Continuous Charge Distribution

Because the charge occupies the three-dimensional material of the sphere, the appropriate quantity is volume charge density, ρ = dQ/dV, measured in coulombs per cubic metre. For a uniformly charged solid sphere, the total charge is Q = ρV; if its radius is R, then V = 4πR³/3. Linear density applies to a line, and surface density applies only when charge is confined to the surface. Therefore C is correct.

In electrostatic equilibrium, excess charge on an ideal conductor resides on its surface, while the electric field inside the conducting material is zero. When that surface charge is spread continuously, it is described by surface charge density σ = dQ/dA. The total charge on a selected region is obtained from Q = ∫σ dA. Thus A is correct; volume density is not mandatory, and neither zero charge nor omission of density follows.

Electric potential is a scalar sum, whereas electric field is the vector sum of contributions from all charge elements. Scalar contributions can cancel at a point even when the corresponding vector contributions do not. For example, at the midpoint of equal and opposite charges, potential may be zero while the fields add. Therefore option B is correct; options A and C wrongly equate the two quantities, and D is false.

Electric potential is a scalar quantity, so direction does not cause cancellation between contributions from different elements. Every element of a uniformly charged ring is at the same distance from the centre and has the same charge sign; hence each contribution has the same sign and the scalar contributions add. The electric field may cancel by symmetry, but the potential does not. Thus A is correct.

For a non-uniform surface, the total charge is generally obtained from Q = ∫σ dA. If an area-averaged density is properly defined, σ_avg = Q/A, the same result can be written as Q = σ_avg A. A value measured at one point is only local density and cannot represent the whole surface. A negative density does not by itself make an average useful, and an unknown area prevents this shortcut.

The electric field of a continuous line distribution is found by integrating contributions such as dE = k dq/r², with directions included. Although the total charge Q is known, the positions of the elements, their distances r, and their directions are not known when the spread is unspecified. Thus the exact nearby field cannot generally be determined. Option B is already given, while C and D are irrelevant.

A point-charge approximation is justified when the observation distance R is much greater than the characteristic size a of the distribution, so that R >> a. The details of the charge arrangement then have a relatively small effect and the leading term is determined mainly by the total charge Q, giving approximately E = kQ/R². Near the distribution, shape and density matter strongly, so B and C are unsafe; D does not establish physical validity.

The field from an element depends on both its charge and its distance, approximately as dE = k|dq|/r², along with the direction of the contribution. A distant region may contain a larger dq but still produce a smaller field because the inverse-square factor is much smaller. Contributions must therefore be compared quantitatively and then added vectorially. Hence B is correct; A and C ignore distance, while D claims cancellation without symmetry or calculation.

Electric field is a vector, so each infinitesimal contribution has components along chosen coordinate axes. By integrating or summing corresponding components separately, symmetry can show that opposite transverse components cancel, while components along a symmetry axis add. The remaining components then give the net field. Therefore A is correct. B is false because field is vectorial, and C and D do not follow from component resolution or superposition.

For any closed surface, Gauss’s law gives Φ_total = Q_enclosed/ε₀. Moving a charge within the surface does not alter the amount or sign of charge enclosed, so the total flux remains unchanged. The charge’s new position can make the field and flux density nonuniform, especially near the nearby part of the surface, but the integral over the entire surface is unchanged. Therefore, option A is correct.