Electric Charges and their Conservation

A negative rod repels electrons from the conductor. When the conductor is earthed, some electrons flow into the Earth; after the earth connection is removed, the rod is withdrawn and the conductor is left electron-deficient. Its positive charge therefore results from missing electrons, not from newly created positive charge. Charge conservation remains valid for the conductor, Earth, and rod together, so option A answers what did not happen.

The positive rod attracts electrons toward the conductor. While the rod remains nearby, the conductor is connected to Earth, allowing electrons to flow from Earth into it. The earth connection is then removed first, and the rod is withdrawn afterward, trapping the excess electrons on the conductor. Thus option A is correct. Electrons are not newly created, and ordinary induction does not require proton conversion or neutron breakdown.

First conserve the total charge: Qtotal = (+12e) + (−4e) = +8e. Because the spheres are identical conductors, contact allows charge to flow until both have equal charge. Each sphere therefore receives half the total: Qeach = +8e/2 = +4e. Option A is correct. The total +8e is not the charge on each sphere, and neither zero nor −4e satisfies conservation and equal sharing.

Rubbing does not create charge from nothing; it transfers electrons between the two bodies. If one body loses an amount of negative charge, the other gains the same amount. Consequently, starting from two neutral bodies, the final charges are equal in magnitude and opposite in sign, and their algebraic sum remains zero. The materials determine which body loses electrons, but conservation rules out options A, C, and D. Hence option B is correct.

The law of conservation of charge states that the algebraic total charge of a closed system remains constant. If one part changes by +7e, the rest of the system must change by an equal and opposite amount so that +7e + (−7e) = 0 overall change. Therefore the other part must gain negative seven elementary charges. A positive change would add charge rather than balance it, so option B is correct.

Before contact, the total charge of the two-sphere system is +6e + 0 = +6e. Identical conducting spheres have the same capacitance and, on contact, their charges redistribute until their potentials become equal. Therefore the total charge divides equally: each sphere receives +6e/2 = +3e. Charge is not lost, and the first sphere does not retain all +6e. Hence option B is correct.

Charge conservation states that the total charge of an isolated system remains constant. An open system is not isolated: electrons or other charged particles may cross its boundary. Therefore, the charge measured inside the system alone can increase or decrease, even though the combined charge of the system plus surroundings remains conserved. Option B correctly describes this exchange; the other choices contradict the basic properties of charge and particles.

Charge quantization means that net charge occurs as Q = ne, where n is an integer and e is the elementary charge. Because e is extremely small, individual steps of size e are easier to distinguish when the total charge is small. For very large charges, the spacing between successive allowed values is negligible compared with the total, so the charge appears continuous. Therefore option B is correct.

A scalar quantity is specified by a magnitude and, when appropriate, an algebraic sign; it does not require a direction in space. For electric charge, plus and minus identify opposite charge types and determine how charges interact, but neither sign points north, south, or along a coordinate axis. Therefore charge remains scalar. Option A is correct, whereas B and C incorrectly assign vector direction and D denies the physical meaning of the sign.

Charging by induction depends on preserving the charge supplied through the Earth connection. While the charged rod is nearby and the conductor is earthed, electrons can enter or leave the conductor. If the Earth connection is removed first, this exchanged charge can no longer flow away, so the conductor retains a net charge. The rod is removed afterward, allowing the retained charge to spread. Hence A is correct; the other choices contradict induction and charge conservation.

In induction charging, the nearby rod causes charge separation while the conductor is connected to Earth. If the rod is removed first, the external electric influence disappears, so the separated charges tend to recombine or redistribute. Because the Earth connection is still present, any remaining imbalance can also flow to or from Earth. The conductor therefore usually ends with nearly zero net charge. Thus A is correct; the other options have no general physical basis.

For a truly isolated system, the algebraic total charge must remain constant. An apparent increase in positive charge may be accompanied by an equal increase in negative charge elsewhere, leaving the net charge unchanged. It may also result from an incomplete system boundary, an unaccounted exchange with the environment, or measurement error. Thus option A is the scientifically appropriate check. The other options contradict established charge properties and conservation.

Charge conservation states that the algebraic total charge of an isolated system remains constant; charge cannot simply be created or destroyed. Charge quantisation states that charge occurs in integral multiples of the elementary charge, q = ne, although macroscopic measurements can appear nearly continuous. Thus option A correctly combines two separate properties. The other options contradict conservation, quantisation, or the scalar nature of charge.

When identical conducting spheres touch, charge redistributes until both spheres reach the same potential. Their total charge is (+2e) + (+10e) = +12e. Because the spheres are identical, the final charge is shared equally: +12e ÷ 2 = +6e on each sphere. Option B incorrectly gives the total rather than each share; C uses an incorrect division, and D ignores the conserved positive charge.

First find the conserved total charge: (−8e) + (+2e) = −6e. Since the conducting spheres are identical, contact makes the final charge divide equally between them. Thus each sphere receives (−6e)/2 = −3e. The result remains negative because the initial negative charge has greater magnitude. Option C is the total charge, B reverses the sign, and D incorrectly assumes complete cancellation.

After contact, each sphere has charge −4e. The total final charge is therefore (−4e) + (−4e) = −8e. In an isolated interaction, charge conservation requires the total charge before contact to equal the total charge afterward. Hence the initial total was −8e. Option B counts only one sphere, C ignores the negative charges, and D has the wrong sign.

The governing concept is quantization of electric charge. Electrons and protons carry fixed elementary charges of magnitude e, with opposite signs. When electrons are transferred, a body's net charge is determined by an integer difference between the numbers of positive and negative elementary charges, so Q = ne, where n is an integer. Therefore fixed elementary charge explains why charge appears in discrete multiples; mass, temperature, or force does not determine charge quantization.

The governing principle is conservation of electric charge in an isolated system. When an electron moves from one body to another, the first body loses charge −e while the second gains charge −e. These changes cancel when the system total is calculated: the charge lost by one body is exactly the charge gained by the other. The electron does not lose its charge, protons do not disappear, and neutrons do not become charged.

An isolated system cannot exchange net charge with its surroundings. Earthing connects the conductor electrically to the Earth, which acts as a very large charge reservoir. Electrons may flow from the Earth to the conductor or from the conductor to the Earth until the required electrical condition is reached. Thus the charge of the conductor alone need not remain constant, although charge conservation still applies to the larger conductor–Earth system.

The governing law is quantization of charge, written as Q = ne, where e is the elementary charge and n must be an integer for an ordinary free body. The value +2.5e has a non-integer coefficient, so it cannot represent the net charge of an isolated free body under the school-level quantization model. Positive charge certainly exists, and charges larger than 2e are possible; the issue is only the fractional coefficient.

For two identical conducting spheres, the geometry and capacitance are the same. After they are connected, charge flows until both spheres reach the same potential. Since equal potential and equal capacitance imply Q1/C = Q2/C, the final charges satisfy Q1 = Q2; together they equal the conserved total charge. A wooden sphere is not a conductor, earthing permits exchange with Earth, and unequal spheres generally do not share charge equally.

When two identical conducting spheres touch, their total charge is shared equally. In the first contact, the original sphere and a neutral sphere have total charge −16 C, so each becomes −8 C. The original sphere then touches a second neutral sphere; their combined charge is −8 C, so each becomes −4 C. Therefore option A is correct. The charge is halved at each separate contact, not retained or reduced to zero.

Each time the charged original sphere touches an identical neutral sphere, the total charge of that pair is shared equally, so the original sphere’s charge is halved. The successive values are +20/2 = +10 C after the first contact, +10/2 = +5 C after the second, and +5/2 = +2.5 C after the third. Hence option B is correct; +5 C is only the value after the second contact.

An electron carries charge −e. Losing six electrons removes six negative charges, producing a change of +6e. Gaining ten electrons then adds −10e. The net charge change is +6e − 10e = −4e, so the final charge is four elementary charges negative. Option B is correct. The distractors +4e and ±16e result from reversing the electron sign or adding the magnitudes without considering the opposite signs.

Because each electron has charge −e, losing five electrons increases the particle’s charge by +5e. Starting from −7e, the final charge is −7e + 5e = −2e. Therefore option B is correct. The particle remains negatively charged because the positive change is smaller than the initial negative charge. −12e incorrectly treats electron loss as adding negative charge, while the other positive options reverse the sign too far.