Q.Draw the pattern of electric field lines, when a point charge $-Q$ is kept near an uncharged conducting plate.

Inside a charged conductor E=0, so the whole interior is at the same potential as the surface.

For a conductor in electrostatic equilibrium, the electric field inside the material is zero everywhere ($E=0$). Since $E=-\dfrac{dV}{dr}$, a zero field means the potential does not change from point to point inside the conductor — the entire conductor (interior as well as surface) is an equipotential region. Hence the potential at any point inside equals the surface potential $V_0$.

In electrostatic equilibrium, excess charge on a conductor resides entirely on its outer surface.

Inside a conductor in static (electrostatic) equilibrium, the electric field must be zero everywhere in the bulk material — otherwise free charges would keep moving. Any excess charge placed on a solid conductor therefore redistributes itself until it lies entirely on the outer surface, arranging itself so that the field inside is exactly cancelled and the surface becomes an equipotential. For an isolated conductor this distribution spreads over the whole surface (its precise density depends on the conductor's shape), not just at one point and not through the volume.

For a charged conducting hollow sphere, the electric field inside is zero, so the potential everywhere inside (including the centre) equals the surface potential.

For points inside a uniformly charged hollow conducting sphere, the electric field $E = 0$ (all charge resides on the outer surface, and by Gauss's law the enclosed charge for any Gaussian surface inside is zero). Since $E = -\dfrac{dV}{dr}$, a zero field means the potential does not change with position inside the sphere — the potential remains constant and equal to the potential at the surface, right up to the centre.

Given surface potential $= 5$ volts, the potential at the centre is also $5$ volts.

Just outside a charged conductor the electrostatic field is perpendicular (normal) to the surface, magnitude E = σ/ε₀.

A conductor in electrostatic equilibrium can have no tangential component of field at its surface — if it did, free charges would experience a force along the surface and keep moving, contradicting equilibrium. So the field must be entirely perpendicular to the surface.

Using a Gaussian pillbox at the surface, its magnitude is

$E = \frac{\sigma}{\varepsilon_0}$

where σ is the local surface charge density, directed along the outward normal.

A conductor placed in an external field develops induced surface charges whose field exactly cancels the external field inside, so the net interior field is zero.

When a conductor is placed in an external electric field, its free electrons redistribute almost instantly, accumulating on the surfaces. These induced charges set up an internal field that opposes the applied field.

Equilibrium is reached when the induced field exactly cancels the external field everywhere inside the conductor:

$E_{inside} = 0$

This is why a conductor acts as an electrostatic shield.

Free electrons in a conductor redistribute on its surface until the field they produce exactly cancels the external field everywhere inside, making the net field inside zero — this is electrostatic shielding.

When a conductor is placed in an external electric field, the free electrons inside it experience a force and migrate, piling up on one face (inducing negative charge there) and leaving a deficiency (positive charge) on the opposite face. This induced surface charge distribution itself sets up an electric field inside the conductor, directed opposite to the external field.

This redistribution continues only until the induced field exactly cancels the external field at every point inside the conductor's body — in electrostatic equilibrium there is no further force on the free charges, so the net (total) electric field inside the material of the conductor must be zero.

This is the basis of electrostatic shielding: a hollow conductor shields whatever is enclosed inside it from any external electric field (e.g. a Faraday cage).

Because E = 0 everywhere inside a uniformly charged spherical shell, the potential cannot change from point to point inside — it stays equal to the surface potential.

Using a Gaussian surface inside the shell, the enclosed charge is zero, so by Gauss's law $E_{inside}=0$. Since $E=-\dfrac{dV}{dr}$, if $E=0$ then $V$ does not vary with $r$ inside the shell — it is constant, equal to the potential at the surface, $V=\dfrac{kQ}{R}$.

At the surface of a charged conductor in electrostatic equilibrium, the field is always perpendicular (normal) to the surface at every point.

Inside a conductor in electrostatic equilibrium, the electric field is zero, and any excess charge resides entirely on its outer surface. If the electric field at the surface had any component tangential (parallel) to the surface, free charge carriers on the conductor would experience a force along the surface and would keep moving — contradicting the assumption of equilibrium (electrostatic conditions, no current flow). Therefore, in equilibrium, the field just outside the surface must be entirely normal (perpendicular) to the surface at every point; the surface itself is an equipotential surface.

Inside a conductor in electrostatic equilibrium, the electric field is zero.

In a charged conductor, free electrons redistribute themselves on the surface until the net field inside becomes zero — if it weren't, the free charges would keep accelerating, which contradicts electrostatic equilibrium. This is why the interior of a conductor is field-free and all excess charge resides on its outer surface, matching Column B option (v) Zero.

A negative point charge near an uncharged conductor induces positive charge on the near surface and an equal negative charge on the far surface; field lines originate on the induced positive charges and terminate on $-Q$, meeting the conductor surface perpendicularly, with none inside it.