$(a)$ Show that the normal component of the electrostatic field has a discontinuity from one side of a charged surface to another given by $(E_2 - E_1) \cdot \hat{n} = \frac{\sigma}{\varepsilon_0}$, where $\hat{n}$ is a unit vector normal to the surface 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 just outside a conductor, the electric field is $\frac{\sigma \hat{n}}{\varepsilon_0}$. $(b)$ Show that the tangential component of the electrostatic field is continuous from one side of a charged surface to another.

$(A)$ Consider a small Gaussian pillbox of cross-sectional area $A$ enclosing a small area element of the charged surface. By Gauss's Law, the flux through the pillbox is $\oint E \cdot dA = \frac{q_{enclosed}}{\varepsilon_0}$. The flux through the sides is negligible. The flux through the two faces is $(E_{2n} - E_{1n})A = \frac{\sigma A}{\varepsilon_0}$, where $E_{2n}$ and $E_{1n}$ are the normal components. Thus, $(E_2 - E_1) \cdot \hat{n} = \frac{\sigma}{\varepsilon_0}$. For a conductor, the field inside is $E_1 = 0$, so $E_2 \cdot \hat{n} = \frac{\sigma}{\varepsilon_0}$, which implies $E = \frac{\sigma \hat{n}}{\varepsilon_0}$. $(b)$ Consider a small rectangular loop of length $l$ and width $w$ spanning the surface. The work done by the electrostatic field in moving a charge around a closed loop is zero $(\oint E \cdot dl = 0)$. As the width $w \to 0$, the contribution from the sides vanishes, leaving $(E_{1t} - E_{2t})l = 0$, where $E_{1t}$ and $E_{2t}$ are the tangential components. Thus, $E_{1t} = E_{2t}$, proving the tangential component is continuous.