The surface force on the field differs dramatically in its direction but has the same magnitude in both cases. For this comparison, we consider the total surface force ($\vec{F}$) exerted by the field in the $x_3 > 0$ region on the field in the $x_3 < 0$ region, across the mid-plane surface $x_3 = 0$.
Here is the breakdown:
| Characteristic | Result | Implication |
|---|---|---|
| Total Surface Force ($\vec{F}_{\text{eq}}$) | $\vec{F}_{\text{eq}} = - \frac{q^2}{4 \pi \epsilon_0 (2d)^2} \vec{e}_3$ | The force is attractive (in the $-\vec{e}_3$ direction, pointing toward the upper charge). |
| Physical Role | This attractive force balances the repulsive Coulomb force between the two equal charges, which is directed in the $+\vec{e}_3$ direction. |
The total surface force is attractive, pulling the two field regions toward each other. This is consistent with the repulsive force between the charges, as the field must provide the necessary reaction force to maintain static equilibrium.
| Characteristic | Result | Implication |
|---|---|---|
| Total Surface Force ($\vec{F}_{\text{opp}}$) | $\vec{F}_{\text{opp}} = + \frac{q^2}{4 \pi \epsilon_0 (2d)^2} \vec{e}_3$ | The force is repulsive (in the $+\vec{e}_3$ direction, pushing away from the lower charge). |
| Physical Role | This repulsive force balances the attractive Coulomb force between the two opposite charges, which is directed in the $-\vec{e}_3$ direction. |
The force is repulsive, pushing the two field regions apart. This is consistent with the attractive force between the charges.
The magnitude of the surface force is the same in both configurations, as it must equal the magnitude of the Coulomb force between the charges ($F_{\text{Coulomb}} = \frac{q^2}{4 \pi \epsilon_0 (2d)^2}$).
The direction of the surface force is opposite: