Continuous Charge Distribution

A wire is modeled as a one-dimensional charge distribution, so its relevant density is the linear density λ, defined as charge per unit length. For a uniform wire, dQ = λ dl and integration over length L gives Q = λL. Surface density must be multiplied by area, while volume density must be multiplied by volume. The total charge is not always zero, so option A is correct.

For a surface distribution, the appropriate quantity is surface charge density σ, defined as charge per unit area. For a uniform surface, every small area dA carries dQ = σdA. Integrating over the complete area A gives Q = σA. Linear density applies to a wire and volume density applies to a three-dimensional body; distance alone cannot determine charge. Thus option B is correct.

Charge distributed throughout the interior of a solid forms a volume distribution. Its volume charge density ρ is charge per unit volume. For a uniform density, an element has dQ = ρdV, and integration over the solid volume V gives Q = ρV. Linear density belongs to line distributions and surface density belongs to sheets; temperature alone does not define the charge. Therefore option A is correct.

The governing concept is the direction of the electric field produced by a point charge. A very small element of a continuous distribution can be treated as an elemental point charge, dQ. For dQ > 0, the field vector points radially outward, away from the element. Therefore option A is correct. Option B describes a negative charge, while options C and D are not general properties of the field; the net field may depend on all elements.

The governing rule is that the electric field points away from a positive charge and toward a negative charge. A sufficiently small charge element can be represented by dQ, like a point charge. For dQ < 0, the field vector at a surrounding point is directed toward the element, so option B is correct. Option A applies to positive charge. The field is not necessarily axial or zero unless a special symmetric arrangement produces that result.

For a uniform linear charge distribution, the linear charge density λ is constant and charge is proportional to length: Q = λL. If the selected length is L/2, its charge is q = λ(L/2) = Q/2. Therefore option B is correct. Option A would require the entire length, option C would correspond to one-fourth of the length, and option D is impossible for a nonzero uniform density.

In a uniform surface distribution, the surface charge density σ is constant, so charge is proportional to area: Q = σA. For a region whose area is A/4, the charge is q = σ(A/4) = Q/4. Hence option B is correct. Half the charge would require half the area, the whole charge would require the complete area, and zero would not follow from a nonzero uniform surface density.

For a uniform volume distribution, the volume charge density ρ is constant and charge is proportional to volume: Q = ρV. If the selected region has volume V/2, then its charge is q = ρ(V/2) = Q/2. Thus option A is correct. It cannot contain twice the total charge, and it is not zero or Q/4 unless the stated volume fraction or density is different. Uniformity is the essential condition.

A wire is modeled as a one-dimensional object, so its charge is described by linear charge density λ = dq/dl. In a uniform linear distribution, equal lengths contain equal charges and λ remains constant. Here equal-length halves carry different charges, so λ changes with position; therefore option B, non-uniform linear distribution, is correct. Surface and volume distributions describe charge spread over two- or three-dimensional regions, not along an ideal wire.

The governing classification depends on the geometrical region occupied by charge. When charge is confined to the two-dimensional outer surface of a sphere, it is described by surface charge density σ = dq/dA. Therefore option B is correct. A linear distribution applies to a wire or curve, a volume distribution applies when charge fills the interior, and a point distribution idealizes all charge at one location.

The governing distinction is the region occupied by charge. A surface distribution has charge only on the boundary and is described using σ = dq/dA. In a uniformly charged solid sphere, charge exists throughout the three-dimensional interior, so it is a volume distribution described by ρ = dq/dV. Therefore option A is correct. A sphere can hold charge, its total charge need not be zero, and it certainly has a surface area.

Linear charge density is defined by λ = Q/L, where Q is the total charge and L is the length. Initially λ = Q/L. After both quantities double, λ' = 2Q/2L = Q/L = λ. Thus the linear density remains unchanged, so option C is correct. It would double only if the charge doubled while the length stayed fixed.

Charge density carries the algebraic sign of the charge being distributed. For example, positive linear, surface, or volume density represents positive charge, while negative density represents negative charge. Thus option A is correct. The sign does not directly describe colour, temperature, or mass; those are different physical properties and are not encoded by charge density.

For a uniform surface distribution, the surface charge density σ is constant, meaning every equal area contains the same charge per unit area. The charge on a part is Q = σA. Since all three parts have the same A and the same σ, each has the same Q. Therefore option A is correct; unequal charge would require unequal areas or nonuniform density.

Uniform linear charge density means λ = dQ/dl has the same value at every point. For a small segment of length Δl, its charge is ΔQ = λΔl. Equal segments have equal Δl and experience the same λ, so their charges are equal. Therefore option A is correct. Different charges would indicate a nonuniform density or unequal segment lengths.

The governing relation for a uniform volume charge distribution is Q = ρV, where Q is total charge, ρ is volume charge density, and V is volume. Since ρ remains constant while V changes to 4V, the new charge is Q′ = ρ(4V) = 4Q. Therefore option A is correct. Option B would imply an inverse dependence, option C ignores the change in volume, and option D has no physical basis here.

The essential first step is to determine the geometry of the charge distribution: line, surface, or volume. This choice determines the appropriate density—λ = dQ/dl, σ = dQ/dA, or ρ = dQ/dV—and therefore the correct integration or calculation method. Option A is correct. Colour, sound, and classroom temperature do not define the electrostatic charge model in this problem.

For a uniform linear distribution, linear charge density is defined as charge per unit length: λ = Q/L. Substituting Q = 6 C and L = 3 m gives λ = 6/3 = 2 C m⁻¹. Therefore option B is correct. Option A results from an incorrect division, option C confuses density with length, and option D simply repeats the total charge without accounting for the 3 m length.

Surface charge density is the charge distributed per unit area, so σ = Q/A for a uniform surface. With Q = 8 C and A = 4 m², σ = 8/4 = 2 C m⁻². Hence option A is correct. Option B uses half the required division result, option C copies the total charge without dividing by area, and option D incorrectly multiplies instead of dividing.

For a uniform volume distribution, volume charge density is total charge divided by volume: ρ = Q/V. Using Q = 12 C and V = 3 m³, we obtain ρ = 12/3 = 4 C m⁻³. Therefore option B is correct. Option A uses an incorrect quotient, while options C and D fail to apply the required division by the given volume.

For a uniform linear charge distribution, the governing formula is Q = λL, because total charge equals charge per unit length multiplied by the length. Substituting λ = 2 C/m and L = 5 m gives Q = (2 C/m)(5 m) = 10 C; the metre units cancel. Thus option D is correct. Options A and B use only one of the supplied values, while option C is an unsupported sum rather than the required product.

For a uniform surface charge distribution, total charge is calculated from Q = σA, where σ is surface charge density and A is area. Here Q = (3 C/m²)(4 m²) = 12 C, with the square-metre units cancelling. Therefore option B is correct. Options C and D merely repeat the individual data values, while option A adds 3 and 4 instead of multiplying density by area, so it does not represent total charge.

For a uniform volume distribution, the total charge is given by Q = ρV. Substituting the volume density ρ = 2 C/m³ and volume V = 6 m³ gives Q = 2 × 6 = 12 C; the cubic-metre units cancel. Thus option C is correct. The other choices either use only one datum or give an unrelated arithmetic result rather than applying the volume relation.

For a uniformly charged ring, the linear charge density λ is constant. The charge on any arc is q = λl, so charge is directly proportional to arc length. If the selected arc has half the circumference, l = (1/2)(2πR), then q = λπR = Q/2, where Q is the charge on the complete ring. Hence the required fraction is one half, not one fourth or zero.

For a uniform surface charge distribution, surface density σ is constant and charge on a patch is q = σA. If the selected patch has area A/4, its charge is q = σ(A/4) = (σA)/4 = Q/4, where Q is the charge on the entire spherical surface. Thus the fraction is one fourth. The radius or shape does not alter this area-ratio result.