Electric Charges and Fields Class 12: Notes, Derivations and Numericals

What Electric Charges and Fields is about

Weightage: This chapter is part of Units I and II, Electrostatics; Current Electricity, which carry 16 of the 70 theory marks together with Electrostatic Potential and Capacitance and Current Electricity.Source: CBSE Physics syllabus 2026-27, Units I and II Electrostatics; Current Electricity

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 6.3 marks per paper. See the questions

Key concepts

Electric charge and its basic properties

NCERT section 1.2 and 1.4

Charge is a property of matter that makes it feel electric forces. There are two kinds, positive and negative: like charges repel and unlike charges attract. When you rub a glass rod with silk, electrons move from glass to silk, so the glass becomes positive and the silk equally negative. Rubbing does not create charge; it only moves electrons.

Three properties matter for the board:

• Additivity: charge is a scalar, so the total charge of a system is the algebraic sum, for example $(+2)+(-5)+(+1)=-2$ units.
• Conservation: the total charge of an isolated system never changes. Charge can move from one body to another, and particles can be created or destroyed in pairs of equal and opposite charge, but the net charge stays the same.
• Quantisation: every charge is a whole-number multiple of the basic charge $e$, so $q=ne$ with $n=0,\pm 1,\pm 2,\dots$ and $e=1.6\times 10^{-19}C$. Because $e$ is so small, quantisation is not noticed for everyday charges of microcoulombs; they behave as if continuous.

Conductors (metals, the human body, the earth) let charge move freely through them; insulators (glass, plastic, nylon) do not.

Worked example

A plastic comb rubbed with dry hair acquires a charge of $-4.8\times 10^{-7}C$. (a) How many electrons did it gain? (b) What mass did the comb gain? (Mass of an electron $=9.1\times 10^{-31}kg$.) [2 marks]

Try it first, then show the solutionSolution

1. The comb is negative, so it gained electrons. $n=\frac{q}{e}=\frac{4.8\times 10^{-7}}{1.6\times 10^{-19}}$
2. $n=3\times 10^{12}$ electrons
3. Mass gained $=n{m}_e=3\times 10^{12}\times 9.1\times 10^{-31}kg$
4. $=2.73\times 10^{-18}kg$, far too small to notice

Answer: (a) $3\times 10^{12}$ electrons (b) $2.73\times 10^{-18}kg$

Coulomb's law

NCERT section 1.5

The force between two point charges $q_1$ and $q_2$ a distance $r$ apart in vacuum acts along the line joining them, is proportional to the product of the charges and inversely proportional to the square of the distance:

$F=\frac{1}{4\pi {\epsilon }_0}\frac{|q_1{q}_2|}{r^2}$, with $k=\frac{1}{4\pi {\epsilon }_0}\approx 9\times 10^{9}N{m}^2{C}^{-2}$ and ${\epsilon }_0=8.854\times 10^{-12}{C}^2{N}^{-1}{m}^{-2}$ (the permittivity of free space).

Vector form. The force on $q_2$ due to $q_1$ is ${\vec{F}}_{21}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r_{21}^2}{\hat{r}}_{21}$, where ${\hat{r}}_{21}$ is the unit vector from $q_1$ to $q_2$. The sign takes care of itself: if $q_1{q}_2>0$ the force on $q_2$ points away from $q_1$ (repulsion). Also ${\vec{F}}_{12}=-{\vec{F}}_{21}$, so Coulomb forces obey Newton's third law.

One coulomb is the charge that, placed 1 m from an equal charge in vacuum, repels it with a force of $9\times 10^{9}N$.

In a medium of dielectric constant (relative permittivity) $K$, the force falls to $F/K$, because ${\epsilon }_0$ is replaced by $\epsilon =K{\epsilon }_0$.

Worked example

Charges of $+2\mu C$ and $-3\mu C$ are 30 cm apart in air. (a) Find the force between them. (b) What is the force if they are put in a liquid of dielectric constant 3 at the same separation? [2 marks]

Try it first, then show the solutionSolution

1. $F=k\frac{|q_1{q}_2|}{r^2}=\frac{9\times 10^9\times 2\times 10^{-6}\times 3\times 10^{-6}}{(0.30{)}^2}$
2. $=\frac{5.4\times 10^{-2}}{0.09}=0.6N$, attractive (unlike charges)
3. In the liquid, $F^{\prime }=\frac{F}{K}=\frac{0.6}{3}$
4. $F^{\prime }=0.2N$, still attractive

Answer: (a) 0.6 N, attractive (b) 0.2 N, attractive

Forces between multiple charges: the superposition principle

NCERT section 1.6

When several charges act on one charge, the force from each pair is found with Coulomb's law as if the others were not there, and the total force is the vector sum:

${\vec{F}}_1={\vec{F}}_{12}+{\vec{F}}_{13}+\cdots +{\vec{F}}_{1n}$

This is the superposition principle. In practice: draw each force as an arrow on the charge, find the magnitudes, then add the arrows. For charges on one line, add with signs. For forces at an angle, use $F=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos \theta }$ or resolve into components. Symmetry often cancels components, for example at the centre of a square of equal charges the net force is zero.

Worked example

Charges $q_1=+4\mu C$ and $q_2=-1\mu C$ are fixed on the x-axis at $x=0$ and $x=0.1m$. Find the net force on $q_3=+1\mu C$ placed at $x=0.3m$. [3 marks]

Try it first, then show the solutionSolution

1. Force from $q_1$ (distance 0.3 m, repulsive, along $+x$): $F_{31}=\frac{9\times 10^9\times 4\times 10^{-6}\times 1\times 10^{-6}}{0.09}=\frac{0.036}{0.09}=0.4N$
2. Force from $q_2$ (distance 0.2 m, attractive, towards $q_2$, along $-x$): $F_{32}=\frac{9\times 10^9\times 1\times 10^{-6}\times 1\times 10^{-6}}{0.04}=\frac{0.009}{0.04}=0.225N$
3. Both act along the x-axis, so add with signs: $F=0.4-0.225=0.175N$
4. The result is positive, so it points along $+x$, away from the two fixed charges

Answer: 0.175 N along the positive x-axis

Electric field and the field of a point charge

NCERT section 1.7

The electric field at a point is the force per unit positive test charge placed there: $\vec{E}=\frac{\vec{F}}{q_0}$ (strictly, with $q_0$ very small so it does not disturb the source). Its SI unit is $N{C}^{-1}$ (equal to $V{m}^{-1}$). Once you know $\vec{E}$, the force on any charge $q$ at that point is $\vec{F}=q\vec{E}$: along $\vec{E}$ for a positive charge and opposite to it for a negative charge.

For a point charge $Q$, $E=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}$, pointing radially outward for positive $Q$ and inward for negative $Q$. The field of a system of charges is the vector sum of the fields of each charge (superposition again).

Why talk of a field at all? The field is defined at every point whether or not a charge is placed there, and it carries the effect of one charge to another. In time-varying situations (Chapter 8) the field is real enough to carry energy across space.

Worked example

Find the electric field 20 cm from a point charge of $+8nC$ in air, and the force on a proton placed there. ($e=1.6\times 10^{-19}C$) [2 marks]

Try it first, then show the solutionSolution

1. $E=\frac{9\times 10^9\times 8\times 10^{-9}}{(0.2{)}^2}=\frac{72}{0.04}$
2. $E=1800N{C}^{-1}$, directed away from the charge
3. $F=eE=1.6\times 10^{-19}\times 1800$
4. $F=2.88\times 10^{-16}N$, also away from the charge

Answer: $E=1.8\times 10^{3}N{C}^{-1}$ outward; $F=2.88\times 10^{-16}N$ outward

Electric field lines

NCERT section 1.8

A field line is a curve drawn so that the tangent at any point gives the direction of the field there. Their density (lines per unit area) shows the strength of the field. The properties to learn:

• They start on positive charges and end on negative charges (or start or end at infinity for a single charge).
• In a charge-free region they are continuous curves without breaks.
• Two field lines never cross: at a crossing the field would have two directions at one point, which is impossible.
• Electrostatic field lines do not form closed loops, because the electrostatic field is conservative.

Pictures to know: radial lines outward from $+q$ and inward to $-q$; for a dipole, lines leave $+q$ and curve round into $-q$; for two equal positive charges, lines push apart with a neutral point midway; for a uniform field, equally spaced parallel lines.

Worked example

In a region, the field lines are parallel but get closer together as you move to the right. (a) Is the field uniform? (b) A proton is released in this region. Will its acceleration stay constant? Give reasons. [2 marks]

Try it first, then show the solutionSolution

1. (a) Closer lines mean a stronger field, so the field strength increases to the right: the field is not uniform.
2. (b) The force on the proton is $F=eE$ and its acceleration is $a=\frac{eE}{m}$.
3. As the proton moves into regions of larger $E$, its acceleration grows; it is not constant.

Answer: (a) No, the field is stronger where the lines crowd together. (b) No, the acceleration $eE/m$ changes because $E$ changes.

Electric flux

NCERT section 1.9

Electric flux measures how much field passes through a surface, in a way similar to the flow of water through a ring. For a small flat area $\Delta S$ with unit normal $\hat{n}$, the area vector is $\Delta \vec{S}=\Delta S\hat{n}$ and the flux is

$\Delta \varphi =\vec{E}\cdot \Delta \vec{S}=E\Delta S\cos \theta$

where $\theta$ is the angle between $\vec{E}$ and the normal (not the surface). Flux is largest when the field is along the normal ($\theta =0$) and zero when the field lies along the surface ($\theta =90^{\circ }$). For a closed surface, the normal points outward, so field lines leaving give positive flux and lines entering give negative flux. Flux is a scalar; its SI unit is $N{m}^2{C}^{-1}$ (also written V m).

Worked example

A uniform field $\vec{E}=3\times 10^3\hat{i}N{C}^{-1}$ exists in a region. Find the flux through a square of side 10 cm when (a) its normal is along the x-axis, (b) its normal makes $60^{\circ }$ with the x-axis. [2 marks]

Try it first, then show the solutionSolution

1. Area $S=(0.1{)}^2=1\times 10^{-2}{m}^2$
2. (a) $\varphi =ES\cos 0^{\circ }=3\times 10^3\times 10^{-2}=30N{m}^2{C}^{-1}$
3. (b) $\varphi =ES\cos 60^{\circ }=30\times 0.5$
4. $=15N{m}^2{C}^{-1}$

Answer: (a) $30N{m}^2{C}^{-1}$ (b) $15N{m}^2{C}^{-1}$

Electric dipole and its field

NCERT section 1.10

An electric dipole is a pair of equal and opposite charges $+q$ and $-q$ a small distance $2a$ apart. Its dipole moment is $\vec{p}=q\times 2a\hat{p}$, directed from $-q$ to $+q$; unit C m. The net charge is zero, but the field is not zero because the two charges are at different places.

On the axis (point P at distance $r$ from the centre, on the side of $+q$): the field of $+q$ is $\frac{kq}{(r-a{)}^2}$ along $\hat{p}$ and that of $-q$ is $\frac{kq}{(r+a{)}^2}$ against $\hat{p}$. Subtracting, $E=kq\frac{4ar}{(r^2-a^2{)}^2}$, which for $r\gg a$ becomes $\vec{E}=\frac{1}{4\pi {\epsilon }_0}\frac{2\vec{p}}{r^3}$, along $\vec{p}$.

On the equatorial line (perpendicular bisector): the two fields have equal magnitude $\frac{kq}{r^2+a^2}$; their components along the bisector cancel and those parallel to the axis add, giving $E=\frac{2kqa}{(r^2+a^2{)}^{3/2}}$, which for $r\gg a$ becomes $\vec{E}=-\frac{1}{4\pi {\epsilon }_0}\frac{\vec{p}}{r^3}$, opposite to $\vec{p}$.

So at the same large distance, the axial field is twice the equatorial field, and both fall off as $\frac{1}{r^3}$ (faster than the $\frac{1}{r^2}$ of a single charge). Molecules such as water have a permanent dipole moment (polar molecules); in others, a dipole moment is induced by a field.

Worked example

Charges of $\pm 5\mu C$ are 4 mm apart. Find (a) the dipole moment, (b) the field at 20 cm from the centre on the axis, (c) the field at 20 cm on the equatorial line. [3 marks]

Try it first, then show the solutionSolution

1. (a) $p=q(2a)=5\times 10^{-6}\times 4\times 10^{-3}=2\times 10^{-8}Cm$, from $-q$ to $+q$
2. $r=0.2m\gg a=2mm$, so use the short-dipole formulas; $r^3=8\times 10^{-3}{m}^3$
3. (b) $E_{axial}=\frac{2kp}{r^3}=\frac{2\times 9\times 10^9\times 2\times 10^{-8}}{8\times 10^{-3}}=\frac{360}{8\times 10^{-3}}=4.5\times 10^{4}N{C}^{-1}$, along $\vec{p}$
4. (c) $E_{eq}=\frac{kp}{r^3}=2.25\times 10^{4}N{C}^{-1}$, opposite to $\vec{p}$

Answer: (a) $2\times 10^{-8}Cm$ (b) $4.5\times 10^{4}N{C}^{-1}$ along $\vec{p}$ (c) $2.25\times 10^{4}N{C}^{-1}$ opposite to $\vec{p}$

Dipole in a uniform external field

NCERT section 1.11

Place a dipole in a uniform field $\vec{E}$ at angle $\theta$ to the field. The force on $+q$ is $q\vec{E}$ and on $-q$ is $-q\vec{E}$: they cancel, so the net force is zero. But the two forces act at different points, so they form a couple. The torque is force times perpendicular distance between the lines of action: $\tau =qE\times 2a\sin \theta =pE\sin \theta$, or in vector form $\vec{\tau }=\vec{p}\times \vec{E}$.

The torque turns the dipole to line up with the field. It is maximum ($pE$) at $\theta =90^{\circ }$ and zero at $\theta =0^{\circ }$ (stable equilibrium) and at $\theta =180^{\circ }$ (unstable equilibrium).

In a non-uniform field the forces on the two ends are unequal, so there is also a net force. If $\vec{p}$ is along the field, the dipole is pulled towards the stronger field; this is why a charged comb attracts small bits of paper (it induces dipoles in them and the field near the comb is non-uniform).

Worked example

A dipole of charges $\pm 3nC$ separated by 2 mm is placed in a uniform field of $5\times 10^{4}N{C}^{-1}$ at $30^{\circ }$ to the field. Find (a) the torque on it, (b) the maximum torque it can experience in this field, (c) the net force on it. [3 marks]

Try it first, then show the solutionSolution

1. $p=q(2a)=3\times 10^{-9}\times 2\times 10^{-3}=6\times 10^{-12}Cm$
2. (a) $\tau =pE\sin \theta =6\times 10^{-12}\times 5\times 10^4\times \sin 30^{\circ }=3\times 10^{-7}\times 0.5=1.5\times 10^{-7}Nm$
3. (b) Maximum at $\theta =90^{\circ }$: ${\tau }_{max}=pE=3\times 10^{-7}Nm$
4. (c) The field is uniform, so the forces $+qE$ and $-qE$ cancel: net force is zero

Answer: (a) $1.5\times 10^{-7}Nm$ (b) $3\times 10^{-7}Nm$ (c) zero

Continuous charge distribution

NCERT section 1.12

When charge is spread over a body, we describe it by a density:

• Linear charge density $\lambda =\frac{\Delta Q}{\Delta l}$ (C m⁻¹) for a wire or rod;
• Surface charge density $\sigma =\frac{\Delta Q}{\Delta S}$, charge per unit area (C m⁻²), for a sheet or shell;
• Volume charge density $\rho =\frac{\Delta Q}{\Delta V}$ (C m⁻³) for a solid.

The field is found by dividing the body into small elements, treating each as a point charge, and adding the fields as vectors (an integral). The densities are 'macroscopic': $\Delta l$, $\Delta S$ or $\Delta V$ is small on the scale of the body but large enough to contain very many electrons, so quantisation can be ignored. Uniform densities give easy totals: $Q=\lambda L$, $Q=\sigma A$ and $Q=\rho V$.

Worked example

A thin spherical shell of radius 10 cm carries a uniform surface charge density of $2\times 10^{-6}C{m}^{-2}$. Find the total charge on it. [1 mark]

Try it first, then show the solutionSolution

1. Surface area $=4\pi R^2=4\times 3.14\times (0.1{)}^2=0.1256m^2$
2. $Q=\sigma \times 4\pi R^2=2\times 10^{-6}\times 0.1256=2.51\times 10^{-7}C$

Answer: $Q\approx 2.5\times 10^{-7}C$ (about $0.25\mu C$)

Gauss's law

NCERT section 1.13

Statement: the total electric flux through any closed surface equals $\frac{1}{{\epsilon }_0}$ times the net charge enclosed by it:

$\oint \vec{E}\cdot d\vec{S}=\frac{q_{enc}}{{\epsilon }_0}$

The syllabus asks you to state it, not prove it. Points examiners look for:

• The closed surface (the Gaussian surface) can have any shape; the flux depends only on the charge inside, not on the shape or size of the surface or where the charge sits inside.
• Charges outside contribute zero net flux (every line that enters also leaves), although they do change the field $\vec{E}$ at points on the surface.
• $\vec{E}$ in the law is the total field due to all charges, inside and outside.
• The law follows from the inverse-square form of Coulomb's law; it would not hold for any other power of $r$.
• A Gaussian surface should not pass through a discrete charge.

If the enclosed charge is zero, the net flux is zero. Gauss's law is most useful when symmetry lets you take $E$ out of the integral.

Worked example

A point charge of $4.0nC$ sits at the centre of a cube of side 10 cm. Find (a) the flux through the whole cube, (b) the flux through one face. (${\epsilon }_0=8.854\times 10^{-12}{C}^2{N}^{-1}{m}^{-2}$) [2 marks]

Try it first, then show the solutionSolution

1. (a) $\varphi =\frac{q}{{\epsilon }_0}=\frac{4.0\times 10^{-9}}{8.854\times 10^{-12}}$
2. $\varphi \approx 452N{m}^2{C}^{-1}$ (the side length does not matter)
3. (b) By symmetry the six faces share it equally: $\frac{452}{6}$
4. $\approx 75.3N{m}^2{C}^{-1}$

Answer: (a) about $452N{m}^2{C}^{-1}$ (b) about $75.3N{m}^2{C}^{-1}$

Applications of Gauss's law: wire, plane sheet and spherical shell

NCERT section 1.14

Choose a Gaussian surface that matches the symmetry so that $E$ is the same everywhere on the part that carries flux.

Infinitely long straight wire (linear density $\lambda$): take a coaxial cylinder of radius $r$ and length $l$. The field is radial, so the flat ends carry no flux; the curved surface gives $E\times 2\pi rl$. The enclosed charge is $\lambda l$. So $E\cdot 2\pi rl=\frac{\lambda l}{{\epsilon }_0}$, giving $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$, radially outward for positive $\lambda$. $E\propto \frac{1}{r}$.

Infinite plane sheet (surface density $\sigma$): take a cylinder of cross-section $A$ passing through the sheet. The field is perpendicular to the sheet, so only the two flat ends carry flux: $2EA$. The enclosed charge is $\sigma A$. So $2EA=\frac{\sigma A}{{\epsilon }_0}$, giving $E=\frac{\sigma }{2{\epsilon }_0}$, independent of distance.

Thin spherical shell (radius $R$, total charge $q$): take a concentric sphere of radius $r$. Outside ($r>R$): $E\cdot 4\pi r^2=\frac{q}{{\epsilon }_0}$, so $E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}$, as if all the charge were at the centre. Inside ($r<R$): no charge is enclosed, so $E=0$ everywhere inside. Just outside the surface, $E=\frac{\sigma }{{\epsilon }_0}$ (since $q=\sigma \cdot 4\pi R^2$).

Worked example

A long straight wire carries a uniform linear charge density of $3\mu C{m}^{-1}$. Find the field at 9 cm from it. [2 marks]

Try it first, then show the solutionSolution

1. $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}=\frac{2k\lambda }{r}$, since $\frac{1}{2\pi {\epsilon }_0}=2k$
2. $E=\frac{2\times 9\times 10^9\times 3\times 10^{-6}}{0.09}$
3. $=\frac{5.4\times 10^4}{0.09}=6\times 10^{5}N{C}^{-1}$, radially outward

Answer: $6\times 10^{5}N{C}^{-1}$, radially away from the wire

Physics in real life: Electric Charges and Fields

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Coulomb's Law and Superposition
• Electric Dipole
• Gauss's Law and Its Applications

Electric Charges and Fields formula sheet

Quantisation of charge

$q=ne,e=1.6\times 10^{-19}C$

n is a whole number. Use it to count electrons gained or lost: $n=q/e$.

Coulomb's law

$F=\frac{1}{4\pi {\epsilon }_0}\frac{|q_1{q}_2|}{r^2},\frac{1}{4\pi {\epsilon }_0}=9\times 10^{9}N{m}^2{C}^{-2}$

Convert µC to C and cm to m first.

Coulomb's law, vector form

${\vec{F}}_{21}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r_{21}^2}{\hat{r}}_{21}=-{\vec{F}}_{12}$

${\hat{r}}_{21}$ points from $q_1$ to $q_2$.

Force in a medium

$F_{medium}=\frac{F_{vacuum}}{K},\epsilon =K{\epsilon }_0$

K is the dielectric constant (relative permittivity).

Permittivity of free space

${\epsilon }_0=8.854\times 10^{-12}{C}^2{N}^{-1}{m}^{-2}$

Electric field (definition)

$\vec{E}=\frac{\vec{F}}{q_0},\vec{F}=q\vec{E}$

Unit N C⁻¹ = V m⁻¹.

Field of a point charge

$E=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r^2}$

Radially outward for +Q, inward for −Q.

Dipole moment

$\vec{p}=q\times 2a\hat{p}$

Unit C m; points from −q to +q.

Dipole field on the axis (r ≫ a)

${\vec{E}}_{axial}=\frac{1}{4\pi {\epsilon }_0}\frac{2\vec{p}}{r^3}$

Exact form: $E=\frac{1}{4\pi {\epsilon }_0}\frac{2pr}{(r^2-a^2{)}^2}$.

Dipole field on the equatorial line (r ≫ a)

${\vec{E}}_{eq}=-\frac{1}{4\pi {\epsilon }_0}\frac{\vec{p}}{r^3}$

Exact form: $E=\frac{1}{4\pi {\epsilon }_0}\frac{p}{(r^2+a^2{)}^{3/2}}$; opposite to $\vec{p}$.

Torque on a dipole

$\vec{\tau }=\vec{p}\times \vec{E},\tau =pE\sin \theta$

Net force is zero in a uniform field.

Electric flux

$\Delta \varphi =\vec{E}\cdot \Delta \vec{S}=E\Delta S\cos \theta$

θ is measured from the normal. Unit N m² C⁻¹.

Gauss's law

$\oint \vec{E}\cdot d\vec{S}=\frac{q_{enc}}{{\epsilon }_0}$

Charge densities

$\lambda =\frac{Q}{L},\sigma =\frac{Q}{A},\rho =\frac{Q}{V}$

Field of an infinitely long straight wire

$E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$

Radial; falls as 1/r.

Field of an infinite plane sheet

$E=\frac{\sigma }{2{\epsilon }_0}$

Same at every distance; perpendicular to the sheet.

Field of a thin spherical shell

$E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}(r\ge R),E=0(r<R)$

Just outside: $E=\sigma /{\epsilon }_0$.

Electric Charges and Fields: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Electric Charges and Fields

The question patterns to expect, and the method that works for each.

Tricks to remember Electric Charges and Fields

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Axial is double

  At the same large distance, axial field = 2 × equatorial field: 'axis gets two, equator gets one, and equator points back'. Both go as $1/r^3$.

  Use it when: Dipole MCQs and checking the result of a derivation.

• Wire, sheet, shell: 1/r, constant, 1/r²

  Line charge gives $\frac{1}{r}$, sheet gives no $r$ at all, shell (outside) gives $\frac{1}{r^2}$. Think of the shapes as 1-D, 2-D and a point seen from outside.

  Use it when: Graph-based MCQs on how E varies with distance.

• Gauss counts only the guests inside

  Flux through a closed surface is like counting the people inside a room: anyone standing outside, however close, is not counted.

  Use it when: Flux questions with charges both inside and outside the surface.

• Cos from the normal

  In $ES\cos \theta$, θ is between the field and the NORMAL. Field along the surface: no flux; field through the surface square-on: full flux.

  Use it when: Avoids using sine by mistake when the angle is given from the plane.

• Minus to plus

  The dipole moment arrow walks from minus to plus, the opposite of how field lines run between the charges.

  Use it when: Direction of $\vec{p}$ and of the equatorial field.

• Zero point near the small one

  The null point always lies nearer the smaller charge: between them if like, outside beyond the smaller one if unlike.

  Use it when: Point-of-zero-field questions.

Electric Charges and Fields: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Electric Charges and Fields

The slips examiners see most often, and what to write instead.

• ✗Mistake: Measuring θ from the surface instead of the normal in flux

  The question gives the angle the field makes with the plane.

  Wrong working

  Field at $30^{\circ }$ to a plane: $\varphi =ES\cos 30^{\circ }$

  Right working

  Angle with the normal is $60^{\circ }$: $\varphi =ES\cos 60^{\circ }=ES\sin 30^{\circ }$

  ✓Do this: Always convert to the angle with the normal (area vector) before using cos.

  Usually costs: 1 mark in a 2-mark numerical.

• ✗Mistake: Using the far-field dipole formula when r is not much larger than a

  Students learn only the $2kp/r^3$ form.

  Wrong working

  Charges ±q at 2a = 6 cm, point 5 cm on the axis: $E=2kp/r^3$

  Right working

  $r$ is comparable to $a$, so use $E=\frac{2kpr}{(r^2-a^2{)}^2}$

  ✓Do this: Check $r\gg a$ before applying the short-dipole result; otherwise use the exact expression.

  Usually costs: Usually the final 1-2 marks.

• ✗Mistake: Not naming or drawing the Gaussian surface in a derivation

  Writing the final formula from memory.

  ✓Do this: Write: 'Consider a cylindrical Gaussian surface of radius r and length l, coaxial with the wire', draw it, and say why the flat ends contribute zero flux.

  Usually costs: 1 mark of the 3 in a Gauss's law derivation.

• ✗Mistake: Forgetting the 2 in the plane-sheet result or the wire result

  Mixing $\frac{\sigma }{2{\epsilon }_0}$ (sheet) with $\frac{\sigma }{{\epsilon }_0}$ (just outside a conductor) and $\frac{\lambda }{2\pi {\epsilon }_{0}r}$ with $\frac{\lambda }{4\pi {\epsilon }_{0}r}$.

  Wrong working

  Sheet: $2EA=\frac{\sigma A}{{\epsilon }_0}$ written as $EA=\frac{\sigma A}{{\epsilon }_0}$

  Right working

  The sheet has two faces of the cylinder carrying flux: $2EA=\frac{\sigma A}{{\epsilon }_0}\Rightarrow E=\frac{\sigma }{2{\epsilon }_0}$

  ✓Do this: Count the faces that carry flux every time.

  Usually costs: 1 mark.

• ✗Mistake: Adding magnitudes of forces that act in different directions

  Superposition treated as plain addition.

  Wrong working

  Two 0.4 N forces at $60^{\circ }$: net = 0.8 N

  Right working

  $F=\sqrt{0.4^2+0.4^2+2(0.4)(0.4)\cos 60^{\circ }}=0.4\sqrt{3}\approx 0.69N$

  ✓Do this: Draw arrows, then resolve or use the parallelogram law.

  Usually costs: 1-2 marks in a 3-mark question.

• ✗Mistake: Saying a charge outside a Gaussian surface does not affect the field on it

  Confusing net flux with the field.

  ✓Do this: Outside charges give zero NET flux, but they still change $\vec{E}$ at points on the surface.

  Usually costs: 1 mark in assertion-reason or reasoning items.

• ✗Mistake: Wrong direction for the equatorial field

  Students state only the magnitude.

  ✓Do this: Write the vector result $\vec{E}=-\frac{\vec{p}}{4\pi {\epsilon }_0{r}^3}$ and say 'opposite to the dipole moment'.

  Usually costs: ½ to 1 mark.

NCERT exercise guide: Electric Charges and Fields

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.