Why do we always get electric field intensity near a charged metallic surface of magnitude twice of the electric field intensity near a thin sheet of charge with same surface charge density as of metal plate?

Explanation

To understand why the electric field intensity near a charged metallic surface is twice that of a thin sheet of charge with the same surface charge density, we can break down the explanation into clear steps:

### Step-by-Step Solution:

1. **Understanding Electric Field Due to a Single Sheet of Charge**:
- For a single infinite sheet of charge with surface charge density \( \sigma \), the electric field \( E \) produced by this sheet is given by the formula:
\[
E = \frac{\sigma}{2\epsilon_0}
...