Continuous Charge Distribution

Correct answer: B. Non-uniform charge distribution

Explanation: The governing distinction is whether density remains constant throughout the distribution. If charge density changes from one position to another, it is a non-uniform or variable charge distribution. Mathematically, the relevant density function, such as lambda(x), sigma(x), or rho(x), depends on position. A uniform distribution has constant density, a point charge is localized, and neutrality concerns total charge rather than spatial variation. Therefore option B is correct.

Correct answer: A. Summing the contributions of infinitesimal charge elements

Explanation: In a continuous charge distribution, charge is divided into infinitesimal elements such as \(dq\). To find the total charge, electric field, or potential, the infinitesimal contribution from each \(dq\) is summed over the entire distribution by integration. Thus, integration is the continuous form of summation. Option B is a possible approximation in special cases, not the meaning of integration.

Correct answer: A. When object size is very small compared with distance

Explanation: The point-charge approximation neglects the physical size of a charged body and treats all its charge as concentrated at one point. It is therefore reliable when the body's dimensions are much smaller than the distance at which the field or force is being evaluated. A long wire, large plate, or position-dependent density requires a distributed-charge treatment, usually involving integration.

Correct answer: A. When charge is on an extended object

Explanation: A continuous charge model is used when charge is spread over an extended body rather than concentrated at a single point. The body may be represented by linear density λ, surface density σ, or volume density ρ, and small elements are added through integration. A negligible-sized object is better approximated as a point charge, while no charge or mass alone does not define this model.

Correct answer: A. Total charge

Explanation: For a uniform line charge, the linear charge density is λ = Q/L, where Q is total charge and L is the length. Rearranging gives Q = λL. Thus, knowing the constant linear density and the length directly determines the total charge. Colour and temperature are unrelated to this electrostatic relation, and mass cannot be obtained without separate material information.

Correct answer: A. Total charge

Explanation: For a uniformly charged surface, surface charge density is defined as σ = Q/A, where Q is the total charge and A is the area carrying the charge. Therefore Q = σA. The given density and area are exactly the required quantities for calculating total charge. Linear length, speed, and sound are not consequences of the surface-charge relation.

Correct answer: A. Total charge

Explanation: For a uniformly charged volume, the volume charge density is ρ = Q/V, with Q denoting total charge and V denoting the occupied volume. Hence the total charge follows immediately as Q = ρV. Area and length are geometric quantities that are not determined alone by ρ and V, and colour has no connection with the electrostatic density equation.

Correct answer: A. Uniform linear charge distribution

Explanation: A thin circular ring has negligible thickness, so its charge is modeled as lying along one-dimensional circumference rather than throughout an area or volume. If the charge is uniform, its linear density is constant: λ = Q/(2πR). Therefore it represents a uniform linear charge distribution. A surface model would apply to a thin sheet, and a volume model to charge filling a three-dimensional body.

Correct answer: B. Surface distribution

Explanation: A spherical shell is a hollow object whose charge is spread over its two-dimensional spherical surface, not along a single line and not throughout the enclosed volume. Its uniform surface charge density is σ = Q/(4πR²). Thus the appropriate classification is uniform surface distribution. A point model may approximate the external field under special conditions, but it is not the actual distribution described here.

Correct answer: C. Volume distribution

Explanation: A solid sphere occupies a three-dimensional region, and a uniformly charged solid sphere has charge distributed throughout that volume. Its volume charge density is ρ = Q/V; for a sphere of radius R, V = 4πR³/3. Therefore it is a volume distribution. A surface distribution describes a shell, while a point distribution ignores the body's internal charge structure and is only an approximation in limited situations.

Correct answer: A. To simplify calculation and identify direction

Explanation: Symmetry allows equivalent charge elements to be grouped and reveals which components of the electric field cancel. For example, in a symmetric ring or spherical distribution, transverse contributions can cancel in pairs, leaving only the component permitted by the symmetry axis. This reduces vector addition and often converts a difficult integral into a simple expression. It neither removes charge nor changes mass or colour.

Correct answer: A. Zero

Explanation: Electric field is a vector, so both magnitude and direction must be considered. Represent the two equal opposite components as +E and −E along the same axis. Their vector sum is E_resultant = E + (−E) = 0. Hence they cancel exactly. The result would be double only for equal components in the same direction; infinity and negative mass are unrelated to vector addition.

Correct answer: A. Superposition rule

Explanation: The governing concept is the principle of superposition. A continuous distribution is divided into very small charge elements, and each element produces a small electric-field vector. The resultant field is obtained by adding all these vectors, often through integration: dE is summed over the distribution. Therefore option A is correct; heat, sound, and friction rules do not determine electrostatic field addition.

Correct answer: A. More charge in one unit area

Explanation: Surface charge density is defined as charge per unit area, written as σ = dQ/dA, or Q/A for a uniform surface. If σ is higher, a chosen equal area contains more charge than an area with lower σ. Thus option A correctly describes the meaning. Higher density does not imply that the entire surface is uncharged, that only mass increases, or that the area becomes zero.

Correct answer: A. Less charge in one unit length

Explanation: For a line distribution, linear charge density is defined by λ = dQ/dl, and for a uniform wire it is λ = Q/L. A lower λ means that an equal length contains less charge, or that charge is less concentrated along the line. Hence option A is correct. It does not mean the line has zero length, that its charge is infinite, or that the line is merely a magnet.

Correct answer: A. It can affect the direction of the field

Explanation: The sign of each charge element determines the direction of the electric field produced by that element: a positive element produces a field away from it, while a negative element produces a field toward it. In a continuous distribution, these signed vector contributions are integrated, so cancellation or reinforcement depends on their signs. Therefore option A is correct; the sign does not specify colour, remove mass, or make length zero.

Correct answer: B. No, positive and negative parts may balance

Explanation: Total charge is the algebraic sum of the charges of all elements: Q = ∫dq. It can be zero even when individual elements carry nonzero charge, because positive and negative contributions may cancel; for example, +5 μC and −5 μC give Q = 0. Thus option B is correct. Zero net charge does not mean zero charge density at every point, and it certainly does not imply infinite charge.

Correct answer: A. Equal

Explanation: A uniform continuous distribution has constant charge density throughout the relevant line, surface, or volume. Therefore equal lengths, equal areas, or equal volumes contain equal charge, provided the compared elements have equal size and orientation where relevant. Mathematically, dq = λdl, σdA, or ρdV; constant density makes equal measures produce equal dq. Hence option A is correct, while zero, always different, and infinite are unsupported.

Correct answer: B. It can be different

Explanation: In a non-uniform distribution, the charge density changes with position. For equal-sized elements, dq = λdl, σdA, or ρdV; although dl, dA, or dV are equal, the density factor may differ from one location to another. Consequently, the elements can carry different charges. Option B is correct. Equal charge is not required, and non-uniformity does not mean that every charge is zero or infinite.

Correct answer: A. Because how charge is spread can also be important

Explanation: The electric field depends not only on the net charge Q but also on the positions, shape, and density of the charge elements. Two distributions can have the same total charge while producing different fields because their distances and directions relative to the observation point differ. The field is found from a sum or integral of elemental contributions. Thus option A is correct; the other statements contradict electrostatic principles.

Correct answer: A. Because charge spread can depend on shape

Explanation: The geometry of a charged object determines where its charge elements lie and how their field vectors combine. A line, sphere, disk, or irregular surface can have a different density pattern and symmetry, leading to different electric fields even when the total charge is the same. Shape also determines the appropriate element, such as dl, dA, or dV. Therefore option A is correct; shape does not erase charge or mass.

Correct answer: A. To add the contribution of each small part

Explanation: A charged wire is modelled as a continuous line distribution. Dividing it into small elements dl allows the charge on an element to be written as dq = λdl and its electric-field contribution to be calculated using Coulomb’s law. The total field is then obtained by integrating, E = ∫dE, over the entire wire. Hence option A is correct; the division is mathematical and does not change the wire’s mass, charge, or colour.

Correct answer: A. Charged wires, plates and spheres

Explanation: The governing idea is that charge may be distributed continuously over an extended body instead of being concentrated at one point. A wire is treated with linear density, a plate with surface density, and a sphere with volume or surface density, depending on the model. Therefore option A is correct; the other choices are unrelated to charge distribution.

Correct answer: A. Equal charge is present in equal-area parts

Explanation: Surface charge density is defined as charge per unit area: σ = Q/A. If σ is constant, any two portions having equal area must contain equal charge, because Q = σA for each portion. Thus option A correctly describes a uniform surface distribution. The other choices either describe zero or concentrated charge, not constant nonzero density.

Correct answer: A. Whether charge is spread along a line, surface or volume

Explanation: The first governing step is to identify the geometrical support of the charge: a line, a surface, or a volume. This determines the appropriate density, respectively λ = dQ/dl, σ = dQ/dA, or ρ = dQ/dV, and therefore the correct integration element. Option A is correct; colour, temperature, and sound do not select the electrostatic model.