Continuous Charge Distribution

Correct answer: A. Electric field is a vector and potential is a scalar

Explanation: Electric field is a vector, so contributions from symmetrically opposite elements of a uniformly charged ring cancel in direction at the centre. Electric potential is a scalar, so contributions add algebraically; for a positively charged ring they produce V = kQ/R, which is generally non-zero. Thus option A correctly identifies the essential distinction; the other statements are false.

Correct answer: B. Along the perpendicular bisector because components along the rod cancel

Explanation: Take two equal charge elements symmetrically placed about the midpoint of the rod. Their electric fields at a point on the perpendicular bisector have equal magnitudes. The components parallel to the rod point in opposite directions and cancel, whereas the perpendicular components point in the same direction and add. Hence option B is correct for a positively charged rod; for a negative rod the direction reverses along the same line.

Correct answer: A. Because charges on opposite parts may not be equal

Explanation: The zero field at the centre of a uniformly charged ring follows from symmetry: opposite elements are equally charged, equally distant, and produce equal opposite field vectors. If the linear charge density varies around the ring, opposite elements may carry different charges, so their contributions need not cancel pair by pair. A non-zero resultant field is therefore possible, making option A correct.

Correct answer: B. Add the contributions with their signs

Explanation: Linear charge density is signed: positive λ represents positive charge per unit length and negative λ represents negative charge per unit length. The total charge must therefore be calculated using Q = ∫λ dl, preserving the sign of each portion. Opposite contributions may partially or completely cancel, but zero cannot be assumed without the lengths and density values. Hence option B is correct.

Correct answer: C. Zero

Explanation: Let each half have length L. The first half contributes Q₁ = (+2 C/m)L, while the second contributes Q₂ = (−2 C/m)L. The total is Q = Q₁ + Q₂ = 2L − 2L = 0. Equal lengths and equal opposite densities make the positive and negative charges cancel exactly. Therefore option C is correct; the result is finite and not positive or negative.

Correct answer: A. Zero

Explanation: The governing idea is addition of charge over a continuous surface: Q = ∫σ dA. The two regions have equal areas, A/2 each, with surface densities +4 C m⁻² and −4 C m⁻². Thus Q = 4(A/2) − 4(A/2) = 0. Equal positive and negative contributions cancel, so option A is correct; the net charge is neither positive nor negative.

Correct answer: B. Add charges of small volume elements

Explanation: For a non-uniform volume distribution, the volume charge density ρ depends on position, so one cannot generally use Q = ρV with a single constant value. Divide the body into differential elements dV; each contributes dQ = ρ(r)dV, and the total is Q = ∫ρ(r)dV. Therefore adding small volume contributions, option B, is correct.

Correct answer: A. Opposite elements are equally distant and carry equal charge

Explanation: The governing concept is symmetry and vector addition. For every small charge element on a uniformly charged ring, an element diametrically opposite it has the same charge and is equally distant from an axial observation point. Their transverse electric-field components have equal magnitudes but opposite directions, so they cancel. The axial components reinforce; hence option A is correct.

Correct answer: A. Components parallel to the surface cancel by symmetry

Explanation: A uniformly charged infinite sheet has translational symmetry in every direction parallel to its surface. No parallel direction is preferred, so contributions from symmetrically opposite surface elements cancel their tangential electric-field components. The components normal to the sheet add and determine the field direction, with magnitude σ/(2ε₀) on either side for an isolated infinite sheet. Thus option A is correct.

Correct answer: A. More charge exists in the same small measure

Explanation: Density measures charge per unit length, area, or volume, depending on the distribution. For a small element, dQ = λdl, σdA, or ρdV. If the element’s size is fixed and the local density is higher, its charge contribution dQ is correspondingly larger. The resulting field contribution also depends on distance and direction, but the charge element itself is greater; therefore option A is correct.

Correct answer: A. When the distance is much larger than the size of the distribution

Explanation: The point-charge approximation is justified when the observation distance R is much greater than the characteristic size a of the charge distribution, R ≫ a. From that distance, details of the spread produce only small corrections, and the leading electric field is approximately E = (1/4πε₀)Q/R², directed from the net charge. Therefore option A is correct, although higher multipole effects may matter closer in.

Correct answer: A. Because distances and directions of different parts become very important

Explanation: The point-charge model keeps only the leading effect of the total charge and assumes the observation point is far from the entire distribution. Near the distribution, different elements can have substantially different distances and directions from the point, so their individual fields vary in magnitude and direction. Integration over the actual distribution is then required; the total charge does not disappear. Hence option A is correct.

Correct answer: A. A shell may have surface charge density, whereas a solid sphere may have volume charge density

Explanation: Charge density depends on how charge is distributed, not only on the total charge. For a thin spherical shell, charge is described by surface density, σ = Q/A, because charge is confined mainly to its surface. For a solid sphere, charge distributed throughout the material is described by volume density, ρ = Q/V. Thus, even for the same Q, the density type and numerical value can differ. Option B contradicts the condition; C and D are physically meaningless.

Correct answer: A. Along the axis of the ring

Explanation: For a uniformly charged ring, consider pairs of charge elements located symmetrically about the axis. Their components perpendicular to the axis are equal in magnitude and opposite in direction, so they cancel. The components parallel to the axis add, producing a net axial field. Far from the ring its magnitude may be approximated by the field of a point charge, but its direction is still along the axis. Therefore A is correct; the field is not generally zero.

Correct answer: A. The field of the uniform distribution

Explanation: A uniform distribution often possesses clear geometrical symmetry, such as spherical, cylindrical, or planar symmetry. Symmetry allows many field components to cancel and can make integration or the use of Gauss’s law much simpler. A non-uniform distribution may still be solvable, but it generally requires more detailed integration because charge density changes with position. Therefore A is the usual conclusion, while B, C, and D make unjustified absolute claims.

Correct answer: B. It decreases from section to section

Explanation: Linear charge density is defined as λ = dQ/dl, the charge per unit length. For two sections having the same length Δl, their charges are approximately ΔQ = λΔl when the density is nearly constant within each section. Since Δl is the same but λ decreases with distance, the charge in later equal-length sections also decreases. Thus B follows directly from the definition; A reverses the trend, while C and D are unsupported.

Correct answer: B. Non-uniform surface distribution

Explanation: Surface charge density is defined as σ = dQ/dA, charge per unit area. A uniform surface distribution requires σ to have the same value at every point on the surface. If σ changes with position, even along only one direction, different surface elements carry different amounts of charge per unit area. The distribution is therefore non-uniform surface charge. Option C describes a line distribution, and D describes charge spread through volume, so neither matches the given geometry.

Correct answer: A. The density may differ in each volume element

Explanation: When volume density depends on radial distance, it is non-uniform and should be written as ρ(r). A small volume element at radius r contains dQ = ρ(r)dV, so elements at different distances can contribute different amounts of charge. The total charge is obtained by integrating over the complete volume: Q = ∫ρ(r)dV. Therefore A is correct. B ignores the given density, C replaces a volume calculation by a surface one, and D can destroy the physical sign of charge.

Correct answer: A. Because electric field is a vector and components can add or cancel

Explanation: The electric field is a vector quantity. For a continuous distribution, each small charge element produces a field contribution dE with a magnitude and a direction. The total field is the vector sum of all contributions, so parallel components add while equal opposite components cancel. Ignoring direction can give a wrong result even when every individual magnitude is calculated correctly. Thus A states the superposition principle; B, C, and D contradict basic electrostatics.

Correct answer: A. The fields of all elements cancel vectorially

Explanation: Every small charge element of the ring produces a non-zero electric field at the centre. However, for each element there is a diametrically opposite element with equal charge and the same distance from the centre. Their fields have equal magnitudes and opposite directions, so they cancel pair by pair. The complete vector sum is therefore zero, even though individual contributions are not zero. A is correct; B, C, and D are not required by the symmetry.

Correct answer: A. No; the electric potential may still be non-zero

Explanation: At the centre of a uniformly charged ring, electric-field vectors from opposite elements cancel, so the net electric field is zero. Electric potential is different: it is a scalar, and the potentials of all positive charge elements add. For a ring of radius R and total charge Q, the centre potential is V = kQ/R, which is positive for Q > 0. Hence zero field does not imply zero potential, making A correct.

Correct answer: A. The surface with larger area

Explanation: Surface charge density is defined as charge per unit area: σ = Q/A. When the total charge Q remains fixed, increasing the area A spreads the same charge over more surface. Therefore σ decreases for the larger surface. The smaller surface has greater charge concentration and hence greater density, so option A is correct; option C would apply only if the areas were also equal.

Correct answer: A. Smaller

Explanation: Linear charge density is charge per unit length, λ = Q/L. Since both wires have the same total charge Q, the longer wire has a larger value of L in the denominator. Consequently, its charge is distributed more thinly along the wire and λ is smaller. Option B would require a shorter wire or a larger total charge, while option C ignores the different lengths.

Correct answer: A. Greater

Explanation: Volume charge density is defined by ρ = Q/V, where Q is total charge and V is volume. For equal Q, reducing V makes the charge more concentrated and increases the value of ρ. Thus the smaller solid has greater volume charge density, so option A is correct. A smaller density would result only if its charge were reduced proportionally or its volume were larger.

Correct answer: A. Equal

Explanation: A uniformly charged rod has constant linear charge density λ throughout its length. The charge of a small element is dq = λ dl. Because the two elements have equal lengths dl and the same λ, both carry equal charge. Their positions at equal distances are not the essential reason; uniform density and equal element lengths are. Thus option A is correct, while opposite signs would require a changing or signed distribution.