Class 12 — Electric Charges and Fields

Chapter Overview & Weightage

Electric Charges and Fields kicks off the Class 12 Physics syllabus. It builds the language we’ll use for the next four chapters (potential, capacitance, current, and EMI). For CBSE boards, this chapter typically carries 5-8 marks across 1-mark MCQs, a 2- or 3-mark short answer, and sometimes a 5-mark numerical or derivation.

Year	Marks
2024	7
2023	6
2022	8
2021	5
2020	7

The 5-mark question almost always involves either (a) Gauss’s law application to a symmetric distribution, or (b) electric dipole field derivation.

Key Concepts You Must Know

Coulomb’s law: $F = kq_1q_2/r^2$ where $k = 1/(4\pi\varepsilon_0) = 9 \times 10^9$ N·m²/C².
Electric field: $\vec{E} = \vec{F}/q$ , force per unit positive test charge.
Superposition: total field is the vector sum of fields from each charge.
Electric dipole: two equal and opposite charges separated by distance $2a$ . Dipole moment $\vec{p} = q \cdot 2\vec{a}$ , pointing from $-q$ to $+q$ .
Field of a dipole on the axial line: $E_{axial} = 2kp/r^3$ (along $\vec{p}$ ).
Field of a dipole on the equatorial line: $E_{eq} = -kp/r^3$ (anti-parallel to $\vec{p}$ ).
Torque on a dipole: $\vec{\tau} = \vec{p} \times \vec{E}$ .
Gauss’s law: $\oint \vec{E} \cdot d\vec{A} = q_{enc}/\varepsilon_0$ .
Electric flux: $\Phi = \oint \vec{E} \cdot d\vec{A}$ .

Important Formulas

$\vec{F}_{12} = \frac{kq_1q_2}{r^2}\hat{r}_{12}$

When to use: force between point charges in vacuum. For continuous distributions, integrate.

$E = \frac{kQx}{(R^2 + x^2)^{3/2}}$

When to use: ring of total charge $Q$ , radius $R$ , point at distance $x$ on the axis. Maximum field at $x = R/\sqrt{2}$ .

$E = \frac{\lambda}{2\pi\varepsilon_0 r}$

When to use: infinite straight wire with linear charge density $\lambda$ . Derive using a cylindrical Gaussian surface.

$E = \frac{\sigma}{2\varepsilon_0}$

When to use: infinite sheet with surface charge density $\sigma$ . Field points perpendicular, independent of distance.

Outside: $E = kQ/r^2$ (acts as if all charge at centre).

Inside: $E = 0$ .

When to use: spherical conductor or shell. The “shell theorem” simplifies to point-charge behaviour outside.

Solved Previous Year Questions

PYQ 1 (CBSE 2023, 3 marks)

Two point charges $+5\,\mu\text{C}$ and $-5\,\mu\text{C}$ are placed 10 cm apart. Find the electric field at a point 10 cm from each charge (on the perpendicular bisector).

Solution: This is the equatorial position of a dipole. Each charge produces a field of magnitude:

$E_1 = E_2 = \frac{kq}{r^2} = \frac{9 \times 10^9 \times 5 \times 10^{-6}}{(0.1)^2} = 4.5 \times 10^6\text{ N/C}$

The vertical components cancel; the horizontal components add. Geometry: the components along the dipole axis are $E_1 \cos\theta + E_2 \cos\theta = 2E_1 \cos\theta$ , where $\cos\theta = 5/10 = 0.5$ .

$E_{net} = 2 \times 4.5 \times 10^6 \times 0.5 = 4.5 \times 10^6\text{ N/C}$

Direction: from $+q$ towards $-q$ (along the dipole axis, antiparallel to $\vec{p}$ ).

PYQ 2 (CBSE 2022, 5 marks)

Using Gauss’s law, derive the electric field due to an infinite plane sheet of charge.

Solution: Consider a cylindrical Gaussian surface of cross-section $A$ piercing the sheet symmetrically. By symmetry, $\vec{E}$ is perpendicular to the sheet on both sides.

Flux through the curved surface = 0 (E parallel to surface).

Flux through both flat ends = $2EA$ (E perpendicular to ends).

By Gauss’s law:

$2EA = \frac{\sigma A}{\varepsilon_0} \implies E = \frac{\sigma}{2\varepsilon_0}$

The field is independent of distance from the sheet.

PYQ 3 (CBSE 2024, 2 marks)

A dipole of moment $p$ is placed in a uniform field $E$ at angle $\theta$ . Find the work done in rotating it from $\theta_1$ to $\theta_2$ .

Solution: Potential energy of dipole: $U = -\vec{p} \cdot \vec{E} = -pE\cos\theta$ .

Work done = $\Delta U = pE(\cos\theta_1 - \cos\theta_2)$ .

Difficulty Distribution

Easy (40%): Coulomb’s law, simple field/force calculations, definitions.
Medium (40%): Dipole problems, applications of Gauss’s law to standard symmetries, flux through specific surfaces.
Hard (20%): Combined geometry (e.g., field at the centre of a square of charges), non-uniform charge distributions requiring integration.

Expert Strategy

For any electric field problem, ask three questions in order:

Can I use Coulomb’s law directly? (Discrete charges, simple geometry.)
Is there symmetry I can exploit with Gauss’s law? (Spherical, cylindrical, planar.)
Do I need to integrate? (Continuous distribution, no symmetry.)

This decision tree saves 70% of your time.

Always draw a clean diagram with vectors before plugging numbers. Components are easier to track when you can see them.

For the 5-mark Gauss’s law derivation, memorise the three standard cases (line, sheet, sphere) cold. Examiners ask for derivations word-for-word from NCERT.

Common Traps

Trap 1: Sign of dipole moment. $\vec{p}$ points from $-q$ to $+q$ . Many students reverse this. The convention matters for torque and potential energy formulas.

Trap 2: Field inside a charged conductor. Always zero in electrostatic equilibrium, regardless of external fields. Students sometimes write $E = kQ/r^2$ inside — wrong.

Trap 3: Flux depends only on enclosed charge. External charges contribute to $\vec{E}$ everywhere, but their net flux through a closed surface is zero. So if a question asks for flux through a Gaussian surface, count only enclosed charges.

CBSE often tests the dipole field formula at non-axial, non-equatorial points. The general formula:

$E = \frac{kp}{r^3}\sqrt{1 + 3\cos^2\theta}$

where $\theta$ is the angle from the dipole axis. Worth memorising for board exams.