Electric Charges and their Conservation

The governing concept is electrostatic repulsion in an electroscope. When charge is present, both leaves acquire charges of the same sign, so they repel one another and diverge. However, the same observation occurs whether both leaves are positive or both are negative. Thus divergence proves the presence of charge but does not identify its sign by itself. A known reference charge or another comparison method is needed; options B and C incorrectly restrict the effect to one sign.

The governing concept is electrostatic induction or polarization. A nearby charged body causes the positive and negative charges inside the neutral ball to rearrange slightly, although the ball remains electrically neutral overall. The side nearer the charged body develops an induced opposite charge and experiences a stronger attractive force than the repulsive force on the farther side, because it is closer. Therefore induction produces a net attraction. Option B wrongly claims a net charge appears immediately.

The governing concept is the distinction between electrostatic attraction and repulsion. Attraction can occur between opposite charges, but it can also occur when a charged object polarizes a neutral body. Repulsion, under ordinary electrostatic conditions, requires like charges on the interacting bodies, so it is strong evidence that both bodies are charged and have the same sign. Hence option A is the definite test. The other options incorrectly attribute repulsion to neutrality, charge destruction, or mass.

The governing principle is charge conservation and the sign of subatomic charges. A neutral atom initially has equal positive proton charge and negative electron charge, so its net charge is zero. Removing three electrons removes three negative elementary charges while the proton charge remains unchanged. The imbalance is therefore +3e, where e is the elementary charge. Thus option A is correct. Option B reverses the sign, C ignores the removed electrons, and D doubles the actual charge change.

The governing principle is conservation and quantization of electric charge. A neutral atom begins with net charge zero. Each added electron contributes one negative elementary charge, −e. Adding four electrons therefore changes the net charge by −4e, so the final charge is negative four elementary charges. Option B has the wrong sign, option C would apply only if no charge were transferred, and option D has the wrong magnitude. The nucleus is not changed in this process.

Use signed-charge addition. The initial charge is +5e. Each gained electron contributes −e, so seven electrons contribute −7e. Therefore the final charge is +5e + (−7e) = −2e. The body remains charged, but the electron gain reverses the sign because the added negative charge is larger than the initial positive charge. Option A is correct; B incorrectly adds magnitudes, C reverses the net sign, and D treats the charges as if they were added with the same sign.

The governing idea is signed charge conservation. Initially the body has −6e. Losing one electron removes a negative charge, which is equivalent to adding +e to the body. Losing four electrons therefore changes the charge by +4e. The final charge is −6e + 4e = −2e. Hence option A is correct. Option B incorrectly makes the body more negative, while C and D reverse the sign without applying the actual arithmetic.

The governing concept is the net charge obtained from the imbalance between protons and electrons. Each proton contributes +e and each electron contributes −e. If there are nine more protons than electrons, the positive contributions exceed the negative contributions by 9e. Therefore the net charge is +9e, or positive nine elementary charges. Option B assigns the wrong sign, option C would require equal numbers, and option D counts both proton and electron charges instead of their difference.

Apply the charge-sign rule: protons contribute +e and electrons contribute −e. When electrons outnumber protons by thirteen, the negative contribution exceeds the positive contribution by 13e. Thus the net charge is −13e, or negative thirteen elementary charges. Option B reverses the sign, option C would require equal numbers of protons and electrons, and option D incorrectly doubles the excess instead of using the difference between the two particle counts.

Charge is scalar because it has magnitude but no spatial direction. However, charge exists in two algebraic types: positive and negative. Therefore, signs must be retained while adding charges. For example, (+5 C) + (−2 C) = +3 C, not 7 C. Thus option A is correct; the other options confuse charge sign with direction, units, or decoration.

Net charge is an algebraic sum, not merely the sum of absolute magnitudes. A positive charge and a negative charge partly or completely cancel: (+6 C) + (−4 C) = +2 C, whereas adding magnitudes gives 10 C. Hence option A is correct. Option B is false because negative charge exists, and C and D incorrectly deny charge addition or magnitude.

The conservation of charge states that the total charge of an isolated or closed system remains constant when no charge crosses its boundary. Internal redistribution can change the positions or concentrations of charges, but it cannot change their algebraic total. Therefore the charge remains −7 C. Zero, +7 C, and −14 C would require removal, reversal, or addition of charge.

Zero net charge means that the algebraic sum of positive and negative charges is zero; it does not mean that charged particles are absent. Ordinary matter can contain positively charged protons and negatively charged electrons in balanced amounts. Therefore A is correct. A neutral body may also contain neutrons, but neutrality is not proof that it contains only neutrons or no mass.

The governing concept is that charging usually changes the number of electrons, not the existence of protons. A negatively charged body has more electrons than protons, so its algebraic net charge is negative. However, it can still contain many protons; therefore, saying that it has no protons at all is not certain and is the incorrect conclusion. Options B and C follow directly from excess electrons, while D is a qualitative description of the net negative state.

Net charge is the algebraic sum of all charges, so the signs must be included rather than adding only magnitudes. Taking the positive charge as +20 C and the negative charge as −17 C, net charge = +20 + (−17) = +3 C. Therefore option A is correct. Option C incorrectly adds magnitudes, option B reverses the sign, and option D would require equal positive and negative charges.

The governing concept is conservation of electric charge: in an isolated closed system, the final total charge equals the initial total charge. Let the unknown charge be q. Then 6 + (−3) + q = −4, so 3 + q = −4 and q = −7 coulomb. Therefore, option B is correct. A positive seven would make the final total positive ten, while the other values do not satisfy conservation.

When separate charged drops merge, the total charge is conserved and the charges must be added algebraically. Thus the new charge is (+3e) + (−5e) + (+7e) + (−1e) = 3 − 5 + 7 − 1 = +4e. Therefore option A is correct. Option B reverses the sign, option C adds magnitudes without signs, and option D incorrectly assumes complete cancellation.

Use conservation of charge and add all final charges algebraically. Let the unknown charge be x. Then (+2q) + (−3q) + (+5q) + x = +q. The three known charges sum to +4q, so x = +q − 4q = −3q. Hence option A is correct. The other choices do not make the final total equal to the stated +q.

When charge enters a conductor, its signed value is added to the conductor’s initial charge. Therefore Qfinal = (+5q) + (−9q) = (5 − 9)q = −4q. The magnitude of the incoming charge is larger and negative, so the final sign must be negative. Option B is correct; +14q results from ignoring the sign, while +4q reverses the subtraction.

The governing concept is algebraic addition of signed charges: positive and negative terms must be combined with their signs. For A, (+4q) + (−9q) + (+3q) = (4 − 9 + 3)q = −2q. For B the result is +2q, for C it is +4q, and for D it is +2q. Thus only option A satisfies the required total charge. The distractors result from incorrect addition or sign handling.

A system can have zero net charge while its separate objects carry nonzero charges; this is charge cancellation, not absence of charge. In option A, (+5q) + (−2q) + (−3q) = (5 − 2 − 3)q = 0. Options B, C, and D give +4q, −10q, and +10q respectively. Therefore A uniquely satisfies the condition.

The governing principle is conservation of electric charge: the algebraic sum of charges before a decay equals the algebraic sum afterward. Let the unknown charge be q in units of e. Then +2 = (−1) + (+4) + q = +3 + q, so q = −1e. Therefore option B is correct. Option A gives a final total of +4e, while options C and D do not satisfy the balance.

Charge conservation requires the final algebraic charge to equal the initial charge. The neutral initial particle has total charge 0. If one product carries +3e and the unknown product carries q, then 0 = +3e + q, which gives q = −3e. Thus option B is correct. Choosing +3e would produce +6e, zero would leave +3e, and +6e would make the imbalance even larger.

An electron carries charge −e, whereas a positron carries charge +e. Before annihilation, their algebraic total is (−e) + (+e) = 0. Charge conservation requires the final products, such as photons, to have the same net electric charge, namely zero. Therefore option C is correct. Options A, B, and D incorrectly assign a nonzero final charge.

Charge is a signed scalar quantity, so positive and negative values must be included with their algebraic signs when a total is calculated. For example, +5e and −3e give +2e, not 8e. Therefore option B correctly explains why an algebraic sum is used. Charge is not a vector with spatial direction, only magnitude, or always zero, so A, C, and D are incorrect.