Q.State, in your own words, the principle of superposition for the electric force on a charge due to several other point charges, and explain why the individual forces must be added as vectors rather than by simply adding their magnitudes.

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 coulomb (C) and length is in metre (m), the SI unit of lambda is C/m.

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.

Step 3 — solve: x = d/3 = 0·18/3 = 0·06 m from the smaller charge (and 0·12 m from the larger).

This 'field-zero between like charges, closer to the weaker one' result is standard in the NCERT/CBSE Class 12 Physics chapter on Electric Charges and Fields.

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: $\lambda = \dfrac{q}{L}$.

Taking the SI unit of charge (coulomb) divided by the SI unit of length (metre) gives coulomb/metre (C m⁻¹). Compare with surface density σ (C/m²) and volume density ρ (C/m³).

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).

Surface charge density σ = Q/A.

Surface density of charge is the charge distributed per unit area of a surface:

$\sigma = \frac{Q}{A}$

Its SI unit is coulomb per square metre (C/m²). (Charge per unit length is linear density λ = Q/l, and charge per unit volume is volume density ρ = Q/V.) Hence the correct expression is σ = Q/A.

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:

$\sigma = \frac{q}{A}$

Its SI unit is coulomb per square metre (C/m²). (Compare: linear density λ = q/length in C/m, volume density ρ = q/volume in C/m³.) So surface density equals total charge divided by total area.

Surface charge density is charge per unit surface area, $\sigma = q/A$, measured in $\text{C/m}^2$.

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:

$\sigma = \frac{q}{A}$

If the charge is spread uniformly, $\sigma$ is the same at every point of the surface; if not, $\sigma$ is defined locally as $\sigma = dq/dA$, the charge on a small area element divided by that area.

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 $q_0$ due to a group of charges $q_1, q_2, \dots, q_n$ is the vector sum of the forces exerted individually by each charge on $q_0$, each computed by Coulomb's law as if the other charges were not present:

$\vec{F} = \vec{F}_1 + \vec{F}_2 + \dots + \vec{F}_n = \sum_{i=1}^{n} \frac{1}{4\pi\epsilon_0}\frac{q_0 q_i}{r_i^2}\hat{r}_i$

This works because the presence of a third charge does not alter the mutual interaction between any pair of charges - electrostatic forces obey linear superposition.