Continuous Charge Distribution

The point-charge model is reliable mainly when the observation distance is much larger than the distribution's size. Near a long wire, charge elements have noticeably different distances and directions relative to the observation point, so their electric-field contributions cannot be represented accurately by placing all charge at one point. Therefore A is correct. The total charge, wire length, and charge sign do not disappear; they are simply insufficient by themselves for a nearby field calculation.

The electric field is a vector, so contributions must be added with their directions. In a complete uniformly charged ring, every small element has an oppositely placed partner producing an equal and opposite field at the centre; the net field is therefore zero. A half-ring lacks the complete set of opposite partners, so cancellation is incomplete and a resultant field remains. Thus A is correct; the other choices do not control electrostatic field cancellation.

At the centre of a complete uniformly charged circle, diametrically opposite elements produce equal and opposite electric-field vectors, so their sum is zero. A semicircular arc contains only one half of that pairwise-symmetric distribution. The missing opposite half means the field contributions do not cancel completely, leaving a non-zero resultant directed along the symmetry axis of the arc. Hence A is correct. The arc is charged, and electric field is a vector, not a scalar.

For a uniform linear charge distribution, equal length intervals must contain equal charges because the linear density lambda is constant and dq = lambda dl. If two equal elements have different values of dq, their charge per unit length is different, showing that lambda changes with position. Therefore the distribution is non-uniform, so option B is correct. This observation does not imply zero total charge or zero length, and it directly contradicts uniformity.

The unit identifies the geometrical measure used by the density. C/m represents linear density, so dq = λ dl; C/m² represents surface density, so dq = σ dA; and C/m³ represents volume density, so dq = ρ dV. Choosing the wrong element leads to incorrect integration and dimensions. Therefore the unit is the key first check for selecting the correct continuous-charge model.

At the centre of a uniformly charged ring, every charge element has an opposite element at the same distance. Their electric-field contributions have equal magnitudes and opposite directions, so the vector sum is zero. Changing the total charge from positive to negative reverses each individual field direction but does not disturb the pairwise cancellation. Thus the central field remains zero, not inward, outward, or infinite.

The field at the centre is the vector sum of contributions from all ring elements. For every element, an opposite element at the same distance produces an equal contribution in the opposite direction. Doubling the total charge doubles the magnitude of each paired contribution, but both members of every pair still cancel exactly. Therefore the resultant remains zero. The field is a vector, and neither density nor total charge becomes zero.

Integration represents the limiting sum of contributions from infinitesimal elements of a continuous distribution. For example, total charge is written as Q = ∫dq, with dq = λdx for a line, σdA for a surface, or ρdV for a volume. Adding these differential contributions over the entire distribution gives the physical total. It neither forces charge to zero nor changes units or removes direction.

Linear charge density is defined by λ = Q/L. Initially Q = 120 C and L = 20 m. The new charge is Q′ = 120/2 = 60 C, while the new length is L′ = 20/5 = 4 m. Therefore λ′ = Q′/L′ = 60/4 = 15 C m⁻¹. Option A uses the original length incorrectly; the other numerical choices do not follow the stated changes.

Surface charge density is σ = Q/A. The changed charge is Q′ = 3 × 90 = 270 C, and the changed area is A′ = 15/2 = 7.5 m². Thus σ′ = 270/7.5 = 36 C m⁻². The original density was 90/15 = 6 C m⁻², so the result is six times the original because charge tripled while area was halved. Hence option C is correct.

For a uniformly charged wire, total charge is Q = λL. Let the second wire have density λ and length L, so its charge is Q₂ = λL. For the first wire, λ₁ = 4λ and L₁ = L/3. Therefore Q₁ = (4λ)(L/3) = (4/3)λL = (4/3)Q₂. The density increase is partly offset by the shorter length, so the answer is four thirds, not four or twelve times.

Surface charge density is σ = Q/A. Since both surfaces carry the same total charge, let their charges be Q. If A₁ = 6A₂, then σ₁ = Q/(6A₂) = (1/6)(Q/A₂) = σ₂/6. Thus the first surface has one sixth the density of the second. A larger area spreads the same charge more thinly; it does not increase the density.

The field at the centre is zero because every charge element on the ring has an opposite element producing an equal field in the opposite direction. Uniformly tripling the total charge triples each paired contribution, but the vector cancellation remains exact. Hence the resultant field is still E = 0, making option C correct. Options A and B ignore cancellation, while D has no physical basis here.

For a complete uniformly charged ring, each element has a diametrically opposite partner whose electric-field contribution cancels it at the centre. Keeping only the lower half removes those opposite partners, so the vector cancellation is incomplete. The remaining contributions combine to produce a nonzero field along the symmetry axis. Therefore option A is correct; the other choices contradict charge or field properties.

Take two diametrically opposite charge elements of the ring. At an axial observation point, their components perpendicular to the axis are equal in magnitude and opposite in direction, so they cancel pairwise. Their components parallel to the axis point in the same direction and add. Thus only the axial component remains, so option B is correct. The field is not generally zero away from the centre.

The governing idea is the far-field or point-charge approximation. If the observation distance is much larger than the rod’s length, differences in distance and direction from individual charge elements become relatively small. The leading field therefore depends mainly on the total charge Q and resembles the field of a point charge Q located near the rod’s centre. Hence option A is correct; the sign is not necessarily negative.

A point-charge approximation is reliable when the observation distance is much larger than the size of the charge distribution. Near a long wire, different small elements are at substantially different distances and subtend different directions at the observation point. Their individual field contributions therefore cannot be represented accurately by one central charge. The continuous distribution must be integrated, so option A is correct.

Local linear density λ(x) describes charge per unit length at one position, whereas total charge is obtained by integrating over the complete distribution: Q = ∫ λ(x) dx. A zero value at one point contributes no charge from that single point, but it does not determine the integral over all other regions. Positive and negative contributions may also alter the result. Therefore option B is correct.

For a surface distribution, total charge is obtained by integrating surface density: Q = ∫σ dA. The two regions have equal area, equal density magnitude, and opposite signs. Thus their contributions are +σA/2 and −σA/2, whose algebraic sum is zero. The cancellation is exact under the stated conditions, so option C is correct. The result is not positive, negative, or infinite because neither sign has a larger contribution.

The governing relation is Q = ∫dq, or for a volume distribution Q = ∫ρdV, where contributions carry their signs. A positive total means the algebraic positive contribution exceeds the magnitude of the negative contribution; it does not require every small element to be positive. Negative regions may therefore coexist with positive regions. Option B states this correctly. Options A and C impose unjustified uniformity, while D contradicts the stated positive total charge.

The electric field direction is defined as the force direction on a positive test charge. By Coulomb’s law, a negative source charge attracts that test charge, so the field vector at the observation point points toward the negative charge element. A sufficiently small element may be treated locally like a point charge for this direction question. Hence option B is correct; the field is not automatically zero or tangential.

Electric field direction is the direction of force on a positive test charge. A positive source charge repels such a test charge, and Coulomb’s law therefore gives a field vector directed away from the positive charge element at the observation point. Treating the small element locally as a point charge is sufficient for this directional conclusion. Thus option B is correct; the field is not necessarily zero or perpendicular to an arbitrary axis.

For a uniform surface distribution, total charge is Q = σA; for a nonuniform one, Q = ∫σdA. In either case, both the density and the area contribute to the result. A large magnitude of σ does not by itself guarantee a large Q, because a very small area may make their product modest. Therefore option B is correct. Option A ignores area, C asserts an unwarranted zero, and D ignores magnitude.

The governing idea is that electric field is a vector superposition of contributions from every charge element, not a quantity determined only by total charge. A ring and a rod can have equal total charge but different positions, distances, directions, and symmetry of their elements. Consequently, their vector sums can differ in magnitude and direction. Option A is correct; B contradicts the condition, while C and D ignore geometry and charge distribution.

For a uniformly charged complete ring, each small charge element has a diametrically opposite element at the same distance. Their electric-field vectors cancel at the centre, so the net field is zero. A half-ring lacks the complete opposite pairing, and its contributions generally leave a resultant field along the symmetry axis of the semicircle. Thus B is correct; A overgeneralizes cancellation, and C and D are false.