Continuous Charge Distribution

A square metre is a unit of area. Therefore C m⁻² represents charge per unit area, called surface charge density and commonly denoted by σ. Linear density uses C m⁻¹, whereas volume density uses C m⁻³. Time distribution is not the relevant spatial classification of charge. Hence option B correctly identifies the distribution.

A cubic metre is a unit of volume, so C m⁻³ means that charge is measured per unit volume. This quantity is the volume charge density, usually represented by ρ, and it appears in the relation dq = ρ dV. Linear and surface densities use metres and square metres in the denominator, respectively. Therefore option C is correct.

A spherical shell is a two-dimensional charged surface: its charge is distributed over area, not along only one line or throughout a three-dimensional volume. The appropriate quantity is surface charge density, σ = dq/dA. Hence option B is correct. Linear density applies to a wire, and volume density ρ = dq/dV applies to a solid charged body. “Time density” is not a standard charge-distribution quantity.

A solid sphere occupies a three-dimensional region, and the statement says that charge is present throughout its interior. Therefore the charge is described by a volume distribution, with volume density ρ = dq/dV. Option C is correct. Linear distribution belongs to a line or wire, surface distribution belongs to a shell or plate, and point distribution represents charge concentrated at one location rather than throughout the sphere.

Surface charge density is defined as σ = dq/dA. If equal areas contain equal charges, the ratio dq/dA is the same for each part, so σ is constant across those parts. This is the defining indication of a uniform surface distribution, making option A correct. A non-uniform distribution would give different charges in equal areas, while volume and linear distributions use volume and length rather than area.

Volume charge density is defined as ρ = dq/dV. When equal volumes contain equal charges, dq/dV has the same value in those portions, indicating a uniform volume distribution. Therefore option A is correct. This observation does not describe a line or surface distribution, and equal finite charges in finite volumes do not imply that the total charge is infinite. The word “may” appropriately allows the conclusion for the stated region.

Linear charge density is defined as charge per unit length: λ = Q/L. Suppose the original wire has density λ and length L. Each one-third section contains charge Q/3 and has length L/3, so its density is (Q/3)/(L/3) = Q/L = λ. Thus cutting the wire changes the total charge of each piece, but not its density, so option A is correct.

Surface charge density is σ = Q/A, the charge per unit area. If the original plate has charge Q and area A, each equal quarter has charge Q/4 and area A/4. Therefore its density is σ' = (Q/4)/(A/4) = Q/A = σ. The total charge and area of an individual part decrease together, so the density remains unchanged. Hence option B is correct.

For a uniformly surface-charged plate, total charge is Q = σA. Since both plates have the same σ, their charges are directly proportional to their areas: Q₁/Q₂ = (σA₁)/(σA₂) = A₁/A₂. With A₁:A₂ = 2:5, the charge ratio is Q₁:Q₂ = 2:5. Therefore option B is correct; reversing the ratio would give the distractor 5:2.

Linear charge density is defined as λ = Q/L for a uniform distribution, where Q is total charge and L is the positive length of the distribution. Dividing a negative charge by a positive length gives a negative value of λ. The density is not zero unless the total charge is zero. Therefore the correct answer is option B; the sign follows directly from the sign of the total charge.

A linear distribution occupies one-dimensional length, so its charge density is λ = dq/dl. Rearranging gives the charge of a small element as dq = λ dl. Surface density σ must be multiplied by area, and volume density ρ by volume; those measures do not describe a purely linear element. Therefore option A gives both the correct physical quantities and the correct differential relation.

For a surface distribution, surface charge density is defined by σ = dq/dA, where dA is a small area element. Therefore the charge on that element is dq = σ dA. A linear density must be paired with a length, while a volume density must be paired with a volume. Since this situation is two-dimensional, option B is the only expression with the correct density and geometrical measure.

A volume distribution fills a three-dimensional region, and its volume charge density is defined as ρ = dq/dV. Hence the charge in a small volume element is dq = ρ dV. A linear density requires a length element and a surface density requires an area element. Although pressure may multiply volume in mechanics, it is not an electric charge density. Therefore option C is correct.

Uniform surface distribution means the surface charge density σ is the same everywhere. Therefore equal areas contain equal charges. If the total charge is Q = 40 C and the surface is divided into N = 8 equal parts, the charge in one part is q = Q/N = 40/8 = 5 C. Hence option B is correct; 8, 10, and 40 C result from incorrect division or using the total charge unchanged.

In a uniform linear distribution, the linear charge density λ is constant, so charge is proportional to length. Five equal-length sections must therefore carry equal charges. With total charge Q = 30 C and five sections, q = Q/5 = 30/5 = 6 C for each section. Option B is correct; 5 C, 10 C, and 30 C do not satisfy conservation of the total charge when five equal parts are present.

A uniform volume distribution has constant volume charge density ρ, so equal volumes contain equal amounts of charge. The total charge is Q = 36 C and there are six equal-volume parts. Thus the charge in each part is q = Q/6 = 36/6 = 6 C. Option A is correct. Twelve C would account for only three parts, three C would give too little total charge, and 36 C would incorrectly assign the whole charge to each part.

A surface charge distribution is uniform only when its surface charge density σ has the same value at every position on the surface. If charge is concentrated more strongly near the edges, σ depends on position and is larger there than elsewhere. Therefore the distribution is non-uniform surface distribution. It is not linear because the charge remains spread over an area, and it is not a volume distribution because no three-dimensional region is being specified.

For a uniform volume charge distribution, total charge is Q = ρV. Volume V is a positive geometric measure, while the volume charge density ρ carries the sign of the charge. If ρ is negative and V is positive, their product is negative, so Q is negative. Therefore option B is correct. It is not necessarily zero, and a physical charge has a definite sign rather than being signless.

For a uniform surface charge distribution, surface charge density is defined as σ = Q/A, where Q is total charge and A is the non-zero area. Substituting Q = 0 gives σ = 0/A = 0. Therefore option A is correct. A positive or negative density would produce a non-zero total charge over a non-zero area, while infinite density is not implied by a zero numerator.

Linear charge density is defined as λ = dQ/dl, or for a uniform segment λ = Q/L. If the original continuous distribution is uniform, every portion has the same charge per unit length. Cutting it into pieces changes each piece’s total charge and length in proportion, but not the local density. Therefore option A is correct; zero total charge or unequal lengths is not required.

Surface charge density is σ = dQ/dA. In a uniformly charged plate, every equal area initially carries the same charge per unit area. Cutting the plate separates existing regions; it does not redistribute charge or alter the charge-to-area ratio of the material in each region. Thus a part’s total charge decreases with its area, while σ remains unchanged, making option A correct.

For a uniform volume distribution, the volume charge density ρ is constant, so the charge in any part is Q_part = ρV_part. If the body is divided into equal volumes, each part has the same value of ρV_part and therefore the same charge. If there are n equal parts, each contains Q_total/n. Equal surface area, zero charge, or negative density is not necessary; option A is correct.

Uniform surface charge density is defined by σ = Q/A, so the total charge is Q = σA. If σ remains constant and the area A increases, the number of equal-area surface portions increases, and the total charge increases in direct proportion to A. For example, doubling the area doubles Q when the density is unchanged. Therefore A correctly expresses the governing relation; B, C, and D have no physical basis.

In a uniform linear distribution, the linear density λ is constant, so charge is directly proportional to length: Qpart = λLpart and Qtotal = λLtotal. Dividing gives Qpart/Qtotal = Lpart/Ltotal = 2/5. Thus the part carries two-fifths of the total charge, so option B is correct. The other fractions do not follow from the stated length ratio and would require a non-uniform distribution or a different fraction.

For a uniform surface distribution, surface density σ is constant, so charge is proportional to area. If the part has area Apart = (3/8)Atotal, then Qpart = σApart and Qtotal = σAtotal. Consequently Qpart/Qtotal = Apart/Atotal = 3/8. Option A is therefore correct. The other choices either use the complementary fraction, an unrelated fraction, or an incorrect conversion of the given area ratio.