❓In which direction does the surface force on the field point for two equal charges?

The surface force on the field in the lower region ($x_3 < 0$), exerted by the field in the upper region ($x_3 > 0$) across the plane of symmetry ($x_3 = 0$), points in the downward direction, or the $-\vec{e}_3$ direction. This result is derived from the calculation: $\vec{F}_{\text{computed}} = -\frac{q^2}{4\pi\epsilon_0 (2d)^2} \vec{e}_3$

Context from the calculation:

1. Nature of the Surface Force: Although the equal charges repel each other, the computed surface force on the field is an attractive force between the two field regions (pulling the field in the $x_3 < 0$ region toward the field in the $x_3 > 0$ region).

2. Repulsive Force Direction: The parenthetical note refers to the direction of the repulsive Coulomb force acting on the lower charge ($q$ at $x_3 = -d$), which also points downward ($-\vec{e}_3$).

3. Static Equilibrium: The attractive surface force on the field in the lower region ($\vec{F}{\text{surface}}$) is required to balance the repulsive force the lower charge exerts on the field ($\vec{F}{q \to \text{field}}$), which points upward ($+\vec{e}_3$).

   $\vec{F}{\text{surface}} + \vec{F}{q \to \text{field}} = 0$

   Therefore, the surface force $\vec{F}{\text{surface}}$ must be opposite to the force $\vec{F}{q \to \text{field}}$, resulting in a downward direction.

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In which direction does the surface force on the field point for two equal charges? | FAQsviadean on Notion