Electric Charges and their Conservation

Correct answer: C. In the cavity, \(E=0\) and \(V\) is the same everywhere, but its value is not necessarily zero.

Explanation: In electrostatic equilibrium, the conductor and the inner surface of its cavity are equipotential. Since the closed cavity contains no charge, the unique electrostatic solution for a boundary at constant potential is a constant potential throughout the cavity. Therefore, \(\mathbf{E}=-\nabla V=0\) everywhere. The external charge may change the constant value of the conductor's potential, so that value need not be zero. Option A is incorrect because it assumes that the potential must be zero.

Correct answer: D. Equal-magnitude negative charge \(-q\)

Explanation: In electrostatic equilibrium, the electric field within the conducting material is zero. Consider a Gaussian surface lying entirely in the conductor and enclosing the cavity. The electric flux through this surface is zero, so Gauss's law requires the net enclosed charge to be zero. Therefore, the charge \(+q\) inside the cavity induces a total charge \(-q\) on the inner surface. The induced charge may be distributed non-uniformly, but its total is always \(-q\), not \(-q/2\).

Correct answer: B. In an insulator, bound positive and negative charges undergo a small relative displacement or permanent dipoles align, whereas free charges redistribute in a conductor.

Explanation: In an insulator, charges are normally bound to atoms or molecules. An external field can slightly separate the positive and negative charge centres or align permanent molecular dipoles. In contrast, free electrons or other free charges redistribute through a conductor; at electrostatic equilibrium, the electric field inside the conductor is zero. Unlike option C, polarisation or induction alone does not change the net charge of an isolated object.

Correct answer: A. The desired charge may not be retained

Explanation: The governing concept is the required sequence in charging by induction. The nearby charged object separates charges in the conductor, while the earth connection allows charge to enter or leave. If the external object is removed first, the separation disappears while earthing remains, so charge can flow to or from Earth and the intended net charge may be cancelled. Option A is correct; the other choices violate conservation and conductor behaviour.

Correct answer: B. In charge separation the net charge generally does not change, whereas in actual charging the net charge changes

Explanation: The governing concept is conservation of total charge. In charge separation or induction without contact, positive and negative charges merely redistribute inside the object, so the object’s net charge generally remains unchanged. In actual charging, electrons are transferred to or from another body or Earth, changing the net charge. Option B states this distinction correctly. A reverses it, while C and D deny real charge redistribution or misidentify the materials involved.

Correct answer: A. \(\dfrac{|\sigma|}{\varepsilon_0}\)

Explanation: In electrostatic equilibrium, the electric field inside a conductor is zero, and the field immediately outside is normal to its surface. Applying Gauss's law to a small pillbox straddling the surface gives \(EA=\sigma A/\varepsilon_0\). Hence, \(E=|\sigma|/\varepsilon_0\). Option B applies to the field on either side of an isolated infinite charged sheet, not to the surface of a conductor.

Correct answer: A. \(+q\)

Explanation: The electric field inside the conducting material must be zero. Hence, a total induced charge \(-q\) appears on the inner surface of the cavity. Since the isolated conductor was initially neutral, its total charge remains zero; therefore, the outer surface must carry \(+q\) to balance the \(-q\) on the inner surface. Option \(0\) refers to the net charge of the conductor, not to the charge on its outer surface.

Correct answer: B. Because free charges redistribute on the outer surface and produce an induced field that cancels the external field in the empty interior.

Explanation: In electrostatic equilibrium, free charges in a conductor redistribute until the electric field within the conducting material is zero. If the closed conductor has an empty cavity, the external field induces charges on its outer surface. The field produced by these induced charges makes the net electric field in the empty interior zero. Option D is incorrect because, after equilibrium is reached, charges do not flow continuously.

Correct answer: B. When the electric field inside the conductor becomes zero and the conductor becomes equipotential

Explanation: At electrostatic equilibrium, the electric field inside a conductor is zero. Therefore, the electric force on a free charge, \(F=qE\), is zero, and no force remains to cause further rearrangement. The conductor consequently has the same potential throughout, so it is equipotential. In option C, a non-zero electric field would still exert a force on free charges. A conductor need not have zero net charge to be in electrostatic equilibrium.

Correct answer: A. The attraction of induced negative bound charge on the nearer side of the sphere is greater than the repulsion of induced positive bound charge on the farther side.

Explanation: The positive point charge slightly displaces bound charges in the atoms or molecules of the insulator; this is polarisation. The nearer side of the sphere becomes relatively negative and the farther side relatively positive, while the sphere remains overall neutral. The attraction on the nearer negative bound charge is stronger because electric force decreases with distance. Hence, it exceeds the repulsion on the farther positive side, producing a net attraction. Option B is incorrect because the forces on the two sides are not equal: their distances from the point charge are different.

Correct answer: B. Because any tangential component of the electric field would keep moving free charges along the surface, so charges redistribute until that component becomes zero.

Explanation: In electrostatic equilibrium, the free charges in a conductor are at rest. If the electric field had a component parallel to the surface, it would exert a force on free charges and make them move along the surface. Charges redistribute until the tangential component becomes zero. Thus, just outside the surface only the normal component remains, so the electric field is perpendicular to the surface. Option A states the correct fact that the field inside a conductor is zero, but it does not by itself explain why the field at the surface has no tangential component.

Correct answer: A. Negative charge (electrons)

Explanation: The positively charged rod attracts the conductor’s free electrons toward the side nearer the rod. When the conductor is earthed, electrons flow from the earth into the conductor and gather on the side facing the rod. Therefore, negative charge enters the conductor from the earth. In a metal, positive ions are fixed in the lattice, so positive charge does not flow through the conductor as in option B.

Correct answer: A. Polarisation

Explanation: The electric field of the charged rod slightly displaces or reorients the bound positive and negative charges in the insulator. The induced charge of opposite sign on the nearer side is closer to the rod, so its attraction is stronger than the repulsion due to the like induced charge on the farther side. No charge is transferred to or from the insulator, so its net charge remains zero. This is polarisation, not earthing or destruction of charge.

Correct answer: A. The electric field inside the conducting material is zero.

Explanation: In electrostatic equilibrium, free charges redistribute until the electric field inside the conducting material becomes zero. Since \(\mathbf{E}=-\nabla V\), \(\mathbf{E}=0\) means there is no change of potential within the conductor; hence the conductor and its surface are equipotential. Option B is not necessary: the electric field outside a conductor may be nonzero even though the conductor is equipotential.

Correct answer: A. An isolated spherical conductor kept away from external influences

Explanation: Because of the rotational symmetry of an isolated spherical conductor, every point on its external surface is equivalent. Hence, in electrostatic equilibrium, charge is distributed uniformly and the surface charge density is \\(sigma=Q/(4\\pi R^2)\\). In contrast, a nearby point charge makes the distribution non-uniform by induction, and irregular or sharp regions generally have greater charge density.

Correct answer: A. The external electric field redistributes the conductor’s free charges by induction, while its total charge remains unchanged.

Explanation: The electric field of the external charge exerts forces on the conductor’s free charges. They redistribute until the conductor reaches electrostatic equilibrium and remains equipotential. The external charge breaks spherical symmetry, so the surface charge density need not remain uniform. This is redistribution by induction; the conductor’s total charge does not change without charge transfer. Option C is incorrect because an equipotential conductor need not have uniform surface charge density.

Correct answer: A. They will redistribute along the surface until the tangential electric field becomes zero.

Explanation: A tangential electric field exerts a force along the surface on free charges, so they move and redistribute, reducing the electrostatic potential energy of the system. This redistribution continues until electrostatic equilibrium is reached, at which point the tangential component of the electric field at the surface is zero. In contrast, the normal component need not be zero; it can be associated with surface charge density.

Correct answer: A. When it is isolated on a dry insulating stand

Explanation: Charge can redistribute freely in a metal, but it does not disappear by itself. A conducting path to Earth is needed for charge to leak away. A dry insulating stand keeps the metal object isolated from Earth and minimises leakage. In contrast, a hand or an earthing wire provides a conducting path to Earth, while a damp floor can increase leakage because of moisture.

Correct answer: A. When both conductors attain the same electric potential

Explanation: A potential difference between conductors in contact causes net charge transfer. The transfer stops when the connected system reaches electrostatic equilibrium and both conductors are at the same potential. Equal potential does not imply equal total charge; conductors of different size or capacitance can retain different final charges.

Correct answer: A. No, the conductor may have net excess charge on its outer surface.

Explanation: In electrostatic equilibrium, free charges redistribute until the electric field within the conducting material becomes zero. The conductor can still have net excess charge, which resides on its outer surface. Therefore, a zero internal field does not prove that the conductor is neutral. Option C is incorrect because excess charge does not remain uniformly distributed throughout the conductor’s volume in electrostatic equilibrium.

Correct answer: A. Charge can flow between the conductor and Earth, while Earth’s enormous capacitance makes the change in its potential negligible.

Explanation: Electrons can flow between an earthed conductor and Earth until the conductor reaches Earth’s potential. Earth has an extremely large effective capacitance, so accepting or supplying an ordinary amount of charge produces a practically negligible change in Earth’s potential. Therefore, Earth acts as a huge charge reservoir. Option B is incorrect because charge must actually flow for the conductor’s charge to change, and the conductor’s final charge is not necessarily zero in every situation.

Correct answer: A. \(-q,\; Q+q\)

Explanation: At electrostatic equilibrium, the electric field inside the conducting material is zero. Hence, a Gaussian surface just within the conductor around the cavity must enclose zero net charge, so the inner surface acquires an induced charge \(-q\). Because the isolated conductor initially had total charge \(Q\), its total charge must remain \(Q\). Therefore, the outer surface must carry \(Q-(-q)=Q+q\). Option C has the correct inner-surface charge but does not preserve the conductor's total charge \(Q\).

Correct answer: A. Because sharing depends on final equilibrium and size

Explanation: When conducting bodies touch, electrons redistribute until their electrical potentials reach equilibrium. Charge conservation fixes the sum of the final charges, but it does not require the two individual charges to be equal. For unequal bodies, size, shape, and capacitance can differ; at equal potential, charge generally divides in proportion to capacitance. Equal sharing is a special result for identical conductors, not a universal consequence of conservation.

Correct answer: A. Define the system boundary and find total initial charge

Explanation: The first step in a conservation problem is to define the system and decide whether it is isolated. Then identify every relevant charge and add algebraically, preserving positive and negative signs, to obtain the initial total charge Qinitial. Conservation requires Qfinal = Qinitial for that chosen system. Adding magnitudes, ignoring electron transfer, or considering only the final state can destroy the cancellation and produce an incorrect result.

Correct answer: A. Because electrons can move back to earth

Explanation: Induction depends on the order of operations. A positive rod attracts electrons toward the near side, and earthing allows additional electrons to enter the conductor from Earth. If the rod is removed while the earth connection is still present, the external electric influence disappears. The conductor can then exchange electrons freely with Earth and return toward neutrality before the connection is broken. Thus option A gives the relevant mechanism; the other choices violate ordinary electrostatic behavior.