Continuous Charge Distribution

Volume charge density is the charge contained per unit volume, defined by ρ = Q/V. Since charge is measured in coulombs and volume in cubic metres, the SI unit is C/m³, or coulomb per cubic metre. C/m is used for linear charge density and C/m² for surface charge density. Joule per coulomb is volt, a unit of electric potential, not charge density.

A thin rod has one dominant geometrical dimension—its length—while its thickness is neglected in the idealized model. Charge spread uniformly along that length is therefore represented by a constant linear charge density λ = Q/L. A surface distribution applies to a sheet or plate, and a volume distribution applies to a three-dimensional body. A rod is not a point charge because its charge is extended along a line.

A plate is modeled as a two-dimensional object because its thickness is usually neglected compared with its length and breadth. When charge is spread uniformly over its surface, the appropriate quantity is surface charge density σ = Q/A, so it is a uniform surface distribution. Linear density is used for a wire or rod, volume density for a body with appreciable thickness, and a plate is not a point charge.

In a solid sphere, the charge is spread through the entire three-dimensional volume rather than being confined to a line or a surface. Therefore, it is a volume charge distribution. It is described by the volume charge density

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ho=rac{dQ}{dV}

ext{.}

ext{In contrast, a surface charge distribution has charge confined only to a surface.}

Linear charge density is defined as λ = Q/L, where Q is total charge and L is the length of the charged wire. If the original value is λ = Q/L, then after the charge doubles it becomes λ′ = 2Q/L = 2λ because L is unchanged. Thus the density doubles. It would remain unchanged only if charge and length changed by the same factor, so option D is not applicable.

Surface charge density is
\(\sigma=\frac{Q}{A}\). When the area \(A\) is constant, \(\sigma\) is directly proportional to the total charge \(Q\). Therefore, doubling \(Q\) doubles \(\sigma\). It would become half only if the total charge were halved while the area remained constant.

A continuous charge distribution is divided into infinitesimal charge elements, \,\(dq\). Since the size of each element is negligible compared with its distance from the observation point, its electric field is calculated as that of a point charge. The electric-field contributions of all elements are then added vectorially using integration. The entire distribution cannot generally be treated as one point charge because its different parts are at different distances and directions from the observation point.

Electric field is a vector quantity, so the field produced by a continuous distribution is obtained by vectorially adding the infinitesimal contributions dE from all small charge elements. In mathematical form, E = ∫dE, with both magnitude and direction included. Adding only magnitudes is valid only in special collinear cases and is not the general rule. Multiplication or refusing to add the contributions violates superposition.

Two diametrically opposite small elements of a uniformly charged ring carry equal charge and are at the same distance from the centre. Therefore, the electric fields they produce at the centre have equal magnitudes. Their directions are opposite, so their vector sum is zero. Unlike option B, equal magnitudes do not mean that the two fields point in the same direction.

For a symmetric continuous charge distribution, electric-field components due to charge elements at corresponding positions may be equal and opposite, so they cancel. This makes it easier to determine the magnitude and direction of the net electric field by adding only the non-cancelling components. Symmetry does not necessarily make the total charge zero; total charge depends on the given distribution. Exam tip: identify the axis or plane of symmetry first, then eliminate the field components that cancel.

For a uniform linear distribution, the charge per unit length is λ and the length is L. Therefore the total charge is obtained from Q = λL, or more generally Q = ∫λ dl. Since λ is constant, the integral reduces to λ times the full length. Surface density with area and volume density with volume belong to different geometries, while electric field and distance do not directly determine this total charge.

For a uniform surface charge distribution, the surface charge density \(\sigma\) is constant. Hence, the total charge is given by \(Q=\sigma A\), where \(A\) is the area of the charged surface. Linear charge density with length applies to a line distribution, while volume charge density with volume applies to a volume distribution. Exam tip: For charge spread over a surface, identify \(\sigma\) and the area.

For a uniform volume charge distribution, the total charge is \(Q=\rho V\), where \(\rho\) is the volume charge density and \(V\) is the volume of the object. Therefore, \(\rho\) and \(V\) are required. In contrast, a linear distribution uses \(Q=\lambda L\), while a surface distribution uses \(Q=\sigma A\).

The charge is distributed along a line, so it is a linear charge distribution. Since the linear charge density

extstyle\lambda

varies with position, it is not the same at every point. Therefore, it is a non-uniform linear charge distribution. In a uniform linear distribution,

extstyle\lambda

is constant.

Because the charge is located on a surface, the distribution is classified as a surface distribution. The additional statement that the charge density has the same value at every point means the surface density σ is constant, so it is uniform. A non-uniform distribution would have σ varying from place to place. Linear and point distributions describe different geometrical models and do not match the given surface description.

Electric field is a vector quantity, so it has both magnitude and direction. The net electric field of a continuous distribution is obtained by vectorially adding the fields due to its infinitesimal charge elements. Fields from different elements may have different directions, and some components may cancel by symmetry. Therefore, option D is not generally correct.

The governing concept is vector addition of electric-field contributions. Electric field is not added as an ordinary scalar; both magnitude and direction must be considered. If two components are equal, say E, but point in opposite directions, their vector sum is E + (-E) = 0. Thus symmetry causes cancellation. A magnitude of 2E would occur only for equal components in the same direction, so option A is correct.

The total charge is the signed sum of all infinitesimal charge elements: \(Q=\int dq\). For line, surface, and volume distributions, respectively, \(Q=\int \lambda\,dl\), \(Q=\int \sigma\,dA\), and \(Q=\int \rho\,dV\). The signed integral of positive and negative contributions can make \(Q\) positive, negative, or zero. Geometrical shape may affect the region of integration, but shape alone does not determine the sign of the total charge.

Linear charge density is defined as charge per unit length: lambda = dQ/dl, or Q/L for a uniformly charged line. Its sign indicates the sign of the charge distributed along the line. Therefore, positive linear charge density means that the line carries net positive charge per unit length. It does not mean that charge is absent, that only mass is present, or that the electric field must be zero. Hence option A is correct.

The correct choice follows from the geometric dimension over which the charge is distributed. A very thin wire is treated as a one-dimensional object, so charge is described per unit length by linear charge density, lambda = dQ/dl or Q/L for uniform distribution. Surface density is used for sheets, volume density for three-dimensional bodies, and a point charge is localized at one position. Therefore option A is correct.

Each charge element of a circular ring lies on its circumference. Every point on the circumference is at a distance equal to the ring's radius from its centre. Therefore, each charge element is at the same distance, namely the radius, from the centre. The diameter is twice the radius, so option B is not correct.

At a point on the perpendicular bisector, equal charge elements on opposite sides of the rod’s centre are at the same distance from the observation point. Their electric-field components parallel to the rod are equal in magnitude and opposite in direction, so they cancel. The components perpendicular to the rod point in the same direction and add; therefore, the electric field is not generally zero.

In a continuous charge distribution, the charge is treated as infinitesimal elements such as \(dq\). The small contribution to electric field or potential due to each \(dq\) is found, and integration adds all these contributions to obtain the total result. Option B describes only division; the essential integration step is summing the small contributions. Exam tip: first express \(dq\) using the appropriate charge density, then integrate the required quantity.

For uniform charge density, charge per appropriate unit measure does not vary with position. Therefore, equal lengths in a linear distribution, equal areas in a surface distribution, and equal volumes in a volume distribution contain equal charge. Hence, option A is correct. Option D is incorrect because a uniform density does not require the total charge to be zero; for example, for a uniformly charged rod, \(\lambda = Q/L\) can be constant while \(Q\) is non-zero.

In real charged wires, plates, and spheres, charge may be distributed over a length, surface, or volume rather than concentrated at one point. Continuous charge distribution is therefore needed to determine the electric field and potential in such cases. Option B can be a useful approximation in some situations, such as far from the object, but it is not always valid.