The computed surface force on the field for two equal charges is verified using Coulomb's Law and the condition for static equilibrium by ensuring the forces on the lower field region cancel out.
The verification process follows these steps:
The Coulomb force exerted by the upper charge ($q$ at $x_3 = d$) on the lower charge ($q$ at $x_3 = -d$) is a repulsive force pointing downward ($-\vec{e}_3$).
$$ \vec{F}{\text{Coulomb}} = \vec{F}{\text{field} \to q} = -\frac{q^2}{4\pi\epsilon_0 (2d)^2} \vec{e}_3 $$
This is the force exerted by the field on the lower charge.
By Newton's Third Law, the force exerted by the lower charge on the field in its vicinity is equal in magnitude and opposite in direction to the force exerted by the field on the charge:
$$ \vec{F}{q \to \text{field}} = -\vec{F}{\text{field} \to q} = -\vec{F}_{\text{Coulomb}} $$
Substituting the result from Step 1:
$$ \vec{F}_{q \to \text{field}} = - \left( -\frac{q^2}{4\pi\epsilon_0 (2d)^2} \vec{e}_3 \right) = \frac{q^2}{4\pi\epsilon_0 (2d)^2} \vec{e}_3 $$
This force points upward ($+\vec{e}_3$).
For the system to be in static equilibrium, the total force on the electromagnetic field in the lower region ($x_3 < 0$) must be zero. This force is the sum of two components:
$$ \vec{F}{\text{total on field}} = \vec{F}{q \to \text{field}} + \vec{F}_{\text{surface}} = 0 $$
Therefore, the expected surface force must exactly balance the force of the charge on the field: