Electric Charges and their Conservation

The governing principle is conservation of total charge in an isolated system. The charge change of the specified part is ΔQ = (+5q) − (−3q) = +8q. Since no charge enters or leaves the complete system, the remaining part must undergo an equal and opposite change: ΔQremaining = −8q. Therefore option B is correct; +2q and −2q do not represent the actual eight-q change.

The guaranteed result follows from conservation of charge, not from equal sharing. The initial total is (+8q) + (−2q) = +6q, and contact only redistributes charge between the conductors if the system is isolated. Since their sizes are unequal, their final charges need not be equal; equal potential, rather than equal charge, is the electrostatic condition. Therefore only option B is guaranteed.

Treat the departing charge with its algebraic sign. The conductor initially has −4q, and removing a charge of −7q means Qfinal = Qinitial − Qdeparting = (−4q) − (−7q) = −4q + 7q = +3q. Equivalently, losing negative charge is a positive change of 7q. Thus option B is correct; simply adding the magnitudes would incorrectly give −11q.

An electron carries charge −e. Giving away five electrons removes charge −5e, which changes the object’s charge by +5e. Receiving two electrons adds −2e. Starting from −6e, Qfinal = −6e + 5e − 2e = −3e. Therefore option A is correct. The result is not −13e because the first step is loss of negative charge, not addition of negative charge.

Electric charge is conserved in particle-creation processes. A particle and its antiparticle have equal charge magnitudes with opposite signs; for example, +e and −e add algebraically to zero. Thus, if the initial system has zero charge, their creation leaves the total unchanged, so option B is correct. Option A gives a nonzero net charge, while C and D contradict the conservation law.

A scalar quantity is described by magnitude, and electric charge additionally has an algebraic sign indicating positive or negative value; it does not possess a spatial direction. Charges therefore combine by ordinary algebraic addition. Option B is correct. Option A incorrectly assigns direction, C confuses charge with force, and D describes neither the defining nature of charge nor the reason it is scalar.

Charge conservation depends on the system boundary. When a conductor is connected to Earth, electrons may flow between the conductor and the Earth, which acts as a huge charge reservoir. The conductor’s charge can therefore change, but the combined charge of conductor plus Earth remains conserved. Option B is correct; isolation would prevent such exchange, and force or electron absence does not define the conservation boundary.

A positively charged conductor has fewer electrons than it has in its neutral state. Earthing provides a conducting path to the Earth, which can supply electrons to the conductor. Electrons therefore move from Earth toward the positive conductor until its potential and charge are balanced. Option B is correct. The reverse direction applies to a negatively charged conductor, while protons do not flow through the metal.

A negatively charged conductor contains an excess of mobile electrons. When it is connected to Earth, the Earth provides a path for these excess electrons to leave the conductor and spread into the larger system. Hence electrons move from the conductor to Earth, making option B correct. Electrons do not move toward protons or neutrons as a direction rule, and the flow is not from Earth into an already negative conductor.

Charge conservation states that the algebraic total charge of an isolated system remains constant; charge can only be transferred between bodies. Quantization states that observable free charge is an integral multiple of the elementary charge, q = ne, where n is an integer. Option B contains both principles. The other options deny conservation, allow fractional elementary charge, or give incorrect claims about sign and units.

For an isolated system, charge conservation requires the sum of all parts to equal the stated total. Let the remaining charge be x. Then −4 C + 7 C + x = 9 C, so 3 C + x = 9 C and x = 6 C. Hence option A is correct. The other choices arise from ignoring the negative sign or using addition instead of the required difference.

When identical conducting spheres touch, charge redistributes equally while their total charge is conserved. Initially the total is q + 0 = q, so after the first contact each sphere has q/2. On the second contact, both spheres already have equal potentials and equal charges; no net transfer occurs. Each therefore remains at q/2, making option B correct. The other choices ignore the first equal sharing or conservation.

Charge is shared equally whenever identical conducting spheres touch. First, the total charge of the first pair is 8q, so after their contact each has 8q/2 = 4q. The second sphere, carrying 4q, then touches the neutral third sphere; their total is 4q and each receives 2q. Thus the third sphere ends with +2q, so option B is correct.

When the drops combine, their total charge is conserved: (+10e) + (−2e) = +8e. Splitting the combined drop into four equal-charge drops distributes this total equally, so each receives (+8e)/4 = +2e. Therefore option A is correct. Option C is the unsplit total, while B has the wrong sign and D uses an incorrect division.

The total charge remains −8e after the drop splits. The known three drops have charge (−1e) + (+2e) + (−3e) = −2e. If x is the fourth charge, then −2e + x = −8e, giving x = −6e. Thus option A is correct. The other options do not make the four-drop total equal to the original charge.

Use e = 1.6 × 10^-19 C. The initial charge is 1.28 × 10^-18 C = 8e. Gaining five electrons adds −5e, so the final charge is 8e − 5e = 3e. Therefore q_final = 3(1.6 × 10^-19) = 4.8 × 10^-19 C, which is positive. Hence option A is correct; the other choices use a wrong sign or fail to subtract the electron charge.

With e = 1.6 × 10^-19 C, the initial charge −9.6 × 10^-19 C equals −6e. Losing four electrons removes four negative charges, producing a change of +4e. Thus q_final = −6e + 4e = −2e = −3.2 × 10^-19 C. Option A is correct. Option B reverses the remaining sign, C adds the lost electrons instead of removing them, and D ignores the nonzero remainder.

Charge remains quantized for a large body, so its value is still q = ne. However, the elementary charge e is extremely small, while a macroscopic body can gain or lose an enormous number of electrons. The difference between successive allowed values is therefore too small to notice in ordinary measurements, making charge appear continuous. Option A denies quantization, while C and D use unrelated or false ideas.

The governing principle is conservation of charge. If charges +2q and −2q appear together, their algebraic sum is (+2q) + (−2q) = 0. Thus the pair contributes no net charge, and the total charge of the closed system does not change. Option A incorrectly adds magnitudes without signs, option B reverses the sign, and option D counts only one member of the pair.

Initially the two-object system has total charge zero because both objects are neutral. Charging by friction transfers electrons from one object to the other; it does not create a net charge in the isolated pair. If one object ends with +6e, the other must have q such that +6e + q = 0. Therefore q = −6e, making option B correct; options A and C fail the zero-total condition.

Before rubbing, the two neutral objects together had zero net charge. During ordinary frictional charging, electrons transfer between materials; protons remain bound in their nuclei, and no net charge is created for the combined system. If one object receives −8e, the other loses eight electrons and becomes +8e, so (−8e) + (+8e) = 0. Hence option C is correct.

The correct method is to write conservation as Q_initial = Q_final, treating charge as an algebraic quantity. Positive and negative signs must be included before combining terms, because opposite charges cancel. Adding magnitudes first can produce a wrong result, and counting only positive charge ignores part of the system. Mass conservation is a different principle and cannot replace charge conservation. Therefore option B is the reliable strategy.

This statement expresses conservation of electric charge. When equal positive and negative charges appear together, their algebraic sum is +q + (−q) = 0, so the net charge of the complete system has not changed. Quantization only says that charge occurs in integral multiples of e; it does not by itself state that charge cannot be created or destroyed. Mass conservation and gravitation are unrelated here, making option B correct.

A nearby negative rod causes electrostatic induction: mobile electrons in the conductor are repelled, so charge separates within the conductor. However, merely bringing the rod near does not provide a path for charge to enter or leave. If the conductor is initially neutral and is neither touched nor earthed, its positive and negative charges remain equal in total. Thus its net charge stays zero, although its distribution changes.

A positive rod attracts electrons toward the nearby side of the conductor. While the conductor is earthed, additional electrons flow from Earth into it. The Earth connection must be removed while the positive rod is still present; this traps the extra electrons. When the rod is subsequently removed, the electrons spread over the conductor, leaving a net negative charge. Removing the rod first would allow the charge separation to disappear before isolation.