Conductors and Insulators

Correct answer: A. A tangential component would move charges

Explanation: In electrostatic equilibrium, free charges inside a conductor must experience no force along its surface. An oblique electric field has a tangential component, E_t, parallel to the surface. The force qE_t would drive mobile charges along the conductor, causing redistribution until the tangential component becomes zero. Consequently, the remaining field at the surface is normal. The other options describe no electrostatic consequence.

Correct answer: A. They can start from the positive charge and end on induced negative charge on the inner surface.

Explanation: A positive charge inside the cavity produces an electric field in the cavity. To keep the field inside the conducting material zero, an equal amount of negative charge is induced on the inner surface of the conductor. Field lines may therefore originate at the positive charge and terminate on this induced negative charge. They do not pass through the conductor, and electrostatic field lines are not closed loops. Thus A is correct.

Correct answer: A. Surface charge density can be higher there, making the local field stronger.

Explanation: For a conductor in electrostatic equilibrium, charge is distributed over its surface, and the surface charge density is generally greater where the radius of curvature is smaller. A pointed region therefore can have a large local surface charge density. Since the field just outside a conductor is related to surface charge density by E = sigma divided by epsilon_0, the field is stronger there. Denser field lines represent this larger magnitude, so A is correct; B, C, and D contradict conductor-field rules.