❓How does the surface force on the field differ when the two charges are equal compared to opposite?

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 $x_3 = 0$.

1. Equal Charges ($q$ and $q$)

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.

Brief audio

Proof and Derivation related to FAQ

How does the surface force on the field differ when the two charges are equal compared to opposite? | FAQsviadean on Notion