A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the holes is `(sigma//2epsi_(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.

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 1: Understand the Electric Field of a Charged Conductor
When there is no hole in the charged conductor, the electric field just outside the surface of a charged conductor is given by:
\[ E = \frac{\sigma}{\epsilon_0} \hat{n} \]
where \( \sigma \) is the surface charge density and \( \hat{n} \) is the outward normal unit vector.
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