Continuous Charge Distribution

Correct answer: A. Charge continuously distributed along a line, over a surface, or throughout a volume

Explanation: 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.

Correct answer: B. Linear charge density

Explanation: 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.

Correct answer: C. Surface charge density

Explanation: 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.

Correct answer: D. Volume charge density

Explanation: 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.

Correct answer: A. Charge per unit length

Explanation: 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.

Correct answer: B. Charge per unit area

Explanation: 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.

Correct answer: C. Charge per unit volume

Explanation: 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.

Correct answer: D. Coulomb per metre

Explanation: 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.

Correct answer: A. Coulomb per square metre

Explanation: 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 \(\mathrm{C/m^3}\).

Correct answer: B. Coulomb per cubic metre

Explanation: 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.

Correct answer: C. Remains constant at every point

Explanation: 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.

Correct answer: B. Varies with position on the surface

Explanation: 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.

Correct answer: B. To divide the distributed charge into infinitesimal parts and add the effects of each part

Explanation: 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.

Correct answer: B. Principle of superposition

Explanation: 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.

Correct answer: A. By integrating all infinitesimal charge elements

Explanation: 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.

Correct answer: B. There is net positive charge per unit length on the line.

Explanation: 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.

Correct answer: A. There is a net negative charge per unit area on the surface

Explanation: 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.

Correct answer: B. It doubles

Explanation: 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.

Correct answer: B. sigma

Explanation: 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.

Correct answer: B. It will become half

Explanation: 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.

Correct answer: B. Along the entire circumference of the ring

Explanation: 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.

Correct answer: A. वे केंद्र पर समान परिमाण के तथा विपरीत दिशाओं में विद्युत क्षेत्र उत्पन्न करते हैं।

Explanation: 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.

Correct answer: A. Some field components due to symmetric charge elements are equal in magnitude and opposite in direction

Explanation: 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.

Correct answer: A. Because electric field is a vector quantity, and its direction is also important.

Explanation: 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.

Correct answer: B. Uniform distribution

Explanation: 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.