Q.Unit of linear charge density is
(A) coulomb/metre
(B) coulomb × metre
(C) metre/coulomb
(D) none of these
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When more than two charges are present, the net force (or field) on any one charge is the VECTOR sum of the individual Coulomb contributions from every other charge, each computed as though the rest were absent -- never a simple sum of magnitudes, since the individual contributions generally point in different directions. For a very large number of closely spaced charges (a charged wire, plate or solid body), it is more practical to describe the charge as smeared out continuously using a charge density: linear λ=dq/dl (C/m) for a wire, surface σ=dq/dA (C/m2) for a shee …
Linear charge density measures how much charge is spread along the length of a line, so its unit follows directly from dividing charge by length. …
Linear charge density is charge per unit length, so its unit is coulomb/metre.
Linear charge density is defined as the charge distributed per unit length of a line: λ=Lq.
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- CBSE 2026Set ANNUAL1 markMCQQ.The SI Unit of linear charge density is(a) Cm(b) C/m(c) m/C(d) Cm^2
›Reveal solutionSolution
Linear charge density lambda = charge / length, so its SI unit is coulomb per metre.
Linear charge density is defined as lambda = q/l, the charge distributed per unit length of a line/wire. Since charge is in …
- CBSE 2026Set SEM31 markMCQQ.Two point charges are placed 0·18 m apart in air. One charge is four times the other charge. If the electric field is zero at a point on the line joining the two charges, then the position of the point is(a) on the extended part of the line of the charges and at a distance of 0·06 m from the larger charge.(b) between the two point charges on the line and at a distance of 0·06 m from the smaller charge.(c) between the two point charges on the line and at a distance of 0·04 m from the larger charge.(d) on the extended part of the line and at a distance of 0·04 m from the smaller charge.
›Reveal solutionSolution
For like charges the null point lies between them, nearer the smaller charge. Solving kq/x² = 4kq/(0·18−x)² gives x = 0·06 m from the smaller charge. Option (b).
Let the smaller charge be q and the larger 4q, separated by d = 0·18 m. The neutral point (E = 0) lies between them (both being of the same sign) at distance x from q.
Step 1 — equate field magnitudes:
kq/x² = k(4q)/(d−x)².
Step 2 — cancel kq and take square roots:
(d−x)²/x² = 4 → (d−x)/x = 2 → d − x = 2x → d = 3x.
…
- CBSE 2024Set A1 markMCQQ.Unit of linear charge density is (A) coulomb/metre (B) coulomb × metre (C) metre/coulomb (D) none of these
›Reveal solutionSolution
Linear charge density is charge per unit length, so its unit is coulomb/metre.
Linear charge density is defined as the charge distributed per unit length of a line: λ=Lq.
…
- CBSE 2024Set 55/5/11 markMCQQ.Assertion (A): Equal amount of positive and negative charges are distributed uniformly on two halves of a thin circular ring as shown in figure. The resultant electric field at the centre O of the ring is along OC. Reason (R): It is so because the net potential at O is not zero. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is NOT the correct explanation of A. (C) A is true but R is false. (D) Both A and R are false.
›Reveal solutionSolution
The upper half of the ring (containing +q, near A) and the lower half (containing -q, near C) both produce a field at O pointing from A towards C -- so the resultant field at O genuinely is along OC. Assertion (A) is true. But equal and opposite charges at the same distance from O contribute equal and opposite potentials that exactly cancel, so the net potential at O is zero -- the opposite of what Reason (R) claims. Option (C).
Field and potential at the centre of an oppositely-charged half-ring pair
The ring is split into two halves: the upper half (through A) carries uniformly-distributed charge +q, and the lower half (through C) carries uniformly-distributed charge −q (the figure shows + symbols on the upper arc and ×/plain marks on the lower arc).
Checking Assertion (A) -- the field direction. Each small element of the positive (upper) half repels a test charge at O, pushing it away from the positive arc -- i.e. from A towards C. Each element of the negative (lower) half attracts a test charge at O towards itself -- i.e. also from A towards C. Both halves' contributions point the same way, straight from A to C, so the two add constructively along a single line: the resultant field at O is indeed directed along OC. A is true. …
- CBSE 2022Set I1 markMCQQ.Surface density of charge is (A) σ = Q/A (B) σ = Q/l (C) σ = Q/V (D) σ = Q·A
›Reveal solutionSolution
Surface charge density σ = Q/A.
Surface density of charge is the charge distributed per unit area of a surface:
σ=AQ
…
- CBSE 2021Set A1 markMCQQ.Surface density of charge is equal to (A) Total charge × Total area (B) Total charge / Total area (C) Total charge / Total volume (D) Total charge × Total volume
›Reveal solutionSolution
Surface charge density σ = charge/area.
Charge can be distributed along a line, over a surface, or through a volume, giving three densities. The surface (or superficial) charge density σ is defined as the charge per unit surface area:
σ=Aq
…
- CBSE 2020Set ANNUAL1 markQ.What is surface charge density? Write its SI unit.
›Reveal solutionSolution
Surface charge density is charge per unit surface area, σ=q/A, measured in C/m2.
When electric charge is distributed over a two-dimensional surface (such as the surface of a charged conductor or a charged plate), it is convenient to describe how densely the charge is packed using surface charge density.
It is defined as the charge q present per unit area A of the surface:
σ=Aq
…
- CBSE 2017Set ANNUAL1 markQ.State the principle of superposition of charges.
›Reveal solutionSolution
The net force on a charge due to many other charges is just the vector sum of the pairwise Coulomb forces.
Coulomb's law gives the force between only two point charges. When more than two charges are present, the principle of superposition extends this to any number of charges.
It states that the force on any charge q0 due to a group of charges q1,q2,…,qn is the vector sum of the forces exerted individually by each charge on q0, each computed by Coulomb's law as if the other charges were not present:
F=F1+F2+⋯+Fn=∑i=1n4πϵ01ri2q0qir^i …
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