Electric Charges and their Conservation

Correct answer: A. Addition with signs, conservation, and checking integral multiples of elementary charge

Explanation: Reliable charge calculations require several connected checks. First, use signed algebraic addition to obtain the net charge rather than adding magnitudes blindly. Second, apply conservation to compare the total charge before and after transfer or contact. Third, where microscopic charge is relevant, check that the result is an integral multiple of the elementary charge e. Option A combines all three valid tools; the other choices contain irrelevant or incomplete ideas.

Correct answer: A. Total charge is conserved, but distribution and the number of electrons on a body can change

Explanation: The central rule is conservation of the total charge of an isolated system, not conservation of the charge on every individual body. During rubbing, contact, or induction, electrons can move between bodies, changing their charge distribution and the number of electrons on each body. Their total algebraic charge, however, remains constant. Thus A is the most complete conclusion. B and C deny the rule’s actual meaning, while D wrongly claims that a positive body contains no electrons.

Correct answer: B. Total charge remains constant in an isolated system while charge may cross the chosen boundary in an open system

Explanation: Charge conservation applies to the total charge of a suitably isolated system. No charge crosses its boundary, so the algebraic total remains constant. In an open system, charge can enter or leave across the selected boundary; consequently, the charge contained inside that boundary may change even though charge is conserved in the larger system. Thus option B states the correct distinction. The other options deny conservation or make absolute claims.

Correct answer: A. Negative two elementary charges

Explanation: For an isolated system, mutual contact may redistribute charge among the bodies, but it cannot change the algebraic total. Add the signed charges: (+7e) + (−11e) + (+2e) = (7 − 11 + 2)e = −2e. Therefore the total charge after contact is negative two elementary charges, option A. Options B and C result from sign or arithmetic errors, while D incorrectly assumes complete cancellation.

Correct answer: A. Positive six elementary charges

Explanation: First find the conserved total charge: (+17e) + (−5e) = +12e. Because the spheres are identical conductors, touching allows charge to redistribute until both spheres have equal charge. The total is therefore divided equally: +12e/2 = +6e on each sphere. Hence option A is correct. Option B is the total charge mistakenly assigned to each sphere, while C has the wrong sign and D uses an incorrect average.

Correct answer: A. Because each would get half an elementary charge, which is not possible for a free sphere

Explanation: The total initial charge is +e. A classical equal-sharing calculation for identical spheres would assign +e/2 to each sphere. However, the net charge of an isolated free body is quantized in integral multiples of e, so two independently observable spheres cannot each retain exactly half an elementary charge. This illustrates the microscopic limitation of the macroscopic equal-sharing model; total charge itself remains +e.

Correct answer: A. First decide the isolated system, then add all charges with signs

Explanation: The reliable method follows the conservation equation Q_initial = Q_final for a clearly defined isolated system. First decide which bodies belong to the system; then list every initial and final charge with its positive or negative sign. Add them algebraically and solve for the unknown while keeping the two totals equal. Thus A is correct. Removing signs, guessing, or selecting only the largest charge discards essential physical information.

Correct answer: A. Near the surface the field is nearly uniform and perpendicular to the surface

Explanation: In electrostatic equilibrium, the electric field inside a conductor is zero and the field immediately outside its surface has no tangential component. If a tangential component existed, free charges would move along the surface, contradicting equilibrium; therefore the external field is normal, or perpendicular, to the conductor. For a large plane surface, edge effects and curvature are negligible, so the field is approximately uniform over a small nearby region and the lines are parallel to one another. Thus option A is correct.

Correct answer: A. Because higher curvature there can produce a stronger external field

Explanation: For an electrostatic conductor, the tangential electric field at the surface is zero, but the normal field just outside is related to surface charge density by E = σ/ε₀. At a sharp region, the radius of curvature is small, so charge crowds there and σ becomes large; consequently the external field is stronger. Therefore A is correct. Charge is not destroyed, the internal field remains zero, and flat regions can also carry charge.

Correct answer: A. Because the fields of opposite charges partially cancel

Explanation: A point charge has a monopole field that varies as 1/r². An electric dipole contains equal and opposite charges, so at a far point their leading 1/r² contributions nearly cancel. The remaining dipole term varies as 1/r³, for example Eaxial ≈ 2p/(4πε₀r³). Hence the dipole field decreases faster. Option A identifies the cancellation; B, C, and D contradict the dipole field law.

Correct answer: A. Because field direction is not defined there

Explanation: An electric field line is defined so that its tangent at any point gives the direction of the electric field at that point. At a zero-field point, the vector has zero magnitude and therefore no unique direction. Although field lines in the surrounding region may curve around the point, no definite tangent can be assigned exactly there. Zero field does not necessarily mean zero potential or absence of charge, and field lines need not be straight.