Electric Charges and their Conservation

For an isolated system, no net charge can enter or leave. The conservation law therefore requires Q_initial = Q_final, unless the system boundary has been changed or an external interaction has been included. Here the claimed change is +2 C to +5 C, so the total changes by +3 C without an allowed source. Hence conservation of charge is violated, not quantization or the unit definition.

A body is positively charged when its net charge is positive, usually because it has fewer electrons than the number of positive charges in its matter. It does not follow that every electron has disappeared. A positively charged object can contain many electrons; only the balance is positive. Therefore “it has no electrons” is not necessary. Options B, C, and D express the required net imbalance in suitable terms.

Negative charging means that the total negative charge exceeds the total positive charge. In ordinary matter this commonly results from extra electrons, but it does not require the complete absence of protons. A negatively charged body normally still contains protons in its nuclei. Thus “it has no protons” is not necessary, whereas the other choices describe a negative net charge or its electron-excess interpretation.

Electrostatic repulsion requires interacting net charges of the same sign. A neutral object can be attracted by a charged object because of polarization, but it does not provide the usual basis for mutual repulsion in this school-level situation. Opposite charges attract, not repel. Therefore observing repulsion is the most certain evidence among these choices that both bodies are charged and have the same sign.

The governing principle is conservation of charge. Electrostatic induction only redistributes the conductor’s existing positive and negative charges: one region may become electron-rich and another electron-deficient. Without earthing, there is no conducting path for charge to enter or leave, so the algebraic sum of charge remains unchanged. Since the conductor started neutral, its net charge stays zero. Therefore option A is correct; B and C describe only local induced effects, not the total charge.

Charge conservation governs this situation. Rearrangement moves electrons from one part of a neutral body to another, producing separated positive and negative regions, but it does not create or destroy net charge. If no charge is transferred to or from the body, the total remains zero. Therefore option A is correct. Options B and C confuse a local induced effect with the body’s net charge, and option D violates conservation of charge.

Use charge quantization and add the signed particle charges. Each proton contributes +e and each electron contributes −e. Therefore the total is 7(+e) + 9(−e) = 7e − 9e = −2e. The system consequently has two elementary charges of negative sign, so option A is correct. Option B reverses the sign, option C adds magnitudes without cancellation, and option D incorrectly assumes equal numbers of protons and electrons.

The signed-charge rule gives +e for every proton and −e for every electron. Thus q = 12(+e) + 15(−e) = 12e − 15e = −3e. The system therefore has a net charge of negative three elementary charges, making option A correct. Option B has the wrong sign, option C adds the two populations instead of subtracting opposite charges, and option D would require equal numbers of protons and electrons.

A neutral atom initially has equal positive and negative charge, so its net charge is zero. Removing one electron removes one unit of negative charge; removing two removes −2e from the atom. The remaining atom therefore has q = 0 − (−2e) = +2e and becomes a doubly positive ion. Hence option A is correct. Option B has the wrong sign, C ignores electron loss, and D accounts for only one removed electron.

A neutral atom starts with net charge zero. Every added electron contributes one negative elementary charge, −e. Adding three electrons changes the charge by 3(−e), so q = 0 − 3e = −3e. The atom consequently becomes a negative ion with three elementary charges of net negativity. Option A is correct; B reverses the sign, C ignores the added particles, and D counts only one of the three electrons.

The quantisation of charge gives Q = ne, where e = 1.6 × 10⁻¹⁹ C is the magnitude of the elementary charge. Thus n = |Q|/e = (9.6 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹) = 6. The negative sign indicates an excess of electrons, not a deficit. Therefore, six excess electrons are present; the other options result from incorrect division.

Charge conservation requires the total charge to remain +6 C. When identical conducting spheres touch, charge redistributes until both have the same electric potential; for identical spheres this means equal charges. Therefore each final charge is (+6 C)/2 = +3 C. Option B ignores sharing, option C ignores the conserved charge, and option D doubles the total charge.

The governing principles are conservation of charge and equal sharing by identical conductors. The initial total charge is +9 C, and touching allows it to redistribute among all three spheres. Since the spheres are identical, each receives one-third of the total: Q_final = +9/3 = +3 C. The other choices either fail to divide the charge equally or do not conserve the total charge.

On contact, free charges move through the conducting material until the two spheres reach the same potential. Identical spheres have the same capacitance, so equal potential means equal final charge. If the initial charge is Q and the second sphere is neutral, conservation gives total charge Q; it is shared as Q/2 on each sphere. No charge is destroyed, and quantisation is not the reason.

In an ordinary solid, protons are positively charged but remain tightly bound inside atomic nuclei by the strong nuclear interaction. They cannot move from atom to atom during common charging processes. Electrons in the outer regions are much more mobile, so a positively charged rod removes electrons from the body, leaving a net positive charge. Options B, C, and D contradict the properties of protons.

The conservation of charge principle applies during friction as well. Mechanical work overcomes microscopic attraction and transfers electrons from one material to the other; it does not manufacture or destroy electrons. One body becomes negatively charged and the other equally positively charged, so their algebraic total remains unchanged. Energy is used for transfer, not for creating net charge. Thus A is correct.

A scalar has magnitude without a spatial direction, and electric charge is scalar in that sense. However, positive and negative signs classify the two kinds of charge and determine algebraic addition. For example, +5 C + (−3 C) = +2 C. The signs do not represent vector direction, units, or an unnecessary convention. Therefore option A correctly explains their role in finding net charge.

Net charge is the algebraic sum of all charges, so the signs must be retained: Q_net = (+12 C) + (−15 C) = 12 − 15 = −3 C. The magnitude of the negative contribution is larger by 3 C, so the result is negative. Option B reverses the sign, option C adds magnitudes without signs, and option D incorrectly assumes cancellation.

Each proton contributes +e and each electron contributes −e. If there are fourteen more electrons than protons, the unmatched contribution is 14 × (−e) = −14e. Therefore the body has a net charge of negative fourteen elementary charges, or approximately −2.24 × 10⁻¹⁸ C. A positive result would require fourteen more protons, while zero would require equal numbers.

The net charge equals the positive contribution minus the negative contribution. Since the positive effect exceeds the electron effect by eleven elementary charges, Q_net = (11)e = +11e. The plus sign is required because the surplus is positive. A negative answer would mean electrons dominate, zero would mean exact balance, and +e would represent only one excess elementary charge rather than eleven.

The conservation rule requires the algebraic sum of the charges of all parts to equal the system charge. Let the unknown charge be q. Then (+9e) + q = 0, so q = −9e. Thus the remaining part carries negative nine elementary charges, making option A correct. Positive nine would give a total of +18e, zero would give +9e, and negative eighteen would give −9e overall.

Take the initial charge as +3e. Each gained electron contributes −e, so gaining two electrons changes the charge by −2e. The final charge is +3e − 2e = +e, or one positive elementary charge. Hence option A is correct. Option B incorrectly adds the electron charges as positive, while options C and D reverse the sign or overestimate the magnitude of the change.

The initial charge is −4e. Losing one electron removes a charge of −e, which increases the body's algebraic charge by +e. Losing three electrons therefore changes the charge by +3e. Thus Qfinal = −4e + 3e = −e, so option A is correct. The body remains slightly negative; losing electrons does not make it more negative, and three lost electrons are not enough to reverse the sign.

For an isolated system, contact can redistribute charge between the bodies but cannot change their combined charge. Add the initial charges algebraically: Qtotal = (+8e) + (−3e) = +5e. Therefore the total charge after contact remains positive five elementary charges, making option A correct. The other options result from ignoring the negative sign, subtracting in the wrong order, or assuming contact destroys the net charge.

Electric charge is quantised as q = ne, where n is an integer and e is the elementary charge. Electrons carry negative charge, whereas protons carry positive charge. Therefore, q = −13e means that the body has thirteen more electrons than protons. Neutrons have no net charge, and atomic nuclei are positively charged, so options C and D cannot explain a negative charge. Hence option A is correct.