Continuous Charge Distribution

Correct answer: B. −70 C

Explanation: The governing equation for a uniform volume charge distribution is Q = ρV, where ρ is volume charge density and V is volume. Substitution gives Q = (−5 C/m³)(14 m³) = −70 C. Cubic-metre units cancel, and the negative sign must be retained because the density is negative. Thus option B is correct. Option A loses the sign, option C uses an incorrect arithmetic operation, and option D ignores the stated charge density.

Correct answer: A. The longest part

Explanation: For a uniform linear charge distribution, the governing relation is Q = λL. Since the linear density λ is the same in every part, each part’s total charge is directly proportional to its length L. The longest part contains the greatest length and therefore the greatest amount of charge. Hence option A is correct. The shortest part would contain the least charge, while equal charges would occur only for equal lengths, not for the unequal lengths stated here.

Correct answer: B. The part with the smallest area

Explanation: For a uniform surface charge distribution, the governing relation is Q = σA. Because the surface density σ is identical in all five parts, the total charge in each part is directly proportional to its area A. The smallest-area part contains the least area and therefore the least total charge. Thus option B is correct. The largest-area part would have the greatest charge, and equal charges would require equal areas, contrary to the question.

Correct answer: A. The part with the largest volume

Explanation: The governing relation for a uniform volume charge distribution is Q = ρV. Since the volume density ρ is constant throughout the distribution, total charge is directly proportional to the volume V of each part. A larger volume contains more elementary volume regions and therefore more charge. Consequently, the largest-volume part has the greatest total charge, so option A is correct. The smallest-volume part has less charge, and a zero-volume part would contain no charge.

Correct answer: A. It doubles

Explanation: For a sufficiently small charge element, the contribution can be treated using Coulomb’s law: dE = k|dq|/r². With r fixed, dE is directly proportional to |dq|. Therefore replacing dq by 2dq changes the magnitude to k|2dq|/r² = 2dE. The direction remains determined by the sign and geometry, but the magnitude doubles. Hence A is correct; B reverses the proportionality and C and D are unsupported.

Correct answer: C. It becomes one-fourth

Explanation: For a small element treated as a point-like charge, Coulomb’s law gives dE = k|dq|/r². If the charge is unchanged and the distance changes from r to 2r, the new magnitude is dE' = k|dq|/(2r)² = dE/4. Thus the field becomes one-fourth of its original value. Option A ignores the inverse-square dependence, B uses inverse-first-power dependence, and D ignores distance dependence.

Correct answer: A. By vector addition

Explanation: Electric field obeys the superposition principle as a vector quantity. Consequently, equal magnitudes cannot simply be added unless all contributions point in the same direction. Each contribution must be resolved into components, and the components must be summed with their signs and directions; the resultant magnitude and direction then follow from those sums. Thus A is correct. B neglects direction, C assumes cancellation without proof, and D ignores the actual geometry.

Correct answer: A. No, only rearrangement of charges occurs

Explanation: A nearby charged rod produces electrostatic induction: mobile electrons in the conductor redistribute, creating oppositely charged near and far regions. However, if the conductor has no contact with Earth or another charged body, no charge can enter or leave it. Its total or net charge therefore remains unchanged, although the surface distribution changes. Hence option A is correct.

Correct answer: A. Zero

Explanation: By Gauss’s law, the net flux through a closed surface depends only on the algebraic total charge enclosed, not on the positions of the charges. An electric dipole contains +q and −q, so Q_enclosed = +q − q = 0. Thus Φ = Q_enclosed/ε₀ = 0, and option A is correct. Individual field lines can cross the surface, but their signed flux cancels overall.