Continuous Charge Distribution

Correct answer: A. Total charge equals linear density multiplied by length

Explanation: 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.

Correct answer: B. From surface density and area

Explanation: 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.

Correct answer: A. From volume density and volume

Explanation: 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.

Correct answer: A. Away from that small positive charge

Explanation: 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.

Correct answer: B. Toward that small negative charge

Explanation: 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.

Correct answer: B. Half the total charge

Explanation: 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.

Correct answer: B. One-fourth of the total charge

Explanation: 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.

Correct answer: A. Half the total charge

Explanation: 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.

Correct answer: B. Non-uniform linear distribution

Explanation: 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.

Correct answer: B. Surface distribution

Explanation: 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.

Correct answer: A. Because charge is spread throughout the volume, not only on the surface

Explanation: 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.

Correct answer: C. It remains unchanged

Explanation: 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.

Correct answer: A. The sign of the charge distributed in that region

Explanation: 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.

Correct answer: A. It will be equal in all three parts

Explanation: 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.

Correct answer: A. They are equal

Explanation: 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.

Correct answer: A. It becomes four times

Explanation: 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.

Correct answer: A. Whether the distribution is linear, surface, or volume

Explanation: 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.

Correct answer: B. 2 C/m

Explanation: 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.

Correct answer: A. 2 C/m²

Explanation: 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.

Correct answer: B. 4 C/m³

Explanation: 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.

Correct answer: D. 10 C

Explanation: 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.

Correct answer: B. 12 C

Explanation: 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.

Correct answer: C. 12 C

Explanation: 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.

Correct answer: C. One half

Explanation: 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.

Correct answer: A. One fourth

Explanation: 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.