Continuous Charge Distribution

Volume charge density describes charge distributed throughout a three-dimensional region. It is defined as ρ = Q/V, or locally ρ = dQ/dV. Because volume is measured in cubic metres, its SI unit is coulomb per cubic metre, C/m³; therefore option C is correct. C/m and C/m² represent linear and surface charge densities, while m/C is the reciprocal type of unit.

For a thin rod, its length is the significant dimension compared with its other dimensions, so the charge is treated as distributed along a line. Since the charge is uniform, it is a uniform linear charge distribution. Its linear charge density is
\(\lambda = Q/L\). In contrast, a surface charge distribution has charge spread over a surface, such as a thin sheet.

The governing classification depends on the dimensional region occupied by the charge. A sheet is effectively two-dimensional, so charge spread across it is a surface charge distribution. Its density is σ = dQ/dA, measured in C/m². Therefore option B is correct. A thin wire gives a linear distribution, a filled solid gives a volume distribution, and a point charge is localized at one position.

The charge occupies the entire three-dimensional interior of the solid sphere, not merely its boundary or a single line. Therefore it is a volume charge distribution. Its volume density is ρ = dQ/dV; for a uniform sphere, this density has the same value throughout the occupied volume. Option C is correct, whereas surface, linear, and point distributions occupy lower-dimensional regions.

Linear charge density describes charge per unit length and is defined by λ = Q/L. Here the wire length L is fixed, while the total charge Q changes to 2Q. Therefore the new density is λ′ = 2Q/L = 2λ, so option A is correct. It would become half only if the same charge were spread over twice the length; unchanged density would require charge and length to scale together.

Surface charge density is the charge distributed per unit area, given by σ = Q/A. Since the plate area A does not change and the charge changes from Q to 2Q, the new density is σ′ = 2Q/A = 2σ. Thus option A is correct. It would remain unchanged only if the area also doubled, and it would decrease if the same charge were spread over a larger area.

A continuous distribution is divided into infinitesimal charge elements \(dq\). The size of each element is taken to be negligible compared with its distance from the observation point, so its electric field is treated as that of a point charge. The vector contributions of all such \(dq\) elements are then added or integrated. It is not treated as the entire charged body, because the field of the whole distribution is obtained by combining the contributions of all its elements.

Electric field has both magnitude and direction, so it is a vector quantity. A continuous distribution is divided conceptually into small charge elements, and the field dE due to each element is added using vector superposition. In the limiting process this sum becomes an integral, E = ∫dE. Hence option A is correct; scalar addition ignores direction, while subtraction is required only for particular opposite directions.

In a uniformly charged ring, diametrically opposite small charge elements carry equal charge and are at equal distances from the centre. Hence, the electric fields they produce at the centre have equal magnitudes but opposite directions. Therefore, the resultant field of each such pair is zero. Mass, colour, and temperature do not cause this symmetric cancellation.

In a symmetric charge distribution, some electric-field components produced by equal charge elements are in opposite directions and cancel. Thus, only the required components need to be added, which simplifies the integration. For example, on the axis of a uniformly charged ring, transverse components cancel while axial components add. Therefore, symmetry does not necessarily make the total charge or electric field zero.

In a uniform linear charge distribution, the linear charge density
latex
lambda
latex
is constant. Therefore, the total charge is
latex
Q =
latex
lambda L
latex
, where
latex
L
latex
is the length of the charged line or wire. For surface and volume distributions, the corresponding relations are
latex
Q =
latex
sigma A
latex
and
latex
Q =
latex
rho V
latex
, so options B and C do not apply to a linear distribution.

For a uniform surface charge distribution, the total charge is given by \(Q=\sigma A\), where \(\sigma\) is the surface charge density and \(A\) is the surface area. Therefore, option A is correct. Option B applies to a linear distribution, \(Q=\lambda l\), while option C applies to a volume distribution, \(Q=\rho V\).

In a uniform volume charge distribution, the volume charge density rho is constant. The total charge is obtained by integrating over the volume: Q = rho V, where V is the total volume. Therefore, volume charge density and total volume are required. Option B applies to a linear charge distribution, for which Q = lambda L, not to a volume distribution.

A charge spread along a line is described by linear charge density λ, so the geometric type is linear. If λ changes from one position to another, it is not uniform; mathematically, λ = λ(x) or λ(s). Therefore the distribution is non-uniform linear, making option B correct. A uniform linear distribution would have constant density, while a surface distribution would refer to charge spread over area rather than along a line.

Charge distributed over an area is described by surface charge density σ = dQ/dA. When σ has the same value at every point of that surface, the charge per unit area is constant, so the distribution is uniform surface distribution. Therefore option B is correct. A non-uniform surface distribution would have σ varying with position, a linear distribution belongs to a line, and a point charge is localized at one position.

Electric field is a vector quantity, so it has both magnitude and direction. The field produced at an observation point by each infinitesimal charge element \(dq\) can have a different direction. Therefore, the total electric field must be found by vector addition; adding only the magnitudes is not sufficient.

Electric field is a vector quantity. When two electric-field components have equal magnitudes and opposite directions, their vector sum is zero. Therefore, the resultant electric field due to these two components is zero. In contrast, if the components were in the same direction, they would add to give twice the magnitude.

A continuous charge distribution is considered as a collection of infinitesimal charge elements \(dq\). Its total charge is \(Q=\int dq\), the signed sum of all positive and negative charge elements. Thus, \(Q\) is positive if positive contributions dominate, negative if negative contributions dominate, and zero if they cancel. The size of the distribution or whether it is uniform does not by itself determine the sign of the total charge.

Linear charge density is defined as \(\lambda=\frac{dq}{dl}\), where \(dq\) is the charge on a small length element \(dl\). If \(\lambda>0\), the charge per unit length is positive. Its SI unit is \(\mathrm{C\,m^{-1}}\). Option B represents negative linear charge density, whereas option C represents zero linear charge density.

Surface charge density is defined as σ = dQ/dA, so its sign indicates the sign of the charge distributed over the surface. A negative value of σ means that the charge element dQ is negative for the chosen surface orientation or region; it does not mean that charge is absent. Thus option A is correct. Zero charge would give σ = 0, while mass and surface area are not determined by the sign of σ.

Each infinitesimal charge element on the circular ring is at the ring’s radius $R$ from the centre. Therefore, all charge elements are equally distant from the centre. Equal distance is distinct from zero electric field; the field at the centre is zero because of symmetry.

For a point on the perpendicular bisector, symmetric charge elements on opposite sides of the rod’s midpoint are at equal distances from the point. Their electric-field components parallel to the rod are equal and opposite, so they cancel. However, the components perpendicular to the rod are in the same direction and add; hence the net electric field is generally not zero.

A continuous charge distribution is divided into infinitesimal charge elements \(dq\). Each element produces a small field contribution \(d\vec{E}\), and the vector sum of all such contributions is found by integration: \(\vec{E}=\int d\vec{E}\). Therefore, option A is correct. Option B may be an approximation in special cases, but it is not the general integration method for a continuous distribution.

Constant charge density means that the charge density does not change with position. Therefore, equal lengths in a line distribution, equal areas in a surface distribution, or equal volumes in a volume distribution contain equal charges. Hence, the charge distribution is uniform. Option D instead describes a non-uniform charge density.

The point-charge model is useful for simple idealized problems, but real objects such as wires, plates, rings and spheres usually carry charge over a length, area or volume. Continuous charge distribution represents this spread through linear density λ, surface density σ or volume density ρ, allowing fields and potentials to be calculated with integration. Therefore option A is correct; the other choices contradict the physical meaning of distributed charge.