Conductors and Insulators

Correct answer: A. Electrostatic equilibrium has not been established in the conductor.

Explanation: A conductor contains free charges. If a nonzero electric field exists in its conducting material, the free charges experience an electric force and redistribute. Therefore, electrostatic equilibrium cannot be established until the electric field within the conducting material becomes zero. Option D is incorrect because
\(\mathbf{E}=-\nabla V\); a nonzero electric field implies that the potential changes with position.

Correct answer: B. Surface charges must not flow in electrostatic equilibrium

Explanation: The governing principle is electrostatic equilibrium in a conductor. A tangential electric field would exert a force qE along the surface on mobile charges, causing them to move. Since charges are at rest in electrostatic equilibrium, the tangential component must be zero; only the normal component may remain. Option B is correct. The other options falsely deny surface charge, claim zero net charge, or confuse conductors with insulators.

Correct answer: A. Because surface curvature and shape affect charge distribution

Explanation: The governing concept is surface charge distribution on a conductor in electrostatic equilibrium. Excess charge stays on the outer surface, but its surface density depends on geometry. At regions with smaller radius of curvature, especially sharp points, charges crowd more strongly, producing a larger surface charge density and electric field. Option A is correct; colour, absence of charge and absence of electrons are not physical explanations.

Correct answer: A. Because surface-charge rearrangement can reduce or cancel the internal electrostatic field

Explanation: Electrostatic shielding follows from the behavior of free charges in a conductor. When an external electric field is applied, charges on a closed conducting shell redistribute over its surfaces so that the net field within the conducting material is zero and, under suitable closed-shell conditions, the protected interior is shielded from external electrostatic influence. The effect is unrelated to sound, light amplification, or conversion of charge into mass; therefore A is correct.

Correct answer: A. Cancellation of the field by free charges

Explanation: The governing concept is electrostatic shielding. An external charge produces an electric field, but free charges in the conducting material redistribute on its surfaces. Their induced field opposes the external field, making the resultant electrostatic field inside the conducting region zero; for a properly closed conductor, the cavity is thereby protected from external influence. Charge is not converted into mass, gravity is not stopped, and insulation is not the cause, so A is correct.

Correct answer: A. Free charges redistribute on the surfaces until the field they produce cancels the external field within the conducting material.

Explanation: Free charges in a conductor are mobile. If a nonzero net electric field existed within the conducting material, these charges would experience a force and continue to move, so the conductor would not be in electrostatic equilibrium. Charges therefore redistribute mainly on the surfaces until their induced field cancels the external field within the conducting material. Option D is incorrect because zero total electric flux does not imply that the electric field is zero at every point.

Correct answer: A. Free charges redistribute until the electric field inside the conducting material becomes zero; hence the excess charge resides on the outer surface.

Explanation: Free charges can move in a conductor. If the electric field inside the conducting material were nonzero, these charges would experience a force and continue moving, so electrostatic equilibrium would not be possible. At electrostatic equilibrium, the electric field inside the conductor is zero; therefore, the excess charge redistributes to the outer surface. Option B is incorrect because charge is not destroyed; only its distribution changes.

Correct answer: A. \(-q\)

Explanation: In electrostatic equilibrium, the electric field inside the material of a conductor is zero. Take a Gaussian surface within the conducting material surrounding the cavity. Its electric flux is zero; therefore, by Gauss’s law, the net charge enclosed must be zero. Hence, \(+q+Q_{\text{inner}}=0\), so \(Q_{\text{inner}}=-q\). Option \(0\) is incorrect because it would leave a net enclosed charge \(+q\) for this Gaussian surface.

Correct answer: A. equal in magnitude to the charge placed in the cavity)

Explanation: In electrostatic equilibrium, the electric field within the conducting material is zero. For a Gaussian surface lying in the conductor and enclosing the cavity, the net enclosed charge must be zero. Hence, the total induced charge on the inner surface is −q. Since the conductor was initially neutral and is isolated, its total charge remains zero; therefore, the total charge on its outer surface must be +q. The charge distribution on the outer surface may be non-uniform, but its total is +q. Option B represents the charge on the inner surface, not on the outer surface.

Correct answer: B. Charge on the outer surface can be zero

Explanation: The governing ideas are Gauss’s law, induction, and earthing. A positive charge inside the cavity induces an equal negative charge on the cavity wall so that the electric field inside the conducting material is zero. If the conductor is earthed, charge can flow between it and Earth; the outer surface need not retain a compensating positive charge and can be neutral. Thus B is correct, while A ignores earthing and C and D are physically meaningless.

Correct answer: B. |E_{text{out}}| = |σ|/ε_0; therefore, a larger |σ| gives a larger field magnitude just outside the surface.

Explanation: In electrostatic equilibrium, the electric field inside a conductor is zero. In vacuum, the normal electric field just outside its surface is determined by the surface charge density: |E_{text{out}}| = |σ|/ε_0. Hence, where |σ| is larger, the magnitude of the external field is also larger. The field is normal outward for positive σ and normal inward for negative σ. The factor 1/2 in option D applies to an isolated uniformly charged non-conducting sheet, not to a conductor.

Correct answer: A. The small radius of curvature at the sharp point produces high surface charge density and a very strong local electric field, which can ionise nearby air.

Explanation: In electrostatic equilibrium, charge redistributes over a conductor’s surface. At a region with a small radius of curvature, such as a sharp point, the surface charge density is high and the electric field just outside the surface becomes very strong. This strong field can ionise nearby air molecules, initiating corona discharge or leakage of charge. Option B is incorrect because all points of a conductor in equilibrium are at the same potential, even though the electric field near a sharp point is stronger.

Correct answer: A. Spherical symmetry in the absence of external influence

Explanation: With no external electric field or nearby charge, every direction and every point on the surface of a sphere are equivalent. Therefore, in electrostatic equilibrium, the surface charge density \(\sigma\) cannot depend on direction and must be uniform over the surface. Being at one potential is necessary for a conductor in equilibrium, but it alone does not ensure uniform charge density; an irregularly shaped conductor can be equipotential while having non-uniform surface charge density.

Correct answer: A. Negative induced charge on the nearer surface, positive induced charge on the farther surface; net charge remains zero.

Explanation: The positively charged rod attracts the conductor’s free electrons toward the nearer surface. Thus, the nearer surface acquires induced negative charge, while deficiency of electrons leaves induced positive charge on the farther surface. Since the conductor is isolated, no charge is transferred to or from it, so its net charge remains zero. Option B reverses the actual induced-charge distribution.

Correct answer: A. To retain the final induced charge

Explanation: In the induction sequence, the nearby charged object first maintains separation of charges while earthing allows electrons to enter or leave. The earth connection must be removed while the external object is still present; otherwise the charge can flow back and neutralise when the inducing object is removed. After isolation from Earth, removing the external object only removes the polarising field, so the conductor retains its net induced charge. Option A is correct.

Correct answer: A. The desired charge may not remain

Explanation: Successful induction charging depends on the order of operations. While the external charged object is present, it separates charge in the conductor. If that object is removed first while the conductor is still earthed, the separating influence disappears but the conducting path to Earth remains. Electrons can then flow until the conductor is neutral or reaches the earth-potential condition, so the desired net charge may not be retained. Option A is correct.

Correct answer: A. The opposite charge effect on the near side is closer

Explanation: The governing concept is polarisation of a dielectric. In an insulator, bound positive and negative charges shift slightly in opposite directions, but no net charge is created. When an external charge is nearby, the induced opposite-sign charge effect lies closer to it than the like-sign effect on the farther side. Since electrostatic force increases as separation decreases, the nearer attraction is stronger and a net attraction appears. Option A is correct.

Correct answer: A. Because identical conductors reach equal potential and have equal capacitance

Explanation: When identical conducting spheres touch, free charge flows between them until electrostatic equilibrium is reached, meaning their potentials are equal. Identical spheres have equal capacitance, and Q = CV; with the same final potential and the same C, their final charges are equal. Thus A is correct. Charge is conserved, the spheres are conductors, and gravity is not the mechanism, so B, C and D are incorrect.

Correct answer: A. Because they become equipotential but their capacitances may differ

Explanation: Conductors in contact exchange free charge until their potentials become equal, not necessarily until their charges become equal. Using Q = CV, unequal conductors generally have different capacitances, so at a common final potential their charges satisfy Q1/Q2 = C1/C2 and need not be equal. A is correct. Charge can enter either conductor, both contain electrons, and total charge is conserved; therefore B, C and D are false.

Correct answer: B. No, because excess charge can be on the surface

Explanation: The zero-field result applies to the interior of a conductor in electrostatic equilibrium: free charges have redistributed until the internal force is cancelled. It does not say that the conductor carries no net charge. Any excess charge resides on its surface, while the field inside remains zero. Hence B is correct. A and C confuse field with charge, and D incorrectly places excess charge at the centre.

Correct answer: A. On the outer surface

Explanation: In electrostatic equilibrium, free charges in a conductor are mobile and repel one another. They continue moving until the electric field inside the conducting material is zero; consequently, the volume charge density in the interior is zero and any excess charge appears on the surface. A is correct. It is not confined to the centre, does not fill the volume, and cannot transform into mass, so B, C and D are rejected.

Correct answer: A. Charge density and the electric field can be higher near the sharp tip

Explanation: For a conductor in electrostatic equilibrium, surface charge density is greater where the radius of curvature is smaller. A sharp tip therefore develops a strong local electric field, approximately related to surface charge density by E = σ/ε₀ just outside the surface. This helps initiate ionisation and discharge near the rod rather than the building. Option A is correct; sharpness does not make an insulator or destroy charge.

Correct answer: A. Polarization makes the nearer side effectively opposite, producing attraction

Explanation: A charged comb polarizes the molecules in initially neutral paper. The side nearer the comb acquires an induced charge of opposite sign, while like-sign induced charge is displaced farther away. Because Coulomb attraction is stronger at the smaller separation, the nearer opposite charges produce a net attractive force, even though the paper’s total charge remains zero. Therefore option A is correct; the paper does not become metal or lose charge.

Correct answer: A. Zero internal electric field, equal potential, and excess charge on surface

Explanation: Electrostatic equilibrium means that the conductor’s free charges have no net tendency to move. If an electric field existed inside, charges would continue moving, so the internal field must be zero. Since E = -dV/dr, zero field throughout the conductor means its potential is constant, or the conductor is equipotential. Any excess charge remains on the surface. Hence option A gives the complete identity.

Correct answer: A. Charge motion, distribution, induction, polarisation, and shielding depend on material nature

Explanation: The broad governing idea is that material structure controls how charges respond to electric fields. In conductors, mobile carriers redistribute readily, support electrostatic induction, and enable shielding. In insulators, charges are mainly bound, so charge remains more localized and polarization is important. Hence option A correctly combines the major behaviours. B ignores material differences, C reverses the carrier property of insulators, and D confuses charge with mass.