Conductors and Insulators

Correct answer: B. Because potential difference creates an electric field and free charges will move

Explanation: The governing relation is E = −∇V: a spatial potential difference indicates a non-zero electric field. That field exerts force F = qE on free charges in the conductor, producing motion or current. Electrostatic equilibrium requires no sustained motion and therefore a constant potential throughout a connected conductor. A potential difference does not reduce mass, make a conductor transparent, or destroy charge. Hence option B is correct.

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

Explanation: For a +q charge inside the cavity, electrostatic equilibrium induces −q on the inner surface. Because the conductor is earthed, charge can flow between the conductor and Earth until the conductor reaches Earth potential. In the ideal symmetric arrangement, the outer surface need not retain any charge and can have Q_outer = 0. Thus option B is correct; the other choices incorrectly claim an unavoidable sign or an infinite charge.

Correct answer: C. Because charge density and field can be higher there

Explanation: At a sharp metallic tip, charge tends to concentrate because the local radius of curvature is small. The increased surface charge density produces a large electric field near the tip, according to E = σ/ε₀ for the nearby surface region. If the field becomes sufficiently strong, it can ionize the surrounding air and initiate corona or discharge. Hence option C is correct; the other choices do not explain ionization or field enhancement.

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

Explanation: The governing relation is Q = CV, together with the requirement that connected conductors reach a common potential. Unequal conductors generally have different capacitances, so at the same final potential their charges satisfy Q1/Q2 = C1/C2 rather than being equal. Thus C is correct. Charge can flow to both conductors, the smaller one still has electrons, and total charge is conserved; therefore A, B, and D are incorrect.

Correct answer: C. Charge density and electric field can be higher at a sharp tip

Explanation: On a conductor, charge density is not uniform when the surface has different curvatures. At a sharp tip, the radius of curvature is small, so charge tends to concentrate there and the local surface charge density becomes large. Since the electric field just outside a conductor is related to surface charge density by E = σ/ε₀, the field is strong near the tip. Therefore option C is correct; a tip does not destroy charge or reduce mass.

Correct answer: A. Bound positive and negative charges can shift slightly

Explanation: An external electric field usually cannot make the bound electrons in an insulator flow freely through the material. However, it can produce a small relative displacement between the positive nucleus and the surrounding negative charge cloud, or orient existing molecular dipoles. This microscopic separation creates polarisation and an induced dipole moment. Therefore option A is correct; the insulator need not become a conductor and no charge is destroyed.

Correct answer: B. Polarisation brings an opposite charge effect closer on the nearer side

Explanation: Paper is an insulator, but its molecules can be polarised by the electric field of a charged comb. The positive and negative charge centres shift slightly, so the side nearer the comb acquires an induced charge effect opposite to the comb’s charge. Because this opposite effect is closer, its attractive force is stronger than the repulsive force from the farther side, producing a net attraction. Thus option B is correct without requiring net charge transfer.

Correct answer: C. Charge separation caused by induction

Explanation: Net charge is the algebraic sum of all charge on the conductor, so equal positive and negative induced charges can give a total of zero. An external charged object or electric field can cause the conductor’s mobile electrons to redistribute: one surface region becomes electron-rich and negative, while another becomes electron-deficient and positive. This is electrostatic induction, not charge destruction or a change into an insulator. Therefore option C is correct.

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

Explanation: The broad governing conclusion is that material structure controls the mobility of charge carriers and the response of bound charges. Conductors permit redistribution, induction and effective electrostatic shielding, whereas insulators retain localized charge and can polarise. Hence charge motion, distribution and field response depend on the material. Option B states this complete conclusion. A and C are false generalisations, while D confuses charge behaviour with mass alone.

Correct answer: C. The total outer-surface charge can be zero

Explanation: The charge +q inside the cavity still induces −q on the inner surface, because the field inside the conducting material must vanish. However, earthing fixes the conductor’s potential and allows charge to flow between the conductor and Earth. The outer surface therefore need not retain the charge required for an isolated neutral conductor; in the usual symmetric setup it can have zero total charge. Hence option C is correct.

Correct answer: B. Surface charge density

Explanation: For a conductor in electrostatic equilibrium, the field just outside the surface is related to the local surface charge density by E_normal = σ/ε₀ in vacuum. Thus a region with larger σ has a stronger nearby normal electric field. This is a local relation, not a property determined by colour, name, or sound. Therefore option B is correct, and the wording ‘local value’ appropriately points to local surface charge density.

Correct answer: A. Surface charge density can be higher there

Explanation: A conductor in electrostatic equilibrium can have a nonuniform surface charge distribution. At a sharp tip, the small radius of curvature causes charges to crowd more closely, producing a larger local surface charge density σ. Since the external field near the surface is approximately σ/ε₀, the field becomes stronger there. The conductor does not become an insulator and charge is not destroyed, so option A is correct.

Correct answer: C. Because charge density and electric field can be higher there

Explanation: At a sharp metal tip, charge accumulates more densely because the radius of curvature is small. The resulting large local surface charge density produces a strong electric field, approximately E = σ/ε₀ just outside the conductor. If this field is sufficiently high, it can ionise nearby air and start corona or electrical discharge. Thus option C gives the correct chain of reasoning; the other statements contradict conductor physics.

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

Explanation: On contact, two conductors exchange free charge until their potentials become equal, not until their charges become equal. For an isolated conductor, Q = CV; unequal size or shape generally gives unequal capacitance. At the same final potential, the conductor with larger capacitance can hold more charge. Therefore B is correct, while the other options contradict charge conservation or conductor properties.

Correct answer: C. Charge density and the electric field can be higher at the sharp tip

Explanation: In electrostatic equilibrium, excess charge on a conductor resides on its surface and is more concentrated where the radius of curvature is small. A sharp tip therefore has a relatively high surface charge density σ, and the field just outside the surface is approximately E = σ/ε₀. The stronger local field can promote ionisation of nearby air and a controlled discharge. A tip does not destroy charge, act as an insulator, or reduce mass.

Correct answer: C. Polarisation creates an opposite-charge region nearer the comb

Explanation: Paper is an insulator, but its molecules can become polarised in the electric field of a charged comb. The side nearer the comb acquires an induced charge effect opposite to the comb’s charge, while the like effect is displaced farther away. Because the attractive force is stronger at the smaller separation, the net force is attraction, even though the paper may have zero net charge. The comb neither increases gravity nor destroys charge.

Correct answer: B. They become equipotential but their capacitances may differ

Explanation: Conductors brought into contact exchange charge until their potentials become equal, not necessarily until their charges become equal. For an isolated conductor, the relation is approximately Q = CV. Unequal conductors generally have different capacitances, so at the common potential their charges satisfy Q1/Q2 = C1/C2 rather than Q1 = Q2. Option B is correct. Charge can reach either conductor, electrons exist in both, and total charge is conserved.

Correct answer: A. Excess charge moves to the outer surface

Explanation: A conductor contains mobile charge carriers. If excess charge remained in its bulk, it would create an internal electric field and exert forces that would continue moving the carriers. In electrostatic equilibrium, this redistribution ends with excess charge on the outer surface, so the excess volume charge density inside the conducting material is zero. Option A is correct. Charge is not converted into mass, electrons are present, and the result is not due to lack of space.

Correct answer: B. Such energy is not considered inside

Explanation: The electric-field energy density in vacuum or a linear dielectric is u = 1/2 εE². Therefore, in the conducting material where the electrostatic field is zero, the local energy density associated with that field is also zero. Option B is the intended answer: no electric-field energy is stored in the conductor’s bulk under this ideal condition. This does not rule out energy in fields outside the conductor or in a cavity containing a field.

Correct answer: B. Because surface charges rearrange and cancel the effect inside

Explanation: This is the principle of electrostatic shielding. An external charge exerts forces on the conductor’s free charges, causing them to redistribute over its surfaces. In electrostatic equilibrium, the induced surface-charge field cancels the external field throughout the empty closed cavity, so the net field there is zero. Therefore B is correct. Colour, absence of air, and destruction of the external charge have no role in the result.

Correct answer: A. Because it will move surface charges

Explanation: In electrostatic equilibrium, free charges in a conductor must have no unbalanced force. A component of electric field parallel to the surface would exert tangential force F = qE_parallel on mobile surface charges, causing them to drift along the surface. Their movement would continue until that component became zero. Hence A is correct. The normal component may exist and is related to surface charge density; the other options do not describe electrostatic behavior.

Correct answer: A. Zero internal field, uniform potential, and excess charge on the surface

Explanation: Electrostatic equilibrium means that free charges in a conductor have stopped experiencing a net driving force. Consequently, the electric field throughout the conducting material is zero; since the field is zero, the potential is constant throughout the conductor and its surface. Any excess charge resides on the surface, with its distribution depending on shape. Thus A is the complete statement; the other choices contradict equilibrium and charge behaviour.

Correct answer: B. Few free charges, localised charge possible, and polarisation possible

Explanation: An insulator contains very few mobile charge carriers, so an applied electric field cannot make charge spread freely throughout it. Excess charge can remain localised, while the bound positive and negative charges may shift slightly, producing polarisation. Hence option B correctly combines the characteristic behaviours. Option A describes a conductor, and C incorrectly denies polarisation and assumes a conductor-like surface distribution. D is incompatible with charge conservation and electric fields.

Correct answer: A. When the conductor is in electrostatic equilibrium

Explanation: In electrostatic equilibrium, free charges in a conductor have redistributed until there is no net force driving further motion. The excess charge can remain on the outer surface, while the field inside the conducting material is zero. These facts are not contradictory: surface charge creates the appropriate external field and maintains zero field within the conductor. Hence A is correct. Becoming an insulator, converting charge into mass, or removing the surface is not the relevant condition.

Correct answer: A. Induction, polarisation, shielding, and charge distribution

Explanation: The common governing idea is charge mobility. Free charges in conductors move or redistribute under an external electric influence, producing electrostatic induction, surface charge rearrangement, and shielding. Charges in insulators are comparatively bound, so an applied field mainly produces polarization through small displacements or dipole alignment. Thus option A correctly combines all four related phenomena; the other options describe unrelated or incomplete properties.