An air capacitor is first charged through a battery. The charging battery is then removed and a electric slab of dielectric constant `K=4` is inserted between the plates. Simultaneously, the distance between the plates is reduced to half, then find change, `C, E, V` and `U`.

Explanation

To solve the problem step by step, we need to analyze the changes in capacitance (C), electric field (E), potential difference (V), and stored energy (U) when a dielectric slab is inserted into a charged capacitor and the distance between the plates is halved.

### Step 1: Initial Parameters
Let:
- Initial capacitance \( C_0 = \frac{\epsilon_0 A}{d} \)
- Initial electric field \( E_0 = \frac{V_0}{d} \)
- Initial potential difference \( V_0 \)
- Charge \( Q \) on the capacitor after charging through the battery.
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