Continuous Charge Distribution

Correct answer: B. The density remains the same

Explanation: The governing relation for a uniformly charged wire is linear charge density λ = Q/L, where Q is total charge and L is length. If the charge becomes kQ while the length becomes kL, the new density is λ′ = kQ/(kL) = Q/L = λ. Thus both quantities scale by the same factor and the charge per unit length is unchanged. It would double only if charge doubled without a corresponding increase in length, so option B is correct.

Correct answer: B. It becomes double

Explanation: Surface charge density is defined by σ = Q/A, where Q is total charge and A is area. Initially σ = Q/A. If the area is reduced to A′ = A/2 while the charge remains Q, then σ′ = Q/(A/2) = 2Q/A = 2σ. Therefore the same charge is distributed over half the area, so the density doubles. It does not become half or zero, and halving the area does not produce a fourfold increase. Option B is correct.

Correct answer: C. Total charge to volume

Explanation: Volume charge density describes charge distributed through a three-dimensional region. Its definition is ρ = Q/V for a uniform distribution, or more generally ρ = dQ/dV. Thus the required ratio is total charge divided by volume, making option C correct. Q/L defines linear density, Q/A defines surface density, and Q/t represents current rather than volume density.

Correct answer: B. Because electric field is a vector quantity

Explanation: Electric field is a vector: each element contributes both a magnitude and a direction. The total field is therefore obtained by vector superposition, E = ∫dE, with components added algebraically along chosen axes. Contributions can reinforce or cancel depending on direction. Hence option B is correct; treating E as a scalar would generally give the wrong result.

Correct answer: C. It doubles

Explanation: For a uniformly charged wire, total charge is Q = λL, where λ is the constant linear charge density and L is length. If L changes to 2L while λ remains unchanged, the new charge is Q′ = λ(2L) = 2λL = 2Q. Therefore option C is correct. The charge remains unchanged only if the density decreases inversely with length.

Correct answer: A. Total charge

Explanation: Surface charge density is defined by σ = Q/A. Rearranging gives Q = σA, so knowledge of the surface density and total area determines the total charge directly. Thus option A is correct. Length and volume are not obtained from these two quantities alone, and mass would require additional information such as material density and thickness.

Correct answer: B. Because fields due to opposite elements are equal and opposite

Explanation: At the centre of a uniformly charged ring, every small charge element has an opposite element at the same distance. Their electric-field magnitudes are equal, while their directions are opposite along the joining line. Pairwise vector cancellation therefore gives E_total = 0. Option B is correct; the ring can have nonzero total charge, and the distance is finite.

Correct answer: C. On volume and volume charge density

Explanation: For a continuous volume distribution, charge is present throughout the three-dimensional region. Volume charge density is defined as charge per unit volume, ρ = dQ/dV. If the density is uniform, integration gives Q = ρV. Thus increasing either the volume or the density increases the total charge. Colour, surface length, and time do not determine charge in the stated static situation, so option C is correct.

Correct answer: B. Add the charge of small elements with variable density

Explanation: For a non-uniform line distribution, the linear charge density λ depends on position. A small element dx carries dQ = λ(x)dx, so the total charge is obtained by summing all elements or integrating: Q = ∫λ(x)dx over the complete line. Average distance is irrelevant, area belongs to surface distributions, and the charge is not automatically zero. Therefore option B is correct.

Correct answer: B. Both parts have equal charge

Explanation: Surface charge density is defined by σ = dQ/dA. For a uniform plate, σ has the same value at every point. If two parts have equal area A, each carries Q = σA, so their charges are equal. The charge need not be zero, and equal regions do not have opposite signs unless the problem explicitly states different charge signs. Hence option B is correct.

Correct answer: A. When the object’s size is significant compared with the observation distance

Explanation: A point-charge approximation replaces an extended body by a single charge located at one point. It is reliable when the body’s dimensions are much smaller than the relevant distance. If the dimensions are comparable to that distance, different elements produce noticeably different fields, so the charge distribution must be treated continuously. Therefore option A is correct; being far away or very small generally improves the approximation.

Correct answer: A. Components from opposite equidistant elements are equal and opposite

Explanation: Consider two equal, symmetrically placed elements of the uniformly charged rod. At a point on the perpendicular bisector, both elements are at the same distance and produce fields of equal magnitude. Their components parallel to the rod point in opposite horizontal directions, so these components cancel pairwise. The perpendicular components point in the same direction and add. The cancellation is due to symmetry, not zero total charge or infinite distance; option A is correct.

Correct answer: A. Along the axis

Explanation: For a point on the axis of a uniformly charged ring, each charge element has a diametrically opposite partner. The components of their fields perpendicular to the axis cancel because of symmetry, while their axial components reinforce one another. Consequently, the resultant field is directed along the ring’s axis, with its sense depending on whether the ring is positively or negatively charged and on the side of the point. It is not generally zero, so option A is correct.

Correct answer: C. Fields of symmetric opposite elements cancel

Explanation: At the centre of a uniformly charged ring, every small charge element has an equal element diametrically opposite to it. Both are at the same distance, so their electric fields have equal magnitudes, but the directions are opposite. Pairwise cancellation around the entire ring therefore gives a resultant field of zero. This conclusion follows from geometric symmetry, not from negative charge or a charge placed at the centre. Hence option C is correct.

Correct answer: A. When the way charge is distributed also affects the result

Explanation: The governing concept is superposition for a continuous charge distribution. The total field is obtained by adding contributions from all elements, commonly written as dE = (1/4πε₀)(dq/r²) in the appropriate direction. Thus, the positions, shape, and charge density—not merely total Q—matter. Option A is correct because different distributions with the same Q can produce different fields. The sign of Q alone or distance alone does not make total charge insufficient in every case.

Correct answer: A. When the element is sufficiently small compared with the relevant distances

Explanation: The governing approximation is the differential-element method. A continuous distribution is divided into very small elements dq; if an element’s dimensions are negligible compared with the distances used to calculate its field, its contribution can be approximated as that of a point charge. Hence option A is correct. The total distribution need not have zero charge, color is irrelevant, and zero separation makes the point-charge expression undefined rather than justified.

Correct answer: A. Add the electric fields of all small charge elements vectorially

Explanation: The superposition principle states that the electric field produced by a continuous distribution is the vector sum, or integral, of the fields due to all infinitesimal charge elements. Each element contributes dE = (1/4πε₀)dq/r² with its own direction. Thus option A is correct; the other choices wrongly discard charge contributions or vector direction.

Correct answer: B. The components perpendicular to the axis

Explanation: Consider two charge elements located at opposite ends of a diameter. At an axial point, their field components perpendicular to the axis have equal magnitudes but opposite directions, so they cancel pairwise. Their axial components point in the same axial direction and add. Hence option B is correct, while option A reverses the symmetry result and options C and D are false.

Correct answer: A. When the area remains unchanged

Explanation: Surface charge density is defined as σ = Q/A, where Q is total charge and A is area. If the area remains constant and Q is multiplied by a factor k, then σ' = kQ/A = kσ, so the density changes in exactly the same ratio. Therefore option A is correct. If area also changes, the ratio depends on both changes.

Correct answer: B. The positive and negative contributions add to zero

Explanation: Total charge is the signed integral of the charge density, Q = ∫ρ dV, or the corresponding line or surface integral. Positive and negative regions may therefore cancel in the total even though the local density is nonzero. Hence option B is correct. A zero net charge does not imply zero charge everywhere, nor does it require density to be positive.

Correct answer: A. Because shape can affect charge distribution and the direction and magnitude of the field

Explanation: The electric field is obtained by integrating contributions from all charge elements, and each contribution depends on its position and direction relative to the observation point. The object’s shape determines how charge can spread over a line, surface, or volume and changes the distances and symmetry of those elements. Thus option A is correct; shape neither removes charge nor automatically changes its sign.

Correct answer: A. Linear charge density changes with position

Explanation: The governing concept is linear charge density, defined as charge per unit length: λ = dq/dl. In a non-uniform distribution, λ varies from one position to another. Thus, for equal small lengths dl, the charges dq = λ dl can be different. Option A is correct; the other choices either deny the meaning of length or make unsupported claims about zero charge or area.

Correct answer: A. It allows the charge and field to be analysed using many small elements

Explanation: A continuous model treats the charge on the wire as infinitely many small elements dq = λ dl rather than as a few isolated point charges. Their electric-field contributions can then be integrated over the wire, while uniformity makes λ constant. Option A is correct; the model changes the method of calculation, not the charge, mass, or magnetic nature of the wire.

Correct answer: B. Non-uniform surface distribution

Explanation: Surface charge density is defined by σ = dq/dA. A uniform surface distribution requires the same σ, so equal areas must carry equal charges. Here equal areas contain different charges, meaning σ changes from region to region. Therefore option B is correct. Option A contradicts the observation, while C describes a length-based distribution and D implies zero charge.

Correct answer: A. They are equal

Explanation: The governing quantity is volume charge density, ρ = dq/dV. In a uniform volume distribution, ρ has the same value throughout the region. For two equal volumes, dq₁ = ρdV and dq₂ = ρdV, so dq₁ = dq₂. Hence option A is correct. Uniformity does not require zero or negative charge; it only requires constant density.