Continuous Charge Distribution

For a linear distribution, the local relation is dq = λ dl. If two elements have the same length dl but different charges dq, their values of λ must be different. Therefore the charge density changes from place to place, which defines a non-uniform distribution. Equal total charge does not imply uniformity, and a zero total charge is neither stated nor required. Hence option A is correct.

Linear, surface, and volume charge densities are different physical quantities: λ has unit C m⁻¹, σ has C m⁻², and ρ has C m⁻³. Using the wrong unit usually means using the wrong geometric measure when recovering charge, such as Q = λL, Q = σA, or Q = ρV. This can misidentify the distribution and produce an incorrect total charge, so option A is correct.

The governing idea is that total charge is the signed sum of charges in all small elements, or Q = ∫dq. A distribution may contain both positive and negative charge densities. If the positive contribution has greater magnitude than the negative contribution, the resultant Q is positive. Therefore option A is correct. Option B wrongly equates positive net charge with positive charge everywhere; C confuses charge sign with uniformity; D contradicts the stated positive total.

Integration represents continuous summation. In a continuous distribution, the charge is divided conceptually into infinitesimal elements, such as dq = λdx for a line, dq = σdA for a surface, or dq = ρdV for a volume. The total charge or electric field is obtained by adding all these contributions through an integral. Hence option A is correct; the other choices are unrelated to adding physical charge elements.

For a uniform linear distribution, the linear charge density is λ = Q/L. The new charge is Q′ = 90/3 = 30 C, and the new length is L′ = 15/5 = 3 m. Therefore λ′ = Q′/L′ = 30/3 = 10 C m⁻¹. Option B is correct. Option A results from using the wrong length, C misses the combined scaling, and D uses the original charge with the new length.

For a uniform surface distribution, surface charge density is σ = Q/A. Initially Q = 84 C and A = 12 m². After the stated changes, Q′ = 2(84) = 168 C and A′ = 7(12) = 84 m². Hence σ′ = 168/84 = 2 C m⁻², so option A is correct. The other values come from failing to apply one of the two changes or from using an incorrect ratio.

For a uniform volume distribution, total charge is Q = ρV. The changed density is ρ′ = 6/2 = 3 C m⁻³, while the changed volume is V′ = 3(10) = 30 m³. Thus Q′ = ρ′V′ = 3 × 30 = 90 C. Option C is correct. Thirty ignores the volume increase, sixty ignores the density decrease, and 180 incorrectly uses the original density with the new volume.

For a uniformly charged wire, total charge is Q = λL. Let the second wire have density λ and length L, so Q₂ = λL. The first wire has density 3λ and length L/2; therefore Q₁ = (3λ)(L/2) = (3/2)λL = 1.5Q₂. Hence option C is correct. The half and three-times choices use only one of the given changes, while equal charge ignores both scaling factors.

Surface charge density is defined by σ = Q/A. Let both surfaces carry charge Q, and let the second area be A. The first area is 4A, so σ₂ = Q/A whereas σ₁ = Q/(4A) = σ₂/4. Therefore the first surface has one-fourth the density, making option D correct. A and B reverse the inverse relation, while C would apply only if the areas were also equal.

The zero field at the centre follows from rotational symmetry, not from the particular value of total charge. Each charge element has an opposite element producing an equal and opposite field, so the vector sum is zero. Doubling the charge doubles both opposing contributions, and they still cancel exactly. Thus the field remains zero, so option C is correct; option A would ignore vector cancellation.

For a complete uniformly charged ring, every element has an opposite counterpart. Their electric-field vectors cancel pairwise at the centre. Removing the lower half removes those opposing contributions, so the remaining upper-half elements have no matching vectors to cancel them. A nonzero resultant field therefore remains, directed along the symmetry axis toward or away from the retained arc depending on the sign of charge. Hence option A is correct.

Consider two diametrically opposite elements of the ring. At an axial point, their components perpendicular to the axis are equal in magnitude and opposite in direction, so they cancel. Their axial components point in the same axial direction and therefore add. Repeating this pairing around the ring leaves only an axial electric field. Thus option B is correct; the field is not zero except at the centre.

In the far-field region, the observation distance is much larger than the rod's length. The rod then appears effectively point-like, and the leading electric-field contribution depends mainly on its total charge Q. It can therefore be approximated by a point charge Q located near the rod's centre for this purpose. This is why option A is correct; the rod does not become uncharged or an infinite surface.

A point-charge approximation is reliable when the observation distance is much larger than the distribution's size. Near a long wire, different charge elements are at noticeably different distances and subtend different directions, so their individual field contributions cannot be represented accurately by only the total charge at one point. The continuous integral or an appropriate wire-field model is needed. Therefore option A is correct.

The local linear charge density λ(x) describes charge per unit length at one position, whereas the total charge is obtained by integrating over the entire distribution: Q = ∫λ(x) dx. A zero value at one point contributes no local charge there, but nonzero density over other regions can produce a nonzero total charge. Its exact value and sign depend on all parts of the distribution. Hence option B is correct.

Surface charge is calculated using Q = ∫σ dA. The two regions have equal areas, and their surface densities have equal magnitudes but opposite signs. Thus their contributions are +σA/2 and −σA/2, whose sum is zero. Therefore, option C is correct. The result is neither positive nor negative, and it is not infinite because both area and density are finite.

A zero total charge means only that the volume integral ∫ρ dV is zero. It does not mean that ρ is zero everywhere or that the electric-field vectors cancel at every observation point. For example, separated positive and negative charge regions can have zero net charge but produce a nonzero dipole field. Hence option B is correct; options A and C confuse net charge with local cancellation.

The total charge is the signed sum or integral of all small contributions: Q = ∫dq. A distribution can contain both positive and negative elements, provided the positive contribution has the larger magnitude, giving Q > 0. Therefore, option B is correct. A positive net charge does not require uniform density or positive charge at every point, and it certainly does not imply Q = 0.

The electric field direction is defined as the force direction on a positive test charge. A negative source charge attracts that test charge, so the field vector at the observation point points toward the negative element. A small element may be treated locally like a point charge for this directional rule. Therefore option B is correct; the field is not necessarily zero or tangential.

A positive source charge repels a positive test charge. Consequently, the electric-field vector at an observation point is directed away from the positive charge element along the line joining the source and observation points. Treating a sufficiently small element as locally point-like gives this standard direction rule. Therefore option B is correct; the field is not always zero or necessarily perpendicular to any chosen axis.

For a uniform surface distribution, Q = σA; for a nonuniform one, Q = ∫σ dA. Thus the magnitude of total charge depends on both the density and the area over which it acts. A high density multiplied by a very small area may produce a small, moderate, or large charge depending on the numerical values. Therefore option B is correct, while the other statements make unjustified absolute claims.

The governing idea is superposition: the field is the vector sum of contributions from every small charge element. Although the total charge is the same, the elements may be located at different distances and their field directions may differ because the geometries are different. A ring can produce cancellation in certain directions through symmetry, whereas a rod generally produces another pattern. Therefore option A is correct; B contradicts the given condition, and C and D ignore geometry and charge density.

For a uniformly charged complete ring, every small element has a diametrically opposite element at the same distance from the centre. Their electric-field vectors cancel pairwise, so the net field at the centre is zero. A half ring lacks the complete opposite pairing, so its contributions do not generally cancel and a nonzero field remains. Thus B is correct; A overgeneralizes, while C and D are physically false.

Take any small charge element on the ring and pair it with the element diametrically opposite to it. At an axial observation point, both elements are equally distant and produce fields of equal magnitude. Their components perpendicular to the axis point in opposite directions and cancel. Their axial components point in the same direction and add. Hence A is correct; the ring need not have zero charge, and individual elements do produce fields.

Consider two equal small elements of a uniformly charged rod placed symmetrically on opposite sides of its midpoint. A point on the perpendicular bisector is at the same distance from both elements, so their field magnitudes are equal. The components parallel to the rod point in opposite directions and cancel. The perpendicular components point the same way and add, leaving the net field along the perpendicular bisector. Thus A is correct; the other choices misuse charge or vector concepts.