The surface force $\mathbf{F}$ exerted by the field in one region ($x_3 > 0$) on the field in another region ($x_3 < 0$) across a boundary surface ($x_3 = 0$) is calculated by integrating the surface force element ($\mathbf{dF}$) over the entire boundary surface $S$. $\mathbf{F} = \int_{S} \mathbf{dF} = \int_{S} \mathbf{\sigma} \cdot \mathbf{n} dA$ In this problem, the normal vector $\mathbf{n}$ is $\mathbf{e}3$, and the surface element is $\mathbf{dF} = \sigma{i3} \mathbf{e}_i dA$.
Here is the complete explanation:
The calculation of the surface force on the electromagnetic field is a direct application of the Maxwell Stress Tensor ($\mathbf{\sigma}$) in its integral form. This method views the electromagnetic field itself as a medium under stress, which transmits forces across any imaginary boundary surface.
The surface force is calculated by integrating the momentum flux carried by the electromagnetic field across the imaginary boundary surface $S$.
In electrostatics (where all charges are at rest and the magnetic field $\mathbf{B}=0$), the total electromagnetic force ($\mathbf{F}$) on all charges within a volume is found by taking the surface integral of the Maxwell stress tensor over the closed boundary $S$ of that volume:
$$ \mathbf{F}{\text{on charges}} = \oint{S} \mathbf{\sigma} \cdot \mathbf{n} dA $$
However, the question asks for the surface force on the field itself, specifically the force exerted by the field in region $A$ ($x_3 > 0$) on the field in region $B$ ($x_3 < 0$) across a single boundary surface $S$ (the $x_3 = 0$ plane). This force is the negative of the force on the charges within region $B$ that arises from the stress on the surface:
$$ \mathbf{F} = \int_{S} \mathbf{\sigma} \cdot \mathbf{n} dA $$
The term $\mathbf{\sigma} \cdot \mathbf{n}$ is the vector representing the force per unit area (the stress) acting on the surface element $dA$.
The calculation follows a specific procedure, as demonstrated in the document you are viewing:
Determine the Electric Field ($\mathbf{E}$): Calculate the total electric field $\mathbf{E}$ at every point on the imaginary surface $S$ ($x_3=0$) due to all charges in the system.
Define the Normal Vector ($\mathbf{n}$): The vector $\mathbf{n}$ must be the outward unit normal to the volume $V$ (or the region) for which you are calculating the force. In your specific problem, when calculating the force exerted by the field in the $x_3 > 0$ region onto the field in the $x_3 < 0$ region, the region of interest is $V: x_3 < 0$. The normal vector $\mathbf{n}$ is therefore the outward vector pointing from the $x_3 < 0$ region into the $x_3 > 0$ region, which is $\mathbf{n} = \mathbf{e}_3$.
Apply the Stress Tensor Components: The force element $\mathbf{dF}$ on an area element $dA$ is $\mathbf{dF} = (\mathbf{\sigma} \cdot \mathbf{n}) dA$. Since $\mathbf{n} = \mathbf{e}3$, the force component in the $i$-direction ($dF_i$) is found using the $i3$ component of the stress tensor, $\sigma{i3}$:
$$ dF_i = \sigma_{i3} dA = \epsilon_0 \left(E_i E_3 - \frac{1}{2} E^2 \delta_{i3}\right) dA $$
Perform the Integration: The total force is then obtained by integrating the force element over the entire surface $S$ (the entire $x_3 = 0$ plane):
$$ \mathbf{F} = \int_{S} \mathbf{dF} = \int_{S} \sigma_{i3} \mathbf{e}_i dA $$
In this problem, for the case of two equal charges, $\mathbf{E}_3=0$ on the surface, which simplifies the force element to:
$$ \mathbf{dF} = -\frac{1}{2} \epsilon_0 E^2 \mathbf{e}_3 dA $$
This immediately shows the force is purely along the $\mathbf{e}_3$ axis and is attractive (negative pressure), pulling the two field regions together.