A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the holes is `(sigma//2 in_(0())` `hat(n)`, where `hat(n)` is the unit vector in the outward normal direction, and `sigma` is the surface charge density near ther hole.

Explanation

To show that the electric field in the hole of a hollow charged conductor is given by \( \frac{\sigma}{2 \epsilon_0} \hat{n} \), where \( \hat{n} \) is the unit vector in the outward normal direction and \( \sigma \) is the surface charge density near the hole, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the System**:
- We have a hollow charged conductor with a tiny hole cut into its surface. The conductor is uniformly charged, which means it has a surface charge density \( \sigma \).

2. **Electric Field Without the Hole**:
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