Solution:
(a) Allow $E_1$ to represent the electric field on one side of the charged body and $E_2$ to represent the electric field on the opposite side of the charged body for the sake of simplicity. Suppose the infinite planar charged body has a uniform thickness, and the electric field due to one of the charged body’s surfaces is equal to zero.
$\vec{E}{1}=-\frac{\sigma}{2 \epsilon{0}} \hat{n}—(1)$
here,
$\hat{n}=$ unit vector normal to the surface at a point
$\sigma=$ surface charge density at that point
The electric field due to the other surface of the charged body is
$\vec{E}{2}=\frac{\sigma}{2 \epsilon{0}} \hat{n}$
The electric field at any point due to the charge surfaces
$\vec{E}{2}-\vec{E}{1}=\frac{\sigma}{2 \epsilon_{0}} \hat{n}+\frac{\sigma}{2 \epsilon_{0}} \hat{n}=\frac{\sigma}{\epsilon_{0}} \hat{n}$
Since inside the conductor, $\vec{E}_{1}=0$
$$
\vec{E}{2}-\vec{E}{1}=\frac{\sigma}{\epsilon_{0}} \hat{n}–(3)
$$
Therefore, the electric field just outside the conductor is $\frac{\sigma}{\epsilon_{0}} \hat{n}$
(b) Whenever a charged particle is transferred from one location to another in a closed-loop, the work done by the electrostatic field is equal to zero. Thus, the tangential component of the electrostatic field is continuous from one side of a charged surface to the other side of a charged surface.