Show that the normal component of electrostatic field has a discontinuly form one side of a charged. Surface to another given by `(vec(E_(2)) - vec(E_(1))). hat(n) = (sigma)/(in_(0))`
where `hat(n)` is a unit vector normal to the surface at a point and `sigma` at a point and `sigma` is the surface charge density at that point. (The direction of `hat(n)` is from side 1 to side 2). Hence show that justy outside a conductor, the electric field `sigma hat(n)//in_(0)`.
(b) Show that the tangential componet of electrostatic field is contionous from one side fo a charged surface to another.

Explanation

To solve the problem, we will break it down into two parts as stated in the question.

### Part (a): Show that the normal component of the electrostatic field has a discontinuity given by \((\vec{E_2} - \vec{E_1}) \cdot \hat{n} = \frac{\sigma}{\epsilon_0}\)

1. **Define the Electric Fields**:
- Let \(\vec{E_1}\) be the electric field just inside the charged surface (on side 1).
- Let \(\vec{E_2}\) be the electric field just outside the charged surface (on side 2).

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