Electric Charges and their Conservation

Correct answer: A. Positive charge

Explanation: A negative rod repels the conductor’s mobile electrons. When the conductor is connected to Earth, some of those electrons flow away into the Earth. Removing the Earth connection first isolates the electron-deficient conductor while the rod still maintains the separation. After the rod is removed, the deficit spreads over the conductor, so its net charge remains positive. The rod itself is not transferred to the conductor.

Correct answer: A. Because the cause of charge separation is removed and Earth removes the imbalance

Explanation: Induction charging depends on keeping the conductor connected to Earth while the external rod maintains charge separation. If the rod is removed first, its electric influence disappears, so the separated charges tend to redistribute and the conductor’s potential returns toward that of Earth. Because the Earth connection is still available, electrons can flow until the net charge is usually zero. The elementary charge and conductor’s mass do not change.

Correct answer: A. A positive change of 14 elementary charges

Explanation: For an isolated closed system, the algebraic sum of charge remains constant. If one part changes by ΔQ₁ = −14e, the change in the other part must satisfy ΔQ₂ = −ΔQ₁. Therefore ΔQ₂ = +14e, so the negative gain in the first part is exactly balanced by a positive change in the second. A negative change would reinforce the imbalance, while +28e is twice the required amount.

Correct answer: A. Their charges are equal and opposite, so total charge remains zero

Explanation: An electron carries charge −e, whereas a positron, its antiparticle, carries +e. When the pair is produced from a neutral initial state, the algebraic sum of the produced charges is (−e) + (+e) = 0. Hence the total charge before and after the process is unchanged. Options B and C assign the wrong signs, and D is unnecessary because neither particle loses its charge.

Correct answer: A. Nuclear positive charge increases by one elementary charge

Explanation: In beta-minus decay, a neutron changes into a proton, an electron, and an antineutrino. The emitted electron has charge −e, while the daughter nucleus gains one proton and therefore its positive charge increases by +e. The nuclear change and emitted charge sum to zero, preserving total charge. A decrease would correspond to a different nuclear process, not ordinary beta-minus decay.

Correct answer: A. Total charge remains constant and charge values can occur only in integral multiples of elementary charge

Explanation: Charge conservation states that the algebraic total charge of an isolated system remains unchanged during any allowed process. Quantization states that observable charge is restricted to integral multiples of e, the elementary charge. These principles are compatible: particles may transfer or rearrange charge, but the total remains fixed and each net value follows the quantization rule. The other statements deny one of these principles.

Correct answer: A. Because final distribution depends on the capacitance and size of the bodies

Explanation: When conducting bodies touch, charge flows until their electric potentials become equal, not necessarily until their charges become equal. For a conductor, Q = CV; therefore, at common potential, the final charges depend on the capacitances, which are related to size, shape, and surroundings. Equal sharing occurs only in special symmetric cases, such as identical isolated spheres. Conservation still fixes the sum of charges.

Correct answer: A. Positive eight elementary charges

Explanation: Electrostatic induction separates charges inside a conductor but does not create net charge when the conductor remains isolated and neutral. If one end has an induced charge of −8e, charge conservation requires the other region to have +8e, so the total remains zero. Option A is correct. Option B would make the conductor net negative, C ignores the compensating separated charge, and D violates conservation.

Correct answer: A. Because electrons can move between the conductor and Earth

Explanation: Charge conservation applies to a chosen isolated system, not necessarily to the conductor alone after it is connected to Earth. Earthing provides a conducting path through which electrons can enter or leave the conductor, changing its charge. The combined conductor–Earth system still conserves charge. Hence A is correct; protons and neutrons do not move or change in the ways described by B, C, and D.

Correct answer: A. Three

Explanation: Adding an electron changes a body's charge by −e. The initial charge is +e and the final charge is −2e, so the change is Δq = −2e − (+e) = −3e. A charge change of −3e requires three added electrons. Therefore option A is correct. One or two electrons would produce −0e or −1e respectively, while zero electrons cannot change the charge.

Correct answer: A. Eight

Explanation: Removing an electron increases charge by +e. The charge changes from −3e to +5e, giving Δq = +5e − (−3e) = +8e. Therefore eight electrons must have been removed. Removing the first three electrons neutralizes the body, and removing five more produces +5e. Thus A is correct; B counts only the second stage, C counts only neutralization, and D is insufficient.

Correct answer: A. There may be an error in measurement or interpretation

Explanation: For an ordinary free body, electric charge is quantized: q = ne, where n is an integer and e is the elementary charge. A result of +0.5e is not an allowed isolated-body charge under this model. The scientifically cautious conclusion is that the measurement, calibration, or interpretation should be checked. Thus A is correct; the result does not disprove quantization or imply a freely existing half-electron.

Correct answer: A. Because each would get five-thirds of an elementary charge, not an integral multiple for free charge

Explanation: Electric charge is quantized: an isolated body normally possesses an integral multiple of the elementary charge e. A purely equal division of +5e among three bodies would assign +5e/3 to each, which is not an integral multiple of e. Hence the simple macroscopic sharing model needs microscopic qualification. Option A expresses this issue; the other options incorrectly change the total charge, deny positive charge, or assert an unsupported negative result.

Correct answer: A. Because the boundary determines whether charge remains inside the system or can cross it

Explanation: Charge conservation is applied to a clearly specified system. If the chosen boundary is closed and isolated, the total charge inside remains constant; if charge crosses the boundary, the system’s internal charge can change even though charge is conserved overall. Therefore the boundary must be identified before writing the conservation equation. Option A is correct; a boundary neither changes charge signs nor creates charge, and its absence does not imply zero charge.

Correct answer: A. Because zero is also an integral multiple of elementary charge

Explanation: Charge quantization is expressed as q = ne, where e is the elementary charge and n must be an integer. The integer n may be positive, negative, or zero. Taking n = 0 gives q = 0, so a neutral body has an allowed quantized charge. Option A is therefore correct. Options B, C, and D are wrong because zero is included in the integer set and represents neither negative nor positive net charge.

Correct answer: A. Consider the system and surroundings together and check total charge

Explanation: The governing principle is conservation of charge for a properly chosen closed or isolated system. In an open subsystem, charge can cross the boundary, so the subsystem’s charge may increase or decrease. This is not a violation. Include the surroundings and account for charge entering or leaving; the combined total remains constant. Option A is correct, whereas B uses an incomplete system, C rejects a valid law, and D ignores negative charge.

Correct answer: A. Because charge can suddenly transfer through the body

Explanation: A shock is explained by a potential difference and the movement of mobile electrons. A charged object and a person may be at different electric potentials; on contact, charge can flow rapidly through the body toward a lower-potential path, often the ground. This brief current stimulates nerves and produces the shock sensation. Option A is correct. Charge is not destroyed, protons normally do not flow out of the body, and neutrons do not carry ordinary electric current.

Correct answer: A. Error in finding total algebraic charge

Explanation: Electric charge is a signed scalar quantity, so the net charge must be found by algebraic addition. For example, (+5q) + (−3q) = +2q, whereas adding magnitudes gives 8q and produces a wrong result. Ignoring signs is therefore an error in calculating total algebraic charge, making option A correct. The other options describe correct methods or concepts and do not identify the stated calculation mistake.

Correct answer: A. Outward for positive and inward for negative

Explanation: Electric-field direction is defined using a hypothetical positive test charge. A positive source charge repels that test charge, so the field points radially outward. A negative source charge attracts it, so the field points radially inward. Therefore option A is correct. Options B and C incorrectly apply one direction to both signs, while D ignores the force reversal caused by changing the source-charge sign; the magnitude is treated separately from direction.

Correct answer: A. One-sixth

Explanation: A cube has six equivalent faces, and a charge at its exact centre has identical geometric surroundings with respect to every face. By symmetry, the total flux Φ_total = q/ε₀ is divided equally among the six faces. Therefore, flux through one face is Φ_total/6 = q/(6ε₀), or one-sixth of the total. The result depends on central placement; options B, C, and D violate the sixfold symmetry.