The distance between the plates of the flat capacitor was increased by 3 times. How will the energy of the capacitor

The distance between the plates of the flat capacitor was increased by 3 times. How will the energy of the capacitor change if the capacitor: a) is disconnected from the voltage source; b) remains connected to a constant voltage source?

Initial values.
Capacity C = e₀eS / d, where S is the area of the plates, d is the distance between them, the rest are coefficients.
Energy W = CU² / 2, where C – capacity, U – voltage.
Charge q = UC
1. The source is disabled, d => 3d.
The charge q = UC remained unchanged.
Capacity C₁ = e₀eS / 3d = C / 3.
The potential difference for a capacitor with a capacitance C₁ and a charge q:
U₁ = q / C₁ = UC / (C / 3) = 3U;
W₁ = C₁U₁² / 2 = (C / 3 * (3U) ²) / 2 = 3 * (CU² / 2);
W₁ = 3W.
Answer: The energy of the capacitor has increased by 3 times.