Electric Charges and their Conservation

The governing law is conservation of charge. In an isolated system, electrons can move from one object to another, changing the charge of each object, but no charge enters or leaves the combined system. Therefore, if the initial total charge is Q, the final total charge is also Q. It need not increase, decrease, or reverse its sign. Those changes may occur for an individual object, not necessarily for the isolated total. Hence A is correct.

An electron carries charge -e. When the first object loses three electrons, its charge changes by -(-3e), which is +3e relative to its previous state. Thus the first object acquires a positive charge whose magnitude is three elementary charges, assuming its initial charge was neutral or that the question asks about the transferred contribution. It does not become negative, zero, or acquire only one elementary charge. Therefore A is correct.

Charge conservation gives the total charge before contact as (+6 C) + (−2 C) = +4 C. When identical conducting spheres touch, charge flows until both have the same potential; because their sizes are identical, the final charge divides equally. Thus each sphere receives (+4 C)/2 = +2 C. Option A is correct; +4 C is the total charge, while the negative and zero choices violate the conserved total or equal sharing.

First apply conservation of charge: the combined charge is −10 C + 4 C = −6 C. On touching, identical conducting spheres reach equal potential and share their total charge equally. Therefore the final charge on each sphere is (−6 C)/2 = −3 C. Option A is correct. The value −6 C is the conserved total for both spheres together, whereas positive options have the wrong sign.

In electrostatic contact, mobile charges redistribute between conductors until their electric potentials become equal. Charge is neither created nor destroyed, so the algebraic total charge of the isolated pair remains constant. For identical conductors, equal geometry leads to equal final charges; each receives half of the initial total. Hence option A is correct. The other options contradict charge conservation or the equal-potential condition.

For an isolated object, charge quantization requires q = ne, with n an integer. Therefore 2e, −3e, and 0 are all allowed values because their multipliers are integers. A charge of e/2 has a non-integer multiplier and is not possible as the net charge of an isolated ordinary object under this school-level model. Hence option A is correct; the other three satisfy the quantization rule.

Initially the combined charge of the two neutral objects is zero. Rubbing does not create charge; it transfers electrons from one material to the other. If one object gains electrons and becomes negative by magnitude Q, the other loses the same number of electrons and becomes positive by magnitude Q. Thus option A follows from conservation of charge. Both negative would give a nonzero total, while neutrality or no rule ignores the transfer process.

The conservation statement is applied most directly by choosing an isolated system, whose boundary does not permit net charge exchange with its surroundings. For such a system, total charge remains constant: Q_initial = Q_final. Therefore option A is correct. In an open system the charge of the selected part may change because charge crosses the boundary, although conservation still holds for the larger closed system. Temperature and mass are irrelevant here.

Connecting a conductor to Earth creates an electrical connection with a vast external reservoir. Electrons may flow between the conductor and Earth until the appropriate potential condition is reached, so the conductor by itself is no longer an isolated system and its charge can change. Option A is correct. Charge is not destroyed; it is redistributed, and there is no rule requiring the conductor’s charge to double. Electrons are precisely the mobile carriers involved.

A positively charged conductor has a deficit of electrons compared with its neutral state. When it is connected to Earth, electrons from the Earth can flow into the conductor because Earth acts as a huge charge reservoir. Supplying enough electrons removes the deficit and makes the conductor neutral. Thus option A is correct. Protons do not normally travel through the metal, neutrons are not formed, and positive charge is not destroyed; charge is transferred.

A negatively charged conductor contains an excess of electrons. When it is connected to Earth, these excess electrons can flow from the conductor into the Earth, which acts as a large reservoir. If the excess is removed, the positive and negative charges balance and the conductor becomes neutral. Option A is correct. Protons do not provide the normal current in the metal, charge is not created, and neutron motion is irrelevant to ordinary earthing.

Charging by induction begins when a charged body is brought near, but not touching, a conductor. Its electric field causes mobile electrons in the conductor to shift, producing separated regions of opposite charge called polarization. Thus option A describes the first event. Charge is not destroyed, protons do not normally transfer through the metal, and neutrons are not formed during this electrostatic process.

Apply charge conservation to the closed system. Let the unknown charge be Q. The algebraic sum must equal the stated total: (+3q) + Q = −q. Rearranging gives Q = −q − 3q = −4q. Therefore option A is correct. The sign must be negative because the known +3q must be more than offset to produce the total −q; the other numerical choices do not satisfy the balance equation.

The governing concept is quantisation of electric charge: an isolated body can have charge Q = ne, where n must be an integer. Here n = Q/e = (2.4 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹) = 1.5, which is not an integer. Therefore the stated charge cannot occur for an isolated object under the given model, so option A is correct. Its positive sign, small size, and coulomb unit are not problems.

Conservation of charge states that the total charge of an isolated system remains constant; charge may move between bodies, but the system total does not change. Quantization states that charge occurs in integral multiples of the elementary charge, q = ne. Thus option A correctly distinguishes a conservation rule from a discreteness rule.

For an isolated system, apply the equation total initial charge = total final charge. Every charge must be included with its sign, because positive and negative charges can cancel algebraically. For example, +5q − 2q = +3q. Ignoring signs or replacing charge with mass gives an incorrect conservation equation, so option A is the proper method.

Let the charge on the first body be +Q and that on the second be −Q. The initial total charge is +Q + (−Q) = 0. When the bodies touch, charge may redistribute between them, but the total charge of the combined isolated system is conserved. Thus the final total charge remains zero. Option C is correct; contact cannot change the algebraic total into a positive, negative, or doubled value.

The algebraic nature of charge means that charge has a sign, and positive and negative values must be included with those signs when calculating a net value. For example, +3e + (−2e) = +e. This is why opposite charges can cancel. Option A is correct. Charge is a scalar quantity, so option C is wrong, and the remaining options incorrectly deny ordinary signed addition.

In ordinary charging, the mobile particles are electrons because they occupy the outer regions of atoms and can be transferred between materials. Protons are positively charged particles held tightly by the strong nuclear force inside atomic nuclei. Removing a proton would require a nuclear process, not routine rubbing, contact, or induction. Hence option A is correct; option B reverses the proton’s charge, and the other statements are factually incorrect.

Charge conservation means that ordinary charging does not create or destroy net electric charge. In common materials, the number of protons in each nucleus remains effectively fixed, while electrons can move from one body to another. A body that loses electrons becomes positive and one that gains them becomes negative; the total charge of the complete system is unchanged. Therefore option B is correct.

The elementary charge e is the magnitude of the charge on one proton or electron. A net charge of +2e means the body has two more units of positive charge than negative charge. In ordinary charging this is interpreted as a deficiency of two electrons relative to the neutral state, not as two protons leaving the nucleus. Thus option B is correct, and it also illustrates charge quantization.

A negative elementary charge belongs to an electron, so a net charge of −3e represents three units of excess negative charge. In the ordinary charging model, this means the body has gained three electrons compared with its neutral state. It does not mean that neutrons changed or that three protons were added. Therefore option A is correct, and the result demonstrates that charge is quantized in integral multiples of e.

The relevant principle is charge quantization: the net free charge of an ordinary isolated body is Q = ne, where n is an integer and e is the elementary charge. Since one-half is not an integer, Q = e/2 is not an allowed isolated free charge in this school-level model. Changing the material or using a vacuum does not alter this rule. Therefore option B is correct.

Charge conservation governs this situation. Bringing the charged rod near a conductor changes the positions of the conductor’s mobile electrons, but the rod does not touch it and no charge crosses the boundary. Thus one region may become electron-rich while another becomes electron-deficient, yet their algebraic total remains constant. Option A is correct; electrons are not destroyed, protons do not leave the nuclei, and no new charge is created.

In charging by contact, mobile electrons can flow from one conducting body to another until electrical equilibrium is approached. This flow changes how the charge is distributed, but for the combined isolated system no charge is created or destroyed. Hence the algebraic sum of the two charges before and after contact is the same. Option C is correct; contact does not automatically double, erase, or convert charge into mass.