Electric Charges and their Conservation

Charge is conserved when a drop divides, so the sum of the charges of the two daughter drops must equal the original charge. Let the unknown charge be q. Then (+4e) + q = +11e, giving q = +11e − 4e = +7e. Thus option A is correct. A negative value would make the total only −3e, while +15e would exceed the conserved charge.

The governing concept is quantization of electric charge: Q = ne, where e = 1.6 × 10^-19 C. Since the charge is positive, the object has a deficiency of electrons, meaning electrons were lost. Thus n = Q/e = (9.6 × 10^-19)/(1.6 × 10^-19) = 6. Therefore, six electrons were lost, so option A is correct. The other numbers do not satisfy the charge-to-electron ratio.

The negative sign indicates that the particle has excess electrons. Using charge quantization, the number of excess electrons is n = |Q|/e, where e = 1.6 × 10^-19 C. Therefore n = (1.12 × 10^-18)/(1.6 × 10^-19) = 1.12/0.16 = 7. Hence the particle contains seven excess electrons, making option A correct. The other choices result from an incorrect division or exponent handling.

The governing idea is conservation of charge in an isolated conductor. A nearby charged rod attracts or repels the conductor’s mobile electrons, producing charge separation: one side becomes relatively positive and the other relatively negative. However, no charge crosses the conductor’s boundary because there is no earth connection. The algebraic sum therefore remains zero. Option A or B confuses polarization with net charging, and option D has no physical basis.

In a metal, a negative net charge means that the conductor has an excess of mobile electrons. When it is connected to Earth, the potential difference allows these excess electrons to flow away through the conducting path into the Earth. Their departure reduces the negative charge until neutrality is reached. Conventional current would be described in the opposite direction, but the question asks specifically about electron motion. Therefore option B is correct.

The microscopic model of a metal contains a lattice of positive ion cores and mobile conduction electrons. During ordinary charging, electrons can be removed from or added to the metal, whereas protons remain bound inside nuclei and do not travel through the conductor. A positive net charge therefore means that the number of electrons is smaller than the balancing positive charge of the ion cores. Heating alone does not explain a definite positive net charge. Hence option B is correct.

Charge redistribution in a metal is governed by the mobility of conduction electrons. The positive ion cores, which contain the protons, occupy nearly fixed lattice positions, and neutrons are also confined within nuclei. When an electric influence is applied, the mobile electrons shift until electrostatic equilibrium is established. Consequently, the redistribution of excess or deficient charge occurs mainly through free electrons, not through the movement of protons or entire nuclei. Therefore option C is correct.

Electric charge is an algebraic quantity, and an electron carries charge −e. The initial charge is +2e, while receiving three electrons changes the charge by −3e. Therefore, q_final = +2e − 3e = −e. Hence option C is correct. Option A incorrectly treats the received electrons as positive, whereas options B and D use incorrect magnitudes or signs.

An electron has charge −e, so losing an electron produces a change of +e in the object’s charge. For two lost electrons, the change is +2e. Starting from −5e, the final charge is q_final = −5e + 2e = −3e. Thus option C is correct. The other choices either add the lost negative charge incorrectly or reverse the sign without justification.

The total charge is the algebraic sum of all signed charges. In A, 4q − 6q + 2q = 0. In B, 5q − 2q − q = 2q, which matches the required positive value. In C, −5q + 2q + q = −2q, and in D, q − 2q + 4q = 3q. Thus only option B gives a total charge of +2q.

To find the net charge, add each signed term algebraically. Option A gives 3q − 3q = 0. Option B gives 2q − 5q + 3q = 0, and option D gives 6q − 2q − 4q = 0. In option C, the sum is −4q + q + 2q = −q, which is not zero. Therefore option C is the only correct answer.

The governing concept is algebraic addition of charge, consistent with conservation of charge. The initial charge is +q and the received charge is −2q, so the final charge is q + (−2q) = −q. Therefore option C is correct. Option B ignores the received charge, while options A and D result from adding magnitudes or mishandling the signs.

Electric charge is actually quantized: q = ne, where e is the elementary charge. However, e = 1.6 × 10⁻¹⁹ C is extremely small, and a macroscopic body can contain or exchange an enormous number of electrons. The allowed steps are therefore too tiny to notice in ordinary measurements, making charge appear continuous. Thus D is correct.

Rubbing two different materials can cause electrons to move from one surface to the other because the materials hold electrons differently. It does not create net charge from nothing, break protons, or normally charge neutrons. One object becomes negative and the other positive by equal and opposite amounts, so the total charge of the combined system remains conserved. Hence D is correct.

Electric charge is a signed physical quantity, so the plus or minus sign carries essential information. The safest first step is to write every initial, transferred and final charge with its sign before performing any addition or subtraction. For example, +q + (−2q) = −q, not 3q. Therefore D is correct; ignoring signs or units can invalidate the calculation.

The governing concept is algebraic addition of charge, consistent with conservation of electric charge in an isolated system. Assign signs to the three values: (+5e) + (−8e) + (+2e) = (5 − 8 + 2)e = −1e. Therefore the net charge is one negative elementary charge, so option A is correct. Option B reverses the sign, option C misadds the values, and option D ignores the nonzero result.

Charge is quantised in integral multiples of the elementary charge. The magnitude of one electron’s charge is e = 1.6 × 10⁻¹⁹ C. Therefore, the number of excess electrons is n = |Q|/e = (11.2 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹) = 7. The negative sign indicates an excess of electrons, not a negative number of particles. Hence option A is correct.

For an isolated macroscopic body, charge quantisation is written as Q = ne, where n is an integer and e is the elementary charge. A value of 2.5e would require n = 2.5, which is not an integer, so it is not allowed in the school-level model. The sign, size, and unit do not make the claim invalid. Thus option A is correct.

Charge is conserved during contact, and identical conducting spheres share the final charge equally. Initially, the two-sphere system has total charge +8 C + 0 C = +8 C. After contact, each sphere receives half of this amount: +8/2 = +4 C. The first sphere therefore changes from +8 C to +4 C, making option A correct.

Use conservation and add the charges of the two parts algebraically. Let the remaining charge be Q. Then Q + (−9 C) = +4 C, so Q = +4 C + 9 C = +13 C. The remaining part must therefore carry a positive thirteen-coulomb charge. Option B has the wrong sign, while options C and D result from incorrect subtraction or failure to account for the negative charge. Hence option A is correct.

The relevant principle is conservation of charge. An initially neutral system has total charge zero, and internal separation or transfer cannot change that total. If one part has +11e, the rest must have charge q satisfying +11e + q = 0. Therefore q = −11e. Option A supplies the equal and opposite charge; B would make the total +22e, C would leave +11e, and D would make the imbalance larger.

Charge conservation requires the final total charge to equal the initial total charge, which is zero. If one created particle has charge +e and the other has charge q, then +e + q = 0, giving q = −e. Thus the second particle must carry one negative elementary charge. A second +e would produce +2e, zero would leave +e, and +2e is even less compatible. Hence option A is correct.

Electric charge is quantized in integral multiples of the elementary charge e = 1.6 × 10⁻¹⁹ C. The number of elementary charges is n = Q/e = (4.8 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹) = 3. Because the charge is positive, the body has lost three electrons rather than gained them. Therefore option A is correct. Options B has the wrong sign, C has the wrong count, and D is physically impossible for isolated charge.

The magnitude of one elementary charge is e = 1.6 × 10⁻¹⁹ C. The number of electrons is found from n = |Q|/e = (14.4 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹) = 14.4/1.6 = 9. The negative sign indicates an excess of electrons, not a deficiency. Hence the body has nine excess electrons, making option A correct; the other numbers come from incorrect division or ignoring the given magnitude.

The governing concept is earthing and the movement of excess charge. A negatively charged metal sphere contains an excess of electrons. When it is connected to the Earth, the Earth acts as a very large charge reservoir and accepts those excess electrons until the sphere reaches the same electrical potential as the Earth. Therefore, electrons move from the sphere to Earth. Option B gives the opposite direction, while options C and D do not describe the earthing process.