Step 1: Understanding the Topic:
This question deals with "Electrostatics" and the properties of conductors in equilibrium. When a conductor is in a static situation, the free electrons within it redistribute themselves until all internal forces are balanced, leading to several unique physical properties.
Step 2: Key Formulas and Approach:
Gauss's Law: $\oint E \cdot dA = q_{encl} / \epsilon_0$.
Surface field of a conductor: $E = \sigma / \epsilon_0$.
Relation between field and potential: $E = -dV/dr$.
Step 3: Detailed Explanation:
A (Correct): If an electric field existed inside, the free electrons would experience a force and move. Since we are in a "static" situation, they must be at rest, meaning the net field is zero.
B (Incorrect): The electric field at the surface is directly proportional to the surface charge density ($E = \sigma / \epsilon_0$). Areas with higher curvature (more "pointed" parts) have higher $\sigma$ and thus higher $E$.
C (Correct): According to Gauss's Law, if $E=0$ everywhere inside, then the net flux is zero, and thus the enclosed charge must be zero. Any excess charge must reside on the outer surface.
D (Correct): If the field had a tangential component, electrons on the surface would move along the surface. In equilibrium, the field must be purely perpendicular (normal) to the surface.
E (Incorrect): Since $E = -dV/dr = 0$, the potential $V$ is constant throughout the conductor, but it is not necessarily zero. It only becomes zero if the conductor is connected to the Earth (grounded).
Step 4: Final Answer:
The correct statements are A, C, and D only.