Electric Charges and their Conservation

The governing idea is that an electron carries charge −e. When the object gives away six electrons, it loses six negative charges, so its charge increases by +6e. Starting from +4e, the final charge is +4e + 6e = +10e. Therefore option B is correct. Option A reverses the effect of losing electrons, while options C and D use an incorrect sign or magnitude.

Each received electron contributes −e, so receiving five electrons changes the charge by −5e. The initial charge is −2e. Applying charge addition gives q_final = −2e + (−5e) = −7e. Thus option B is correct. Option C incorrectly changes the sign, and options A and D result from subtracting or combining the magnitudes incorrectly.

The quantization of charge is expressed as Q = ne, where the sign identifies the type of excess charge. Q = −19e means that the body has nineteen excess electrons. Each removed electron increases the body's charge by +e, so removing 19 electrons changes the charge by +19e and gives Qfinal = −19e + 19e = 0. Adding electrons would make it more negative, while neutrons do not affect charge.

The quantization principle states that the observable charge of an isolated body is Q = ne, where n must be an integer (positive, negative, or zero). Here n = 11/3, which is not an integer, so a free body cannot possess exactly this charge under the school-level model. The value need not be positive; negative integral multiples are also allowed, so B and C give false reasons.

The conservation of charge principle says that contact can redistribute charge between conducting spheres but cannot change the total charge of an isolated pair. After contact, the total is (+9e) + (+9e) = +18e. Therefore the initial total charge was also +18e, regardless of how it was divided before contact. The fact that the spheres are identical explains equal sharing, but the requested total follows directly from addition and conservation.

When objects exchange charge only with each other, the pair forms a closed system, so internal transfer cannot change the total. Add the signed charges: Qtotal = +26e + (−8e) = +18e. The negative charge subtracts from the positive charge rather than adding to its magnitude. Thus the final total remains +18e, even if the individual charges redistribute. +34e ignores the sign, and −18e reverses it.

Neutrality describes the net charge, not the absence of charged constituents. Ordinary matter contains positively charged protons and negatively charged electrons, along with neutral particles such as neutrons. In a neutral body, the total positive charge equals the total negative charge, so their algebraic sum is zero. Therefore A is correct; the body does not consist solely of neutrons, protons, or electrons as the other options claim.

Charge quantization means that net charge is an integral multiple of the elementary charge, q = ne. The elementary charge e is extremely small, about 1.6 × 10^-19 C, while a macroscopic object contains an enormous number of charged particles. Consequently, successive allowed charge values differ by an imperceptibly small amount, so the charge appears continuous. Option B is false because macroscopic bodies can be charged; C and D contradict physical facts.

The net charge is +23e, where e is the magnitude of the elementary charge. In ordinary charging of a body, electrons move while the positively charged protons remain bound inside atomic nuclei. Therefore, +23e means that the body has 23 fewer electrons than the neutral state. An electron excess would produce -23e, and neutron deficiency or proton destruction is not the normal mechanism of electrostatic charging.

For an ion, net charge is q = (N_p − N_e)e, where N_p and N_e are the numbers of protons and electrons. Given q = −16e, division by e gives N_p − N_e = −16, or N_e − N_p = 16. Hence there are sixteen more electrons than protons. The negative sign indicates an electron excess; it does not mean the number is zero, half of 16, or twice 16. Therefore option A is correct.

The initial particle is neutral, so its total charge is 0. Charge conservation requires the algebraic sum of the two final charges also to be zero. If one product has charge +5e and the other has charge q, then +5e + q = 0, giving q = −5e. A second positive charge or zero would leave a nonzero total, while −10e would give −5e overall.

The initial charge is −25e, representing an excess of 25 electrons. Removing 10 electrons removes 10 units of negative charge, so the charge changes algebraically as q_final = −25e + 10e = −15e. The body remains negatively charged because fifteen excess electrons are still present. Option B would correspond to adding electrons, while C reverses the sign incorrectly and D ignores the initial charge.

The governing idea is conservation and quantization of charge. A charge of +12e means the body is deficient by twelve electrons. Adding fifteen electrons changes the charge by −15e, so the final charge is +12e − 15e = −3e. Thus option A is correct. Option C has the wrong sign, while B and D incorrectly add the magnitudes without considering that electrons are negative.

For an isolated system, the algebraic sum of the charges of all parts remains constant. Let the remaining charge be Q. Then Q + (−9e) = +4e, so Q = +4e + 9e = +13e. Therefore option A is correct. Option B reverses the required sign, and options C and D result from using an incorrect subtraction or ignoring the negative charge of the first part.

Charging by rubbing involves transfer of electrons between two objects, while the total charge of the pair is conserved. If one object loses eighteen electrons, the other object gains those same eighteen electrons. Since each electron has charge −e, the gained charge is 18(−e) = −18e. Therefore option A is correct; B has the opposite sign, C ignores the transfer, and D doubles it incorrectly.

Initially the two neutral objects together have zero net charge. Rubbing transfers electrons rather than creating charge. If one object becomes +21e, it has lost charge equivalent to twenty-one electrons; those electrons must be gained by the other object. The second object therefore becomes −21e, making the total +21e − 21e = 0. Thus A is correct; B has the wrong sign and C or D violates the balanced transfer.

The law of conservation of charge states that the algebraic sum of charge in an isolated system remains constant with time. Charges may transfer between bodies, so the charge of an individual body can change even though the system total does not. Therefore option A is the precise statement. B incorrectly applies conservation to every body separately, C confuses conservation with neutrality, and D ignores negative charge.

The governing concept is conservation of charge: the algebraic sum of charges in a closed system remains constant. Let the remaining charge be q. Then q + (+4e) = −7e, so q = −7e − 4e = −11e. Therefore, the remaining part has negative eleven elementary charges, as stated in option A. Options B and D result from mishandling the signs, while C has the wrong sign.

Charge conservation gives the total charge before contact as (+9e) + (+3e) + (−6e) = +6e. Because the spheres are identical conductors and are in simultaneous contact, they reach the same potential and share the total charge equally. Thus each receives (+6e)/3 = +2e. Option A is correct; option B is the total charge, not the charge on one sphere, and C and D contradict the calculation.

A deficiency of twelve electrons gives the first body a charge of +12e, because removing negative charge leaves a positive excess. The total charge of the pair is zero, so the second body must carry −12e. In ordinary charging language, −12e means twelve extra electrons. Thus option A is correct. Option B would give two positive charges, while C and D do not provide the required compensating charge.

Start with zero charge. Removing five electrons removes negative charge and leaves the body with +5e. Adding eight electrons contributes −8e, so the final charge is (+5e) + (−8e) = −3e. Therefore option A, negative three elementary charges, is correct. Option B reverses the sign; options C and D incorrectly add the magnitudes without accounting for the opposite effects of removal and addition.

A neutral body initially has zero net charge. Adding ten electrons contributes −10e. Removing four electrons removes part of that negative charge, which is equivalent to adding +4e. Thus the final charge is −10e + 4e = −6e. Option A is correct. Option B has the wrong sign, option C treats both operations as increasing the negative magnitude, and option D ignores the remaining six extra electrons.

The elementary charge has magnitude e, and an electron carries charge −e. Removing one electron therefore leaves a deficiency of one negative charge and produces only +e. To obtain a charge of +2e, the particle must have a net deficiency of two electrons, or an equivalent charge imbalance involving other particles. Thus option A is correct. The other options either reverse the electron’s sign or incorrectly deny charge quantization and positive charge.

The governing concept is conservation and algebraic addition of electric charge. If the two charges are +q and −q, their initial total is +q + (−q) = 0. Contact may redistribute charge between the bodies, but it cannot change the total charge of the combined system in an isolated situation. Therefore, zero is correct; the double-positive and double-negative choices ignore the opposite signs, while indeterminate is unjustified.

The governing idea is that ordinary charging usually changes the number of electrons, not the number of protons in the nucleus. A body is negatively charged when it has more electrons than protons, so its net charge is negative even though protons remain present. Option A states this correctly. Option B wrongly treats negative charge as the absence of protons, option C describes an unrelated nuclear transformation, and option D assigns negative charge to protons.