Continuous Charge Distribution

The governing concept is uniform volume charge density, defined as charge per unit volume: ρ = Q/V. Since ρ is constant, the charge in any part is proportional to its volume. Thus Q_part/Q_total = V_part/V_total = (7/10)V/V = 7/10. Therefore option C is correct. Options A and B use incorrect fractions, while option D would apply only if the part had the entire volume.

Linear charge density is defined as λ = Q/L, the charge per unit length. For a uniform wire, every segment initially has the same λ. Cutting changes the total charge and length of each piece in the same proportion, so their ratio remains unchanged: Q₁/L₁ = Q₂/L₂ = λ. Hence both pieces have equal linear density, making option C correct. Length changes total charge, not density.

Surface charge density is σ = Q/A, or charge per unit area. Because the original plate is uniformly charged, every portion retains the same σ after division, provided the charge distribution is not disturbed. The larger piece contains more total charge because it has greater area, but its charge per unit area is unchanged. Therefore option C is correct; confusing total charge with density leads to option A or B.

Volume charge density is ρ = Q/V, the charge contained per unit volume. In a uniformly charged solid, cutting it into pieces does not alter this local ratio. A larger piece carries more total charge and a smaller piece carries less, but Q/V remains equal for every piece: Q₁/V₁ = Q₂/V₂ = Q₃/V₃ = ρ. Therefore option C is correct, not the option based only on total charge.

For a uniform surface distribution, the total charge is Q = σA. Area is a geometrical measure and is positive for a nonzero surface. Therefore, if σ < 0, multiplication by A > 0 gives Q < 0. Option B is correct. The charge would be zero only if the density or area were zero; it is not signless because charge is an algebraic quantity with a sign.

For a uniform linear distribution, linear charge density is defined as λ = Q/L. Since the total charge Q is negative and the length L is a positive geometrical quantity, their quotient is negative: λ < 0. Thus option B is correct. A zero density would require zero total charge, while the density is not signless because it records the sign of charge per unit length.

For a uniform surface distribution, surface charge density is constant and is related to total charge by sigma = Q/A. Since the given total charge Q is zero while the area A is nonzero, division gives sigma = 0/A = 0. Therefore option A is correct. A positive or negative density would produce a corresponding nonzero total charge over a nonzero area, while an infinite density is neither implied nor physically justified by these conditions.

For a uniform linear charge distribution, total charge is Q = lambda L, where lambda is the constant charge per unit length and L is the length. Substituting lambda = 0 gives Q = 0 multiplied by L = 0, even though L is nonzero. Therefore A is correct. A positive or negative result would require a nonzero signed density, and an infinite charge cannot be inferred from a finite length and zero density.

For a uniform volume charge distribution, total charge is obtained from Q = ρV, where ρ is volume charge density and V is volume. Substituting the values gives Q = (−2 C/m³)(5 m³) = −10 C. The cubic-metre units cancel, leaving coulombs. Therefore option B is correct; option A incorrectly reverses the sign, while C and D do not use the full volume correctly.

For a uniform surface charge distribution, the total charge is Q = σA, where σ is surface charge density and A is the area. Here Q = (−3 C/m²)(4 m²) = −12 C, because the square-metre units cancel. Thus option B is correct. Option A has the wrong sign, option C comes from an incorrect operation, and option D ignores the nonzero density.

Uniform linear charge density is defined as charge per unit length, so λ = Q/L. Using the given values, λ = (−18 C)/(6 m) = −3 C/m. The negative sign indicates that the charge is negative, and division by length gives the required per-metre unit. Therefore option B is correct; A loses the sign, C forgets to divide, and D is unsupported.

A density expressed in coulombs per square metre is surface charge density, written as σ. For a small surface element with area dA, the charge is dq = σ dA. The square-metre unit cancels correctly, leaving coulombs. Therefore option B is correct. A would apply to linear density, C to volume density, and D has no role in defining charge from surface density.

A density given in coulombs per cubic metre is volume charge density, denoted by ρ. For a differential volume dV, the charge element is dq = ρ dV. Multiplying the units C/m³ by m³ leaves C, confirming the relation. Thus option A is correct; the other choices mismatch linear or surface density with the wrong geometric measure.

Charge density measured in coulombs per metre is linear charge density, represented by λ. For a small length dl of the distribution, the corresponding charge is dq = λ dl. The units (C/m)(m) give coulombs, confirming the formula. Therefore option C is correct; area and volume measures belong to surface and volume densities, while time and speed are irrelevant.

The total surface charge is Q = ∫σ dA. Let each half have area A/2, with densities +σ₀ and −σ₀. Then Q = (+σ₀)(A/2) + (−σ₀)(A/2) = σ₀A/2 − σ₀A/2 = 0. Equal areas and equal density magnitudes make the contributions cancel exactly. Thus option C is correct; neither a positive nor a negative net charge remains.

Electric field is a vector quantity, so its contributions must be combined using vector addition. When several field elements point in the same direction, their components along that direction have the same sign and add in magnitude. For example, parallel contributions E1 and E2 give a resultant E = E1 + E2 in that direction. Cancellation occurs only for suitable opposite components, not for equal-direction fields. Hence B is correct.

The governing rule is vector cancellation. If two field contributions have equal magnitudes E and opposite directions, they can be represented as +E and −E along one chosen axis. Their resultant is E_total = E + (−E) = 0. Such pairwise cancellation commonly follows from symmetry, although the total field is zero only for the specified contributions or complete symmetric arrangement. Therefore C is correct; doubling would require the same direction.

Charge density is classified according to the dimensional region in which charge is distributed, not merely according to the shape of the object. For charge confined to a spherical surface, the appropriate quantity is surface charge density σ = dQ/dA; for a uniformly charged complete spherical surface, σ = Q/(4πR²). Volume density ρ would apply to charge throughout the interior, while linear density λ applies to a line. Hence B is correct.

The name of a charge distribution depends on where the charge exists. Surface distribution means charge is restricted to a two-dimensional boundary and is described by σ = dQ/dA. In this question, charge occupies the entire three-dimensional interior of the solid sphere, so it is a volume distribution described by ρ = dQ/dV; for uniform density, ρ = Q/(4πR³/3). Thus A is correct.

Surface charge density for a uniform distribution is defined as σ = Q/A. The problem states Q = 0 and A is non-zero, so direct substitution gives σ = 0/A = 0. Equivalently, the integral relation Q = ∫σdA becomes Q = σA for uniform σ; with a finite non-zero area, zero total charge requires σ = 0. A positive or negative uniform density would produce a non-zero total charge, so A is correct.

The governing relation for a uniformly charged line is Q = lambda L, where Q is total charge, lambda is linear charge density, and L is length. Here lambda = 0 while L is finite and non-zero, so Q = 0 multiplied by L = 0 C. Thus option A is correct. The length does not create charge by itself; options B and C wrongly assign a sign, while D is impossible for zero density.

For a uniform linear distribution, the linear charge density is lambda = Q/L. Substituting the given values gives lambda = (-40 C)/(8 m) = -5 C m^-1. Therefore option B is correct. The negative sign must be retained because the total charge is negative. Option A loses that sign, option C fails to divide by length, and option D ignores the non-zero charge.

For a uniform surface distribution, total charge is calculated from Q = sigma A, where sigma is surface charge density and A is area. Hence Q = (-5 C/m²)(3 m²) = -15 C. Option B is correct because the area units cancel and the negative sign is preserved. Option A reverses the sign, option C reports the density instead of total charge, and D incorrectly assumes cancellation.

The governing equation for a uniform volume charge distribution is Q = rho V, where rho is volume charge density and V is volume. Substitution gives Q = (-3 C/m³)(7 m³) = -21 C. Therefore option B is correct. The cubic-metre units cancel, and the negative sign indicates net negative charge. Option A changes the sign, C uses incorrect arithmetic, and D disregards the stated density.

A surface charge distribution is called uniform only when its surface charge density sigma has the same value at every position on the surface. If sigma becomes greater near the edge, it varies with position, so the distribution is non-uniform. Therefore option B is correct. Option A contradicts the stated variation, while C describes charge distributed along a line and D describes charge distributed throughout a volume; neither matches a surface whose density changes from place to place.