Solution:

Let’s solve step by step why the energy stored becomes $2U$:

1. Initial Setup

• Capacitance of the capacitor:

  $$C = \frac{\varepsilon_0 A}{d}$$where:

  • $A$ = area of the plates,
  • $d$ = distance between the plates,
  • $\varepsilon_0$ = permittivity of free space.
• When the capacitor is charged by a battery to a voltage $V$, the charge stored is:

  $$Q = CV$$The energy stored in the capacitor is:

  $$U = \frac{1}{2} C V^2$$

2. When the Distance is Doubled

After the battery is removed, the charge $Q$ on the capacitor remains constant because there’s no external connection. However, the capacitance changes due to the increased plate separation.

• New capacitance:

  $$C' = \frac{\varepsilon_0 A}{2d}$$

• Energy stored in a capacitor is given by:

  $$U' = \frac{Q^2}{2C'}$$Substitute $C' = \frac{\varepsilon_0 A}{2d}$:

  $$U' = \frac{Q^2}{2 \cdot \frac{\varepsilon_0 A}{2d}} = \frac{Q^2 \cdot 2d}{2 \varepsilon_0 A}$$Simplify:

  $$U' = \frac{Q^2 d}{\varepsilon_0 A}$$

3. Relating $U'$ and $U$

From the initial setup:

$$U = \frac{1}{2} C V^2$$

Substitute $C = \frac{\varepsilon_0 A}{d}$ and $Q = CV$:

$$U = \frac{1}{2} \cdot \frac{\varepsilon_0 A}{d} \cdot V^2 = \frac{Q^2}{2C}$$

From the doubling of plate separation:

$$U' = 2 \cdot \frac{Q^2}{2C} = 2U$$

Final Answer:

The energy stored in the capacitor after doubling the separation is:

$$\boxed{2U}$$