Electric Charges and their Conservation

Each gained electron contributes a charge of −e. Therefore, gaining eight electrons changes the charge by −8e. Adding this to the initial +3e gives +3e − 8e = −5e. Thus option B is correct. The particle changes from positive to negative because it gains more negative elementary charges than its initial positive excess. +11e incorrectly ignores the electron sign, and −11e adds magnitudes rather than algebraic charges.

The total charge remains conserved during splitting. Let q be the charge on the third drop. The charge equation is (−4e) + (+2e) + q = −13e. The first two drops together have −2e, so q = −13e − (−2e) = −11e. Therefore option A is correct. The positive alternatives have the wrong sign, and −7e would make the total −9e rather than −13e.

First apply conservation of charge during combination: the total charge is +5e + (−9e) = −4e. The combined drop therefore carries −4e. When it splits into two equal-charge drops, this conserved charge is divided equally: −4e/2 = −2e on each drop. Hence option A is correct. −4e is the charge of the combined drop before splitting, not of each final drop.

A free object can possess charge only in integral multiples of the elementary charge e = 1.6 × 10^-19 C. Test each option by dividing by e. For option C, (8.0 × 10^-19)/(1.6 × 10^-19) = 5, an integer, so it is physically allowed. The ratios for A, B, and D are 3.5, 4.5, and 1.5, respectively, and are not integers. Thus option C is correct.

Charge quantization states that the net charge on an observable body is Q = ne, where n is an integer and e is the elementary charge. The values +3e and -4e correspond to integer values of n, while zero corresponds to n = 0 and is also allowed. However, +e/2 would require n = 1/2, which is not an integer. Therefore option C violates charge quantization.

The governing principle is conservation of charge: the sum of the final charges must equal the initial total, +2q. Let the unknown charge be x. Then 5q - 6q + x = 2q, so -q + x = 2q and x = 3q. Equivalently, the known charges total -q, so +3q is needed to reach +2q. Therefore option A is correct; the other signs or magnitudes do not conserve charge.

In a closed system, total charge is conserved, so any increase in one part must be balanced by an equal decrease in the other part. The first part changes by ΔQ₁ = (+4q) - (-2q) = +6q. Therefore the other part must undergo ΔQ₂ = -6q, because ΔQ₁ + ΔQ₂ = 0. Hence option B is correct. Option A gives the same-direction change and would alter the total charge.

Let the charge of the remaining system be x. Conservation of charge requires the sum of the known part and the remaining part to equal the total: 7q + x = -5q. Subtracting 7q from both sides gives x = -5q - 7q = -12q. Thus the remaining system has charge -12q, so option A is correct. The positive alternative incorrectly ignores the sign of the total charge.

Electric charge is conserved during the process, so the final total must remain zero. If the unknown charge is x, then 2e + 3e + x = 0. The known particles together carry +5e, so x must be -5e to cancel this positive charge. Therefore option B is correct. A would make the total +10e, while C and D would leave a nonzero final charge.

Breaking into particles does not destroy or create net electric charge, so the final sum must equal the initial charge, -2e. Let the second particle have charge x. Then +4e + x = -2e, giving x = -2e - 4e = -6e. Thus the other particle carries six negative elementary charges, and option A is correct. The other choices fail to reproduce the initial total charge.

This is charging by induction, governed by conservation of charge and the motion of free electrons in a conductor. The negative rod repels electrons, so while the conductor is earthed some electrons flow into the Earth. The earth connection must be removed first; after that, removing the rod leaves an electron deficiency. Hence the conductor remains positively charged. Option A would require electrons to enter, while option C describes only temporary polarization without earthing.

A positive conductor has fewer electrons than required for electrical neutrality. On connection to Earth, the Earth acts as a huge charge reservoir, so electrons flow from Earth into the conductor until its deficit is compensated and its net charge becomes zero. Charge has not disappeared: the conductor gains negative charge while the Earth loses an equal amount, so the total charge of the conductor–Earth system remains conserved. Protons and neutrons do not undergo these changes.

Charge conservation applies to a suitably chosen complete system, not necessarily to one part of an open system. When a conductor is connected to Earth, electrons can enter or leave it, so the conductor’s individual charge changes. If the conductor and Earth are treated together as the system, every electron transferred is counted: charge lost by one part is gained by the other. Thus total charge remains constant. An isolated conductor cannot exchange charge in this way.

Use algebraic addition of charge, with positive and negative signs retained: Q_final = Q_initial + Q_entering = (+3q) + (−8q). The magnitudes differ by 8q − 3q = 5q, and the larger magnitude is negative, so Q_final = −5q. Option A has the wrong sign, while options C and D incorrectly add the magnitudes without respecting the opposite signs. Thus option B is correct.

A charge that leaves must be subtracted algebraically from the initial charge. Therefore Q_final = Q_initial − Q_leaving = (−9q) − (−4q) = −9q + 4q = −5q. Equivalently, removing negative charge makes the conductor less negative by 4q. Option A would result from adding the two negative magnitudes, while options C and D give an incorrect positive sign. Hence option B is the only consistent result.

In a closed system, the charge of the whole system equals the sum of the charge of the subsystem and the remainder. Let the remainder be Q. Then (−5 C) + Q = +12 C, so Q = +12 C + 5 C = +17 C. Therefore option C is correct. A results from subtracting magnitudes without signs, while B and D have the wrong sign.

An isolated system cannot exchange net charge with its surroundings, so its total charge remains constant. If one part loses +2q, the charge removed from that part must appear in the remaining part. Thus the remaining part undergoes a change of +2q. Option C is correct; A reverses the transfer, B ignores conservation, and D doubles the transferred amount.

The elementary charge is e = 1.6 × 10^-19 C, so the initial charge −3.2 × 10^-19 C equals −2e. When two electrons are lost, the object loses charge −2e; equivalently, its charge increases by +2e = +3.2 × 10^-19 C. Therefore q_final = −3.2 × 10^-19 + 3.2 × 10^-19 = 0. Option C is correct. Options A and B fail to account for the positive change caused by electron loss, while D gives only the change rather than the final charge.

The elementary charge is e = 1.6 × 10^-19 C, so the initial charge +6.4 × 10^-19 C equals +4e. Gaining four electrons adds −4e, or −6.4 × 10^-19 C. Hence q_final = +4e − 4e = 0. Option C is correct; A ignores the gained electrons, B reverses the complete cancellation, and D adds magnitudes incorrectly.

When conducting spheres touch, charge can flow until the appropriate electrostatic condition is reached, but contact does not create or destroy net charge. For unequal spheres, the final charges need not be equal because their sizes and capacitances differ. Nevertheless, the algebraic sum of their charges remains the initial total. Therefore option B is certainly correct.

For identical conducting spheres, touching causes the combined charge to divide equally. Initially the charges are q and 0, so after the first contact each sphere has (q + 0)/2 = q/2. The first sphere then touches an identical neutral sphere, so their combined charge q/2 is shared equally; the first sphere receives (q/2)/2 = q/4. Thus option C is correct.

Charge quantization states that the charge of a free object must be q = ne, where n is an integer and e = 1.6 × 10⁻¹⁹ C. For option D, n = (4.0 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹) = 2.5, which is not an integer. The other choices give 2, 4 and 6 elementary charges, so D is correct.

The negative charge means that the object has gained electrons. The number of excess electrons is found from N = |q|/e, with e = 1.6 × 10⁻¹⁹ C. Thus N = (2.4 × 10⁻¹⁸)/(1.6 × 10⁻¹⁹) = 15. Therefore option D is correct; the negative sign identifies excess electrons rather than changing their count.

A positive charge indicates that the object has fewer electrons than it had originally, so it has lost electrons. Using N = q/e, where e = 1.6 × 10⁻¹⁹ C, N = (2.08 × 10⁻¹⁸)/(1.6 × 10⁻¹⁹) = 13. Hence option D is correct. The positive sign indicates electron deficiency, not positive particles being added.

Conservation of charge applies to the complete isolated system, so the system must first be defined by a boundary. If charge crosses that boundary, the total charge inside the selected system can increase or decrease, even though the charge of the larger closed system remains conserved. Therefore option D is correct; the other choices are unrelated to charge accounting.