Electric Charges and their Conservation

For an isolated or free body, charge occurs in integral multiples of the elementary charge: q = ne, where n must be an integer. Here q = (7/2)e = 3.5e, and 3.5 is not an integer. Thus a free body cannot possess this exact charge under ordinary charge-quantisation conditions. The sign, material, or size of the body does not remove this integral-multiple requirement, so option B is correct.

Each proton contributes +e and each electron contributes −e. If the number of protons exceeds the number of electrons by three, the net charge is q = 3(+e) = +3e. Neutrons do not affect this calculation because they have no net electric charge. Therefore the atom is actually a positively charged ion rather than a neutral atom, and option A is correct. The charge is not doubled, so option D is incorrect.

The net charge equals the positive proton contribution plus the negative electron contribution. Five extra electrons mean an excess charge of 5(−e), so q = −5e. The ion is therefore negatively charged, often called an anion. Equal numbers would give zero charge, while five extra protons would give +5e; neither situation applies here. Thus option B is the only correct answer.

The governing ideas are conservation of charge and equal sharing by identical conducting spheres. Initially, the total charge is −10e + 0 = −10e. Since the spheres are identical and come into contact, charge redistributes equally: each sphere receives (−10e)/2 = −5e. Thus option A is correct. Option C would ignore sharing, while B reverses the sign and D fails to conserve the total charge.

The elementary charge has magnitude e, and electrons carry charge −e. A net charge of +4e therefore means that the body has four fewer electrons than it would have in its neutral state, assuming its protons remain bound in the nuclei. Ordinary charging transfers electrons rather than destroying protons or removing neutrons. Hence option A is correct; option B would produce −4e, and C and D do not explain normal electrification.

The observation can be written as Q = ne, where Q is the measured charge, e is the elementary charge, and n is an integer. This means charge is available in discrete packets rather than arbitrary continuous amounts. That property is called quantization of charge, so option B is correct. Conservation would instead refer to the total charge remaining constant during an interaction; it does not specifically describe integral multiples.

The total charge is the algebraic sum of the individual charges. Let the unknown charge be q. The condition is (+9e) + q = 0, so q = −9e. Therefore the second point charge must be negative nine elementary charges, making option B correct. Two positive charges would give +18e, a zero second charge would leave +9e, and −18e would produce a total of −9e rather than zero.

Each electron carries a charge of −e. Removing one electron removes negative charge from the body, which leaves the body with a net positive charge of +e relative to its initial state. Removing seven electrons therefore gives Q = 7e, or positive seven elementary charges. Option A is correct. The body would become negative only if electrons were added; fourteen is an incorrect multiplication, and zero ignores the transferred charge.

An electron contributes one elementary negative charge, −e. If nine electrons are added to a neutral body, the change in its charge is ΔQ = 9(−e) = −9e. Thus the body acquires negative nine elementary charges, so option B is correct. Adding electrons cannot produce a positive result; zero would mean no net transfer, and −18e incorrectly counts twice the nine transferred electrons.

The elementary charge has magnitude e, and a charge of −12e means the body has twelve units of excess negative charge, normally represented by twelve extra electrons. Removing twelve electrons changes the charge by +12e, giving −12e + 12e = 0. Therefore option A neutralizes the body. Adding electrons would make it more negative, while removing protons or adding neutrons is not the ordinary charging process and does not give the required correction.

A charge of +11e indicates a deficit of eleven elementary negative charges relative to the positive charge in the body. Adding eleven electrons contributes −11e, so the final charge is +11e − 11e = 0. Hence option A is correct. Removing electrons or adding protons would increase the positive charge, and neutrons have no net electric charge, so removing them would not neutralize the body.

The governing principle is conservation of electric charge. Contact may allow electrons to move between the two bodies, changing how the total charge is distributed, but it does not change the algebraic sum when the system is isolated. Since no charge enters or leaves, the initial total charge −6e remains the final total charge. Therefore option A is correct. Contact does not automatically make the system neutral or double its charge.

Ordinary matter contains positively charged protons and negatively charged electrons, along with electrically neutral neutrons in many nuclei. A body becomes positively charged mainly when some electrons are removed; its remaining electrons do not disappear. Thus the positive charge represents an excess of positive charge over negative charge, not the presence of positive particles alone. Option A is correct; the other choices contradict the known charges of electrons, protons, and neutrons.

A neutral object is not necessarily free of charged particles. It generally contains positive protons and negative electrons in equal total amounts. Since each electron has charge −e and each proton has charge +e, equal numbers or equal total magnitudes cancel algebraically, giving Qnet = (+Ne) + (−Ne) = 0. Therefore A is correct. Neutrons are neutral, and removing all protons or electrons would not describe ordinary neutral matter.

The quantisation relation is q = ne, where n is an integer and e is the elementary charge. Here n = 0, so q = 0. A zero net charge means that positive and negative charges balance each other, although the body may still contain charged particles. Therefore, the body is neutral. Options B and C require a non-zero net charge, while D is not a valid description of this quantised value.

The charge of one electron is −e. If the body has net charge −8e and this charge is attributed to excess electrons, the number of extra electrons N follows −Ne = −8e. Cancelling e gives N = 8. Therefore, eight extra electrons are required. Four would produce only −4e, sixteen would produce −16e, and zero extra electrons would give no negative excess charge.

Rubbing two different materials can transfer electrons from one surface to the other because their electrons are held with different strengths. One object gains electrons and becomes negative, while the other loses electrons and becomes positive. The algebraic total charge of the combined system remains unchanged. Therefore rubbing separates or transfers charge rather than creating net charge. The other options incorrectly describe changes in particle identity or mass.

Electric charge is a signed scalar quantity, so its net value is found by algebraic addition. Positive charges contribute with a plus sign and negative charges with a minus sign. For example, (+5e) + (−2e) = +3e, not 7e. Therefore algebraic addition is essential when both signs are present. Mass or temperature alone does not require a charge sum, and no charged particle gives a net charge of zero.

Let the first particle have charge +e and the second have charge −e. Their combined charge is the algebraic sum, (+e) + (−e) = 0. The equal magnitudes cancel because the signs are opposite. Thus the pair has zero net charge, although each particle individually remains charged. Options B and C incorrectly add magnitudes without respecting signs, while D ignores the second particle altogether.

The proton carries charge +e, whereas the electron carries charge −e, where e is the elementary charge magnitude. Thus their charge magnitudes are equal, but their signs are opposite. This relation also explains why a hydrogen atom, containing one proton and one electron, has zero net charge when it is neutral. Option B has the wrong sign, and C and D contradict the known charges of these particles.

Use conservation of charge for the two-object system. If one object gains a charge of −15e during rubbing, the total charge created by the process is zero, so the other object must lose the same number of electrons. Losing 15 electrons gives the second object a charge of +15e. The charges are equal in magnitude and opposite in sign; −15e or zero would not balance the transfer, while +30e is twice the required amount.

The governing principle is conservation of electric charge: in an isolated system, charge can move between bodies but cannot enter or leave the system. Add the signed charges algebraically: (+7 C) + (-2 C) = +5 C. Redistribution may change each body's individual charge, but it cannot change this total. Therefore option A is correct; +9 ignores the negative sign, while -5 reverses the result.

For an isolated system, the total charge is found by adding every charge algebraically, keeping its sign. Thus, (-8 C) + (+3 C) + (+2 C) = (-8 C) + (+5 C) = -3 C. Charge conservation means this total remains unchanged during any internal redistribution. Option B is correct. Option A loses the negative sign, option C adds magnitudes incorrectly, and option D assumes cancellation that does not occur.

The governing concept is conservation of charge and equal sharing by identical conducting spheres. The total charge before contact is (+14) + (−6) = +8 C. Because the spheres are identical, charge redistributes equally when they touch. Thus each sphere receives (+8)/2 = +4 C. Option B is correct; +8 C is the total charge, while −4 C and +10 C do not satisfy charge conservation and equal sharing.

Use conservation of charge together with equal sharing among identical conductors. The total initial charge is +8 + 0 − 4 + 12 = +16 C. Since four identical spheres touch, this charge is distributed equally among them. Therefore, the final charge on each sphere is +16/4 = +4 C. Option A is correct. +16 C is the combined total, not the charge on one sphere; the other choices give incorrect totals per sphere.