Continuous Charge Distribution

In a continuous charge distribution, charge is regarded as spread along a line, over a surface, or throughout a volume. It is analysed by adding very small charge elements such as \(dq\). In contrast, option B describes a point charge concentrated at one location, not a continuous distribution.

The radius of a thin, long rod is taken to be negligible compared with its length, so the charge distribution is modelled as a line. Therefore, the appropriate quantity is linear charge density: \\(\lambda = \frac{dq}{dl}\\). Surface charge density is used for charge spread over a surface, whereas volume charge density is used for charge distributed through a three-dimensional body.

A flat charged sheet is modeled as a two-dimensional distribution because its charge is spread over area rather than only along a line or throughout a volume. The appropriate quantity is surface charge density, sigma = dQ/dA, with SI unit C/m^2. Linear density applies to wires, volume density applies to three-dimensional objects, and mass density describes mass rather than electric charge. Therefore option C is correct.

The charge is distributed throughout the three-dimensional volume of the solid cylinder, so it is described by volume charge density:

displaystyle \(\rho=\frac{dq}{dV}\). Linear charge density applies to a one-dimensional distribution such as a wire, whereas surface charge density applies when charge is spread only over a surface.

Linear charge density \(\lambda\) represents the charge present per unit length on a charged wire or line: \(\lambda = \frac{dq}{dl}\). Its SI unit is \(\mathrm{C\,m^{-1}}\). Charge per unit area refers instead to surface charge density \(\sigma\). Exam tip: connect linear, surface, and volume charge densities with length, area, and volume respectively.

Surface charge density \(\sigma\) is the charge distributed on a surface per unit area. For a uniform distribution, \(\sigma = Q/A\); more generally, \(\sigma = dQ/dA\). Its SI unit is \(\mathrm{C/m^2}\). Option A represents linear charge density, while option C represents volume charge density.

Volume charge density  ext{ ext{ρ}} represents the charge present per unit volume in a region:  ext{ ext{ρ}} = rac{dq}{dV}। Its SI unit is  ext{C m}^{-3}. In contrast, charge per unit length is linear charge density, and charge per unit area is surface charge density.

Linear charge density is given by

template:

lambda = dq/dl

that is, charge per unit length. Hence, its SI unit is coulomb per metre (C/m). Coulomb per square metre is the unit of surface charge density, while coulomb per cubic metre is for volume charge density. Exam tip: divide by m, m² and m³ respectively for linear, surface and volume charge densities.

Surface charge density is defined as charge per unit area:
\(\sigma = \frac{Q}{A}\). Therefore, its SI unit is \(\mathrm{C/m^2}\), or coulomb per square metre. Option B is the unit of linear charge density \(\mathrm{C/m}\), whereas option C is the unit of volume charge density \(\mathrm{C/m^3}\).

Volume charge density describes how much charge is present per unit volume. It is defined by rho = dQ/dV, so its SI unit is coulomb divided by cubic metre, written C/m^3. C/m represents linear charge density and C/m^2 represents surface charge density. Metre per coulomb is the reciprocal type of unit and does not represent charge per volume. Hence option B is correct.

Linear charge density is charge per unit length: \(\lambda = Q/L\). In a uniform linear charge distribution, equal lengths carry equal charges, so \(\lambda\) remains constant at every point. A density that varies with position represents a non-uniform linear charge distribution.

Surface charge density is defined as \(\sigma=\frac{dq}{dA}\). In a non-uniform surface charge distribution, the charge \(dq\) on equal small area elements \(dA\) is not the same at different positions. Therefore, \(\sigma\) varies with position, for example, \(\sigma=\sigma(x,y)\). Option A describes a uniform surface charge distribution.

In a continuous charge distribution, charge is spread over a line, surface, or volume. It is divided into small charge elements such as \(dq\); the electric field or potential due to each \(dq\) is found, and all contributions are added using integration. Reducing total mass is not the purpose. Exam tip: First write a suitable \(dq\), for example \(dq=\lambda\,dl\), before setting up the integral.

A continuous charge distribution is divided into infinitesimal charge elements \(dq\). Each element produces an electric-field contribution \(d\vec{E}\) at the point. By the principle of superposition, these contributions are added vectorially: \(\vec{E}=\int d\vec{E}\). Merely adding charge magnitudes is not sufficient because electric field has direction as well as magnitude.

In a continuous charge distribution, charge is divided into infinitesimal elements \(dq\). Integrating all these elements gives the total charge: \(Q=\int dq\). Adding small charges is the basic idea, but integration is its mathematical form for a continuous distribution. Exam tip: Remember \(dq=\lambda dl\), \(dq=\sigma dA\), and \(dq=\rho dV\) for line, surface, and volume charge distributions respectively.

Linear charge density \(\lambda = \frac{dq}{dl}\) represents charge per unit length along a line. A positive \(\lambda\) means that the line has net positive charge per unit length. The density need not be uniform at every point; its sign is positive. Option D refers instead to surface charge density, which is charge per unit area.

Surface charge density is defined as sigma = dQ/dA, meaning charge distributed per unit area. Since area is a positive geometrical quantity, a negative value of sigma indicates that the net charge associated with each small surface element is negative. It does not imply negative area or zero charge. A positive charge density would give option B, while zero density would give option C. Therefore option A is correct.

Average linear charge density is lambda = Q/L, where Q is total charge and L is the wire length. The charge remains Q, but the new length is L/2. Therefore lambda' = Q/(L/2) = 2Q/L = 2lambda. The density doubles because the same charge occupies half the length. It does not become half, zero, or unchanged. Thus option B is correct.

For charge distributed over a surface, the relevant density is surface charge density, sigma = dQ/dA. The differential area dA shows that charge is measured per unit area. By contrast, lambda = dQ/dl is linear density for a wire, rho = dQ/dV is volume density for a three-dimensional body, and Q is total charge rather than a density. Therefore option B is correct.

The average volume charge density is \(\rho_{\text{avg}}=Q/V\). With \(Q\) constant and \(V'=2V\), \(\rho'_{\text{avg}}=Q/(2V)=\rho_{\text{avg}}/2\). Thus it becomes half. It would double only if the volume were reduced to half while the charge stayed the same.

A ring is a one-dimensional closed curve. In a continuous charge-distribution model, its charge is considered distributed along the entire circumference, so it is described by a linear charge distribution. Treating all the charge as located at the centre would be a point-charge model, not a continuous ring distribution.

In a uniformly charged ring, a small charge element and the diametrically opposite element carry equal charge and are at equal distances from the centre. Therefore, their electric fields at the centre have equal magnitudes but opposite directions. Their vector sum is consequently zero. Option C is incorrect because all elements of a uniformly charged ring have the same sign of charge.

For a point on the perpendicular bisector, charge elements located at equal distances on opposite sides of the rod’s centre are equidistant from the point. Hence, their electric-field components parallel to the rod are equal and opposite, so they cancel. Their components along the perpendicular bisector point in the same direction and add. Exam tip: In symmetry questions, pair equidistant charge elements first and then compare the directions of their components.

Electric field is a vector quantity. Therefore, while combining the fields due to small parts of a continuous charge distribution, both magnitude and direction must be considered. If two fields are in opposite directions, their resultant can be the difference of their magnitudes rather than their sum. Hence, adding magnitudes alone is not always correct. Option B is incorrect because an electric field has both magnitude and direction.

A distribution is called uniform when its charge density has the same value at every position. Depending on the geometry, this may mean constant linear density, surface density, or volume density. Consequently, equal lengths, areas, or volumes contain equal charges under the relevant model. A non-uniform or variable-density distribution changes with position, while a point charge is not a continuous distribution. Hence option B is correct.