are equal in magnitude and sign ($q$ and $q$):
Axial Components Cancel: The axial component ($\mathbf{E}3$) of $\mathbf{E}{\text{lower}}$ points upward (in the $+\mathbf{e}3$ direction), while the axial component of $\mathbf{E}{\text{upper}}$ points downward (in the $-\mathbf{e}_3$ direction). Since the charges and distances are identical, these two components are equal in magnitude and exactly cancel out.
$$ \mathbf{E}3 = E{3, \text{lower}} + E_{3, \text{upper}} = 0 $$
Radial Components Add: Both radial components ($\mathbf{E}_{\rho}$) point outward from the axis. They are equal in magnitude and point in the same direction, so they add up.
$$ \mathbf{E}{\rho} = E{\rho, \text{lower}} + E_{\rho, \text{upper}} $$
The total electric field $\mathbf{E}$ on the equidistant plane is therefore purely radial, pointing perpendicularly outward from the $x_3$-axis:
$$ \mathbf{E} = \frac{q \rho}{2\pi\epsilon_0 r^3} \mathbf{e}_\rho $$
This means that the field lines on the $x_3=0$ plane are parallel to the surface ($\mathbf{E}_3 = 0$). This characteristic—field lines lying parallel to the boundary surface—is what leads to the calculation of an attractive surface force between the two field regions when using the Maxwell stress tensor.