Electric Charges and their Conservation

An electron carries charge -e. Removing an electron removes negative charge from the atom, leaving a positive deficiency. Removing two electrons therefore changes the net charge by -2(-e) = +2e, where e is the elementary charge. Hence the ion has a charge of plus two elementary charges, so option A is correct. The negative choices would correspond to gaining electrons, not removing them.

Each electron contributes a charge of -e. When an atom gains three electrons, it receives three additional negative elementary charges, so the net charge change is 3(-e) = -3e. Therefore the formed ion has minus three elementary charges, making option A correct. A positive charge would result from losing electrons, while zero would require no net electron gain or an exactly compensating positive change.

Quantization means that electric charge is discrete rather than continuously divisible at the elementary level. It is expressed as q = ne, where n is an integer and e is the elementary charge. Thus a charge can be 0, ±e, ±2e, and so on, within the usual macroscopic description. Option A is correct. The other choices describe neither the integral-multiple relation nor the stated property of charge.

The governing concept is conservation of charge: the algebraic total charge of an isolated system remains constant. Equal newly produced charges have opposite signs, so their contribution is (+q) + (−q) = 0. They may be created as a pair, but the system’s net charge does not change. Therefore option A is correct; options B and C ignore cancellation, while D incorrectly suggests that all charge disappears.

Electric charge is a signed scalar, so the signs must be retained during addition. Calculate the algebraic sum: (−5 C) + (+2 C) = −3 C. The magnitudes subtract because the signs are opposite, and the result takes the sign of the larger magnitude, 5 C. Thus option A is correct; +3 reverses the sign, while ±7 incorrectly adds magnitudes without cancellation.

The total charge is found by adding every charge algebraically, including its sign. First combine the positive charges: +2 C + +1 C = +3 C. Then add the negative charge: +3 C + (−5 C) = −2 C. Hence option A is correct. The other choices result from ignoring the negative sign, adding magnitudes incorrectly, or assuming complete cancellation.

Let the unknown charge be q. By conservation and algebraic addition, the two parts must satisfy q + (−2 C) = −6 C. Adding 2 C to both sides gives q = −4 C. Therefore option A is correct. A positive 4 C would produce only −? Actually (+4) + (−2) = +2 C, while −8 C and +8 C do not give the stated total.

In ordinary charging, electrons are the mobile particles, whereas protons remain bound inside atomic nuclei. If an object loses some electrons, its negative charge decreases while the fixed positive nuclear charge is left uncompensated. The net charge therefore becomes positive, so option A is correct. Proton emission is not the usual explanation for electrostatic charging, and neutrons are electrically neutral.

An electron carries negative elementary charge, while a proton carries positive charge and a neutron has zero charge. When an object gains more electrons than it loses, the negative contribution exceeds the positive contribution, producing a negative net charge. Thus option A is correct. Excess protons would make the object positive, and neutron number or total mass does not determine electrostatic sign.

Neutrality refers to zero net charge, not to the absence of charged particles. In a neutral object, the algebraic sum of positive and negative charge is zero; therefore their total magnitudes are equal, such as (+Q) + (−Q) = 0. Option A is correct. Options B and C describe objects with nonzero charge, while D would produce a positive net charge.

Initially, neutrality means that the sphere’s positive and negative charges balance. When electrons leave, the amount of negative charge decreases, but the positively charged nuclei remain in the sphere. The resulting algebraic sum is positive, so option A is correct. Protons normally do not leave a metal during ordinary charging, neutrons do not carry charge, and the sphere’s mass does not become zero.

The governing law is conservation of electric charge: the algebraic sum of charge in an isolated system remains constant. The initial neutral system has total charge 0. If the first particle has +q and the second has charge Q, then +q + Q = 0. Solving gives Q = −q. Thus option A is correct. A second +q would give +2q, while zero would fail to balance the positive charge.

The governing rule is algebraic addition of electric charges, including their signs. The total is (+q) + (−3q). Since the negative magnitude is larger by 2q, the result is −2q. Therefore option A is correct. Adding magnitudes without signs would incorrectly produce 4q, and choosing +2q would reverse the net sign. Charge is a scalar quantity, so no vector-direction operation is needed here.

Charge quantization is written as q = ne, where e = 1.6 × 10⁻¹⁹ C is the magnitude of one elementary charge and n is the number of elementary charges. Hence n = q/e = (3.2 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹) = 2. The powers of ten cancel and the coefficient ratio is 3.2/1.6 = 2. Therefore the object has two elementary charges, so option A is correct.

A negative charge means that the object has an excess of electrons. The number of excess electrons is found from n = |q|/e, where e = 1.6 × 10⁻¹⁹ C. Hence n = 4.8/1.6 = 3. The correct answer is therefore three excess electrons; the negative sign identifies excess electrons rather than changing their count.

A positive charge indicates that the object has a deficiency of electrons, so it has lost electrons. Using q = ne, the number lost is n = q/e = (6.4 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹) = 4. Therefore, the object has lost four electrons. The positive sign indicates loss, while the quotient gives the number.

Electric charge is a scalar because it has magnitude and algebraic sign, but no spatial direction. Charges in a system are combined by ordinary signed addition, such as +q − 2q = −q, rather than by vector addition. Option B incorrectly assigns a direction, while options C and D confuse charge with velocity or force.

Rubbing can separate existing charges by transferring electrons from one material to another; it does not create net charge from nothing. If one object gains electrons and becomes negative, the other loses an equal amount and becomes positive. The total charge of the combined system remains conserved, so options B, C, and D contradict charge conservation.

In ordinary charging, especially by friction or contact, electrons move between bodies while protons remain bound in atomic nuclei. The receiving body becomes more negative and the donating body more positive, but the charge lost equals the charge gained. Consequently, the total charge of the closed system remains unchanged, making option A correct.

Rubbing causes electrons to transfer between the two objects. One object may gain electrons and become negatively charged, while the other loses electrons and becomes positively charged. These changes are equal and opposite, so the algebraic total charge of both objects together remains the same. Therefore, option A expresses conservation of charge correctly.

An isolated system does not exchange charge with its surroundings, so no external charge enters or leaves. Internal processes may redistribute charge among its components, but the algebraic sum of all charges remains constant: Q_initial = Q_final. The total need not be positive or negative; it can also be zero. Hence option B is correct.

The governing concept is conservation of charge during rubbing. Electrons, not protons, are transferred between the glass and silk. If the glass rod becomes positive, it has lost electrons; those electrons are gained by the silk. Therefore, the silk becomes negatively charged. Option A is incorrect because the two objects acquire opposite charges, while zero would imply no net electron transfer.

Quantization of charge means that charge is not normally acquired in arbitrary amounts. It is expressed as q = ne, where n is an integer and e is the elementary charge, approximately 1.6 × 10⁻¹⁹ C. Thus values such as 2e, 5e, or −3e are allowed. This makes option B correct; the other statements confuse quantization with size or temperature effects.

The quantization rule states that charge is q = ne, where n must be an integer and e is the elementary charge. Here q = 5e, so n = 5, which is an allowed integral value. This directly demonstrates quantization of charge. Conservation concerns the total charge of an isolated system, and scalar nature concerns how charge is combined, so neither is the specific rule illustrated here.

A neutral body has equal total positive and negative charge. If the positive charge is +Q and the negative charge is −Q, the algebraic sum is Qnet = +Q − Q = 0. Neutrality therefore refers to zero net charge, not the absence of charged particles. Option C is correct; the body may still contain protons and electrons, while options A, B, and D give nonzero or impossible results.