Two charged capacitors have their outer plates fixed and inner plates connected by a spring of force constant k. The charge on each capacitor is q. Find the extension in the spring at equilibrium.
A. q² / (2Aε₀k)
B. q² / (4Aε₀k)
C. q² / (Aε₀k)
D. Zero

To solve this problem, we need to consider both the electrostatic force between the charges on the capacitor plates and the restoring force of the spring.

### Given:
- The charge on each capacitor: \( q \)
- The force constant of the spring: \( k \)
- The area of each plate: \( A \)
- The permittivity of free space: \( \varepsilon_0 \)

The system is in equilibrium when the electrostatic repulsive force between the charges on the plates is balanced by the restoring force of the spring.

### Electrostatic Force:

The force between two charged plates of a capacitor is given by:

\[
F_{\text{elec}} = \frac{q^2}{2A \varepsilon_0}
\]

This is derived from the formula for the electric field between two plates of a parallel plate capacitor and the interaction of the charges on the plates.

### Spring Force:

The restoring force of the spring is given by Hooke's law:

\[
F_{\text{spring}} = kx
\]

where \( x \) is the extension in the spring.

### Equilibrium Condition:

At equilibrium, the electrostatic force equals the spring force:

\[
\frac{q^2}{2A \varepsilon_0} = kx
\]

Solving for \( x \):

\[
x = \frac{q^2}{2A \varepsilon_0 k}
\]

Thus, the extension in the spring at equilibrium is:

\[
x = \frac{q^2}{2A \varepsilon_0 k}
\]

### Final Answer:
The correct option is **A**:

\[
\frac{q^2}{2A \varepsilon_0 k}
\]