Chapter 1

Electric Charges and Fields

CBSE · 10 subtopics

Formula SheetEvery formula this chapter can be examined on, in one table.

Constants and units CBSE expects

Elementary charge $e = 1.6\times10^{-19}\,\text{C}$ (NCERT quotes $1.602192\times10^{-19}\,\text{C}$ ). Electron carries $-e$ , proton $+e$ .
Permittivity of free space $\varepsilon_0 = 8.854\times10^{-12}\,\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$ ; use $8.85\times10^{-12}$ in numericals.
Coulomb constant $k = \dfrac{1}{4\pi\varepsilon_0} = 9\times10^{9}\,\text{N m}^2\,\text{C}^{-2}$ , so $\dfrac{1}{\varepsilon_0} = 4\pi\times9\times10^{9}\,\text{N m}^2\,\text{C}^{-2}$ .
$1\,\mu\text{C} = 10^{-6}\,\text{C}$ ; $1\,\text{mC} = 10^{-3}\,\text{C}$ ; about $6\times10^{18}$ electrons make up a charge of $-1\,\text{C}$ .
Units to quote: charge in C; electric field in $\text{N C}^{-1}$ (same as $\text{V m}^{-1}$ ); electric flux in $\text{N m}^2\,\text{C}^{-1}$ (same as V m); dipole moment in C m.
Charge densities: linear $\lambda$ in $\text{C m}^{-1}$ , surface $\sigma$ in $\text{C m}^{-2}$ , volume $\rho$ in $\text{C m}^{-3}$ .
Ratio of electric to gravitational force between a proton and an electron, $\dfrac{ke^2}{G\,m_e m_p}\approx 2.4\times10^{39}$ — a pure number, no units.

Basic properties of electric charge

Formula	Symbols, units and conditions
$q = ne,\qquad n = 0,\ \pm1,\ \pm2,\ \ldots$	Quantisation of charge. $q$ is the charge on a body, $n$ an integer, $e$ the elementary charge. Any measured free charge must be an integral multiple of $e$ ; fractional multiples such as $3.5e$ are impossible.
$q_{\text{total}} = q_1+q_2+\cdots+q_n$	Additivity. Charges add algebraically, with their own signs — charge is a scalar.
$\sum q_{\text{before}} = \sum q_{\text{after}}$	Conservation. The total charge of an isolated system is unchanged in any process; charging by friction only transfers electrons.
$q_{\text{gain}} = -ne$	Charge acquired by a body that gains $n$ electrons; a body that loses $n$ electrons acquires $+ne$ .

Coulomb's law

Formula	Symbols, units and conditions
$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2} = k\dfrac{q_1q_2}{r^2}$	Magnitude of the mutual electrostatic force between two point charges $q_1, q_2$ separated by $r$ in vacuum, directed along the line joining them.
$\vec{F}_{21} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{21}^{2}}\,\hat{r}_{21},\qquad \vec{F}_{12} = -\vec{F}_{21}$	Vector form. $\hat{r}_{21}$ points from $q_1$ to $q_2$ . The sign of $q_1q_2$ automatically gives repulsion (positive product) or attraction (negative product); the pair obeys Newton's third law.
$F_{\text{medium}} = \dfrac{1}{4\pi\varepsilon_0 K}\dfrac{q_1q_2}{r^2} = \dfrac{F_{\text{vacuum}}}{K}$	Force in a medium of dielectric constant (relative permittivity) $K = \varepsilon_r$ . Since $K>1$ , inserting a slab of any insulator between two charges reduces the force.
$q_1' = q_2' = \dfrac{q_1+q_2}{2}$	Charge on each of two identical conducting spheres after they are touched and separated. Use it before applying Coulomb's law again.

Forces between multiple charges (superposition)

Formula	Symbols, units and conditions
$\vec{F}_1 = \vec{F}_{12}+\vec{F}_{13}+\cdots+\vec{F}_{1n} = \dfrac{q_1}{4\pi\varepsilon_0}\sum_{i=2}^{n}\dfrac{q_i}{r_{1i}^{2}}\,\hat{r}_{1i}$	Principle of superposition. The force on one charge is the vector sum of the pairwise Coulomb forces, each unaffected by the presence of the others.
$F_{\text{net}} = \sqrt{F_1^2+F_2^2+2F_1F_2\cos\alpha}$	Parallelogram law for combining two of those forces at angle $\alpha$ ; for two perpendicular equal forces $F$ this gives $\sqrt{2}\,F$ .

Electric field

Formula	Symbols, units and conditions
$\vec{E} = \lim_{q_0\to 0}\dfrac{\vec{F}}{q_0},\qquad \vec{F} = q\vec{E}$	Definition: force per unit positive test charge. The limit keeps the test charge from disturbing the source. Force on a negative charge is opposite to $\vec{E}$ .
$\vec{E} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}\,\hat{r},\qquad E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\lvert q\rvert}{r^2}$	Field of a point charge $q$ at distance $r$ : radially outward for $q>0$ , radially inward for $q<0$ ; spherically symmetric.
$\vec{E} = \dfrac{1}{4\pi\varepsilon_0}\sum_{i=1}^{n}\dfrac{q_i}{r_{iP}^{2}}\,\hat{r}_{iP}$	Field at $P$ due to a system of $n$ point charges, by superposition. $r_{iP}$ is the distance from $q_i$ to $P$ and $\hat{r}_{iP}$ the unit vector from $q_i$ to $P$ .
$\dfrac{q_1}{x^2} = \dfrac{q_2}{(d-x)^2}$	Null-point condition on the line joining two like charges $d$ apart; $x$ is measured from $q_1$ . For unlike charges the null point lies outside the pair, beyond the smaller charge, and the condition becomes $\dfrac{q_1}{(d+x)^2} = \dfrac{q_2}{x^2}$ .
$E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{qx}{(r^2+x^2)^{3/2}},\qquad q = \lambda(2\pi r)$	Field on the axis of a uniformly charged thin ring of radius $r$ and linear charge density $\lambda$ , at axial distance $x$ from the centre. For $x\gg r$ it becomes $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{x^2}$ — the ring behaves as a point charge; at the centre ( $x=0$ ) the field is zero.
$\vec{a} = \dfrac{q\vec{E}}{m},\qquad \vec{v} = \vec{u}+\vec{a}t$	Motion of a charged particle in a uniform field: constant acceleration, so the Class 11 kinematic equations apply directly.
$qE = mg\tan\theta$	Equilibrium of a charged bob of weight $mg$ on a string making angle $\theta$ with the vertical in a horizontal field $E$ ; hence $q = \left(\dfrac{mg}{E}\right)\tan\theta$ .

Electric flux

Formula	Symbols, units and conditions
$\Delta\phi = \vec{E}\cdot\Delta\vec{S} = E\,\Delta S\cos\theta,\qquad \Delta\vec{S} = \Delta S\,\hat{n}$	Flux through a small plane area element. $\theta$ is the angle between $\vec{E}$ and the outward normal $\hat{n}$ , not between $\vec{E}$ and the plane of the surface.
$\phi = \oint_S \vec{E}\cdot d\vec{S}$	Total flux through a surface $S$ ; a scalar, SI unit $\text{N m}^2\,\text{C}^{-1}$ (equivalently V m). For a closed surface $\hat{n}$ is the outward normal by convention.
$\phi = 0 \quad\text{when } \vec{E}\perp\hat{n}\ (\theta = 90^\circ)$	Curved side of a coaxial cylinder in an axial field, and the sides of a Gaussian pillbox parallel to $\vec{E}$ , contribute nothing.

Electric dipole

Formula	Symbols, units and conditions
$\vec{p} = q\,(2a)\,\hat{p},\qquad p = q\times 2a$	Dipole moment: magnitude charge times separation, direction from $-q$ to $+q$ . SI unit C m. Net charge of a dipole is zero.
$\vec{E}_{\text{axial}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2pr}{(r^2-a^2)^2}\,\hat{p}$	Exact field at an axial point, distance $r$ from the centre, dipole length $2a$ ; directed along $\vec{p}$ .
$\vec{E}_{\text{axial}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2\vec{p}}{r^3}\qquad (r\gg a)$	Short-dipole limit on the axis.
$\vec{E}_{\text{equatorial}} = -\dfrac{1}{4\pi\varepsilon_0}\dfrac{\vec{p}}{(r^2+a^2)^{3/2}}$	Exact field at a point on the perpendicular bisector, distance $r$ from the centre; antiparallel to $\vec{p}$ .
$\vec{E}_{\text{equatorial}} = -\dfrac{1}{4\pi\varepsilon_0}\dfrac{\vec{p}}{r^3}\qquad (r\gg a)$	Short-dipole limit on the equatorial line.
$\dfrac{E_{\text{axial}}}{E_{\text{equatorial}}} = 2 \qquad (r\gg a)$	Standard one-mark result: at equal distances the axial field is twice the equatorial field and oppositely directed. Both fall off as $1/r^3$ , faster than the $1/r^2$ of a point charge.

Dipole in a uniform external field

Formula	Symbols, units and conditions
$\vec{\tau} = \vec{p}\times\vec{E},\qquad \tau = pE\sin\theta$	Torque on a dipole making angle $\theta$ with a uniform field. $\vec{\tau}$ is perpendicular to both $\vec{p}$ and $\vec{E}$ . Maximum $\tau = pE$ at $\theta = 90^\circ$ ; zero at $\theta = 0^\circ$ and $180^\circ$ .
$\vec{F}_{\text{net}} = q\vec{E}+(-q)\vec{E} = 0 \quad\text{(uniform field only)}$	Net force vanishes in a uniform field, so the dipole only rotates. In a non-uniform field the two forces differ and there is a net force as well as a torque — translation plus rotation.
$U = -\vec{p}\cdot\vec{E} = -pE\cos\theta$	Potential energy of a dipole in a uniform field. $U_{\min} = -pE$ at $\theta = 0^\circ$ (stable), $U = 0$ at $\theta = 90^\circ$ , $U_{\max} = +pE$ at $\theta = 180^\circ$ (unstable).
$p_r = \sqrt{p_1^2+p_2^2+2p_1p_2\cos\alpha}$	Resultant moment of two dipoles inclined at angle $\alpha$ ; for two equal dipoles $p$ at $120^\circ$ , $p_r = p$ .

Continuous charge distribution

Formula	Symbols, units and conditions
$\lambda = \dfrac{\Delta Q}{\Delta l},\qquad \sigma = \dfrac{\Delta Q}{\Delta S},\qquad \rho = \dfrac{\Delta Q}{\Delta V}$	Linear, surface and volume charge densities. Useful conversions: charge on length $l$ of a wire is $\lambda l$ ; on area $A$ of a sheet is $\sigma A$ ; on a shell of radius $R$ is $q = 4\pi R^2\sigma$ .
$\vec{E} = \dfrac{1}{4\pi\varepsilon_0}\sum_{\text{all }\Delta V}\dfrac{\rho\,\Delta V}{r'^2}\,\hat{r}'$	Field of a continuous distribution, by superposition over volume elements; $r'$ is the distance from the element to the field point.

Gauss's law

Formula	Symbols, units and conditions
$\phi = \oint_S \vec{E}\cdot d\vec{S} = \dfrac{q_{\text{enc}}}{\varepsilon_0}$	The total electric flux through any closed surface equals $\dfrac{1}{\varepsilon_0}$ times the net charge enclosed. Independent of the shape and size of the surface and of where inside the charges sit; charges outside contribute to $\vec{E}$ but not to $\phi$ .
$\phi = \dfrac{q_{\text{enc}}}{\varepsilon_0\varepsilon_r}$	Same law when the enclosed region is filled with a medium of dielectric constant $\varepsilon_r$ .
$\phi_{\text{face}} = \dfrac{q}{6\varepsilon_0}$	Flux through one face of a cube with a point charge $q$ at its centre — used whenever a charge sits at distance $d/2$ above the centre of a square of side $d$ .
$q_{\text{enc}} = \varepsilon_0\phi$	Reading a net enclosed charge off a measured flux.

Applications of Gauss's law

Formula	Symbols, units and conditions
$E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$	Infinitely long straight uniformly charged wire, at perpendicular distance $r$ ; radially outward for $\lambda>0$ . $E\propto\dfrac{1}{r}$ .
$W = \dfrac{q\lambda}{2\pi\varepsilon_0}\ln\dfrac{r_1}{r_2}$	Work done in taking a charge $q$ from perpendicular distance $r_2$ to $r_1$ from such a wire.
$\vec{E} = \dfrac{\sigma}{2\varepsilon_0}\,\hat{n}$	Infinite uniformly charged plane sheet; $\hat{n}$ is the outward normal on either side. Independent of distance from the sheet — away from it if $\sigma>0$ , towards it if $\sigma<0$ .
$E_{\text{between}} = \dfrac{\sigma}{\varepsilon_0},\qquad E_{\text{outside}} = 0$	Two large parallel sheets carrying $+\sigma$ and $-\sigma$ : the fields add between the plates and cancel outside.
$E_{\text{between}} = \dfrac{\sigma_1-\sigma_2}{2\varepsilon_0},\qquad E_{\text{outside}} = \dfrac{\sigma_1+\sigma_2}{2\varepsilon_0}$	Two parallel sheets with like densities $\sigma_1>\sigma_2$ : region between the sheets, and the regions beyond them.
$E = \dfrac{q}{4\pi\varepsilon_0 r^2}\quad (r\ge R),\qquad E = 0\quad (r<R)$	Thin uniformly charged spherical shell of radius $R$ and total charge $q = 4\pi R^2\sigma$ . Outside, the shell behaves exactly as a point charge $q$ at the centre; inside, the field is zero everywhere.
$E_{\text{surface}} = \dfrac{\sigma}{\varepsilon_0}\qquad (r = R)$	Value at the surface of the shell — the field jumps discontinuously from $0$ to $\dfrac{\sigma}{\varepsilon_0}$ there.
$E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{qr}{R^3} = \dfrac{\rho r}{3\varepsilon_0}\qquad (r<R)$	Inside a solid sphere of uniform volume charge density $\rho$ ; outside it the field is again $\dfrac{q}{4\pi\varepsilon_0 r^2}$ .
$E(r) = \dfrac{Ze}{4\pi\varepsilon_0}\left(\dfrac{1}{r^2}-\dfrac{r}{R^3}\right)\quad (r<R),\qquad E = 0\quad (r>R)$	NCERT's early atom model: point nucleus $+Ze$ inside a uniform negative cloud of radius $R$ , the whole atom neutral.
$E_{\text{inside conductor}} = 0$	Electrostatic field inside the material of any conductor is zero; all excess charge resides on the outer surface — so the charge a metallic sphere holds does not depend on whether it is hollow or solid.
$\sigma_{\text{inner}} = \dfrac{-q}{4\pi r_1^2},\qquad \sigma_{\text{outer}} = \dfrac{Q+q}{4\pi r_2^2}$	Hollow conducting shell of inner radius $r_1$ , outer radius $r_2$ , carrying its own charge $Q$ , with a point charge $q$ at the centre of the cavity: charge $-q$ is induced on the inner surface and $Q+q$ appears on the outer surface. (For $-q$ at the centre, the inner surface carries $+q$ and the outer $Q-q$ .)
$E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{x^2}\ (r<x<R),\qquad E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q+Q}{x^2}\ (x>R)$	Small sphere of charge $q$ and radius $r$ inside a concentric shell of radius $R$ carrying $Q$ .

Short NotesThe chapter subtopic by subtopic, in the textbook's own order.10 subtopics

The whole chapter in brief

1.1 What electrostatics is

Electrostatics is the study of forces, fields and potentials due to charges at rest. Static means not changing with time.
Frictional electricity (a comb attracting paper, a spark from a sweater) is the everyday evidence: rubbing transfers electrons and leaves the two bodies oppositely charged.

1.2 Electric charge

Two kinds of charge exist. Like charges repel; unlike charges attract. The property distinguishing them is called polarity.
By Franklin's convention, glass rubbed with silk is positive; plastic rubbed with fur (and silk rubbed by glass) is negative.
A charged body attracts a light neutral body as well, by inducing charge on it — so attraction alone never proves the second body is charged, but repulsion is a sure test of charge.
Gold-leaf electroscope: a metal rod in a box with two thin gold leaves at the bottom. Charge given to the knob spreads onto the leaves, which repel and diverge; the divergence measures the amount of charge.
Charging by friction transfers electrons only. A body that loses electrons becomes positive; one that gains electrons becomes negative. No new charge is created.

1.3 Conductors and insulators

Conductors have charges (electrons in metals, ions in electrolytes) free to move; charge given to a conductor spreads at once over its whole surface. Metals, human and animal bodies and the earth are conductors.
Insulators (glass, porcelain, plastic, nylon, wood) resist the flow of charge; charge put on an insulator stays where it is put. Semiconductors are an intermediate third class.
A plastic comb gets electrified on rubbing but a metal spoon held in the hand does not, because charge on the metal leaks away through the body to the earth. Held by an insulating handle, the metal rod does show charging.
Charging by induction: bringing $+Q$ near an uncharged conductor draws a negative induced charge to the near face and leaves an equal positive charge on the far face. The net charge on the conductor stays zero — charge is only redistributed.

1.4 Basic properties of electric charge

Point charge: a charged body whose size is negligible compared with the distances involved, so all its charge is taken to be at one point.
Additivity: charges add algebraically, signs included; charge is a scalar. Total of $+1, +2, -3, +4, -5$ is $-1$ .
Conservation: the total charge of an isolated system is constant. Charged particles may be created or destroyed (a neutron decaying to a proton plus an electron), but always in pairs of equal and opposite charge.
Quantisation: $q = ne$ with $n$ an integer and $e = 1.6\times10^{-19}\,\text{C}$ . A charge of $3.5e$ is impossible. First suggested by Faraday's laws of electrolysis, demonstrated by Millikan in 1912.
Quantisation may be ignored at the macroscopic level: a charge of $1\,\mu\text{C}$ is about $10^{13}$ electronic charges, so the graininess is invisible. At the microscopic level, where charges are a few tens of $e$ , it cannot be ignored.

1.5 Coulomb's law

Statement: the electrostatic force between two point charges is directly proportional to the product of their magnitudes, inversely proportional to the square of the distance between them, and acts along the line joining them.
$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$ , with $\dfrac{1}{4\pi\varepsilon_0} = 9\times10^{9}\,\text{N m}^2\,\text{C}^{-2}$ and $\varepsilon_0 = 8.854\times10^{-12}\,\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$ .
Vector form $\vec{F}_{21} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{21}^2}\hat{r}_{21}$ covers both attraction and repulsion in one equation, and gives $\vec{F}_{12} = -\vec{F}_{21}$ , in agreement with Newton's third law.
Definition of the coulomb that follows: $1\,\text{C}$ is that charge which, placed $1\,\text{m}$ from an equal charge in vacuum, is repelled with a force of $9\times10^{9}\,\text{N}$ . A very large unit in practice.
Coulomb used a torsion balance, and got equal sub-multiples of an unknown charge by touching a charged sphere to an identical uncharged one, halving the charge each time.
In a medium of dielectric constant $K$ the force falls to $F/K$ ; inserting any insulating sheet between two charges reduces the force.

1.6 Forces between multiple charges

Principle of superposition: the force on any one charge is the vector sum of the forces exerted on it by every other charge, taken one pair at a time, each pairwise force being unaffected by the presence of the rest.
$\vec{F}_1 = \vec{F}_{12}+\vec{F}_{13}+\cdots+\vec{F}_{1n}$ ; add by the parallelogram law or by resolving into components.
Two equal perpendicular forces $F$ give a resultant $\sqrt{2}\,F$ at $45^\circ$ — the pattern behind most three-charge MCQs.

1.7 Electric field

Definition: the electric field at a point is the force experienced per unit positive test charge placed there, $\vec{E} = \displaystyle\lim_{q_0\to0}\dfrac{\vec{F}}{q_0}$ . The limit is taken so the test charge does not disturb the source charge.
$\vec{F} = q\vec{E}$ . On a positive charge the force is along $\vec{E}$ ; on a negative charge it is opposite to $\vec{E}$ .
Field of a point charge: $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\lvert q\rvert}{r^2}$ , radially outward for $+q$ , radially inward for $-q$ , with the same magnitude at all points of a sphere centred on the charge (spherical symmetry).
$\vec{E}$ is independent of the test charge, since $\vec{F}\propto q_0$ and the ratio does not depend on $q_0$ . It is a function of position and exists at every point of space.
Superposition holds for fields as it does for forces: $\vec{E} = \dfrac{1}{4\pi\varepsilon_0}\sum_i\dfrac{q_i}{r_{iP}^2}\hat{r}_{iP}$ .
SI unit $\text{N C}^{-1}$ , alternatively $\text{V m}^{-1}$ .

1.8 Electric field lines

An electric field line is a curve whose tangent at every point gives the direction of the net electric field there; an arrow fixes which of the two tangent directions is meant.
Magnitude is carried by the density of lines: lines crowd where the field is strong, spread out where it is weak. Only the relative density matters, since any number of lines may be drawn.
Because the field of a point charge falls as $1/r^2$ while the enclosing area grows as $r^2$ , the number of lines crossing a sphere around the charge is the same at every radius — the picture is self-consistent.
Properties: (i) lines start on positive charges and end on negative charges, beginning or ending at infinity for a single charge; (ii) in a charge-free region they are continuous curves with no breaks; (iii) two lines can never cross; (iv) they never form closed loops.
No crossing, because at the crossing point the field would have two directions at once, which is absurd.
No closed loops, because the electrostatic field is conservative — the work done round a closed path is zero, which a closed field line would violate.
In a uniform field the lines are equally spaced parallel straight lines.

1.9 Electric flux

Electric flux through a surface is the total number of electric field lines crossing it normally: $\Delta\phi = \vec{E}\cdot\Delta\vec{S} = E\,\Delta S\cos\theta$ , with $\theta$ the angle between $\vec{E}$ and the area vector.
Area is treated as a vector, $\Delta\vec{S} = \Delta S\,\hat{n}$ , directed along the normal to the surface; for a closed surface the outward normal is taken, by convention.
Total flux $\phi = \displaystyle\oint_S\vec{E}\cdot d\vec{S}$ . Flux is a scalar. SI unit $\text{N m}^2\,\text{C}^{-1}$ or V m.
$\phi$ is maximum ( $ES$ ) when the surface is held normal to $\vec{E}$ , and zero when the plane of the surface contains $\vec{E}$ ( $\theta = 90^\circ$ ).
A uniform field through a closed surface gives zero net flux: what enters one side leaves the other.

1.10 Electric dipole

An electric dipole is a pair of equal and opposite point charges $+q$ and $-q$ separated by a small distance $2a$ . Its total charge is zero but its field is not.
Dipole moment $\vec{p} = q(2a)\hat{p}$ , magnitude $2qa$ , directed from $-q$ to $+q$ along the dipole axis. Unit C m; it is a vector.
Axial field (short dipole): $\vec{E} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2\vec{p}}{r^3}$ , along $\vec{p}$ .
Equatorial field (short dipole): $\vec{E} = -\dfrac{1}{4\pi\varepsilon_0}\dfrac{\vec{p}}{r^3}$ , opposite to $\vec{p}$ .
The dipole field falls off as $1/r^3$ , faster than the $1/r^2$ of a point charge, because the two fields nearly cancel at large distance. It depends on $q$ and $a$ only through the product $qa$ .
The field of a dipole depends on the angle between $\vec{r}$ and $\vec{p}$ as well as on $r$ — it is not spherically symmetric.

1.11 Dipole in a uniform external field

In a uniform field the two forces $q\vec{E}$ and $-q\vec{E}$ are equal, opposite and non-collinear: net force zero, but they form a couple.
Torque $\tau = qE\times 2a\sin\theta = pE\sin\theta$ , i.e. $\vec{\tau} = \vec{p}\times\vec{E}$ , directed perpendicular to the plane containing $\vec{p}$ and $\vec{E}$ . It tends to align $\vec{p}$ along $\vec{E}$ .
$\tau = 0$ at $\theta = 0^\circ$ and $\theta = 180^\circ$ ; $\tau$ is maximum, $pE$ , at $\theta = 90^\circ$ .
Potential energy $U = -\vec{p}\cdot\vec{E} = -pE\cos\theta$ ; in stable equilibrium ( $\theta=0^\circ$ ) $U = -pE$ , in unstable equilibrium ( $\theta=180^\circ$ ) $U = +pE$ .
In a non-uniform field the forces on the two charges differ in magnitude, so there is a net force as well as a torque: the dipole both translates and rotates. If $\vec{p}$ is parallel to $\vec{E}$ the net force is towards increasing field; if antiparallel, towards decreasing field.
This is why a charged comb attracts a neutral scrap of paper: the comb polarises the paper (induces a dipole moment along its own field) and its field is non-uniform, so the induced dipole is pulled towards the comb.

1.12 Continuous charge distribution

Where charges are too many to list, use smoothed-out densities: $\lambda = \dfrac{\Delta Q}{\Delta l}$ ( $\text{C m}^{-1}$ ), $\sigma = \dfrac{\Delta Q}{\Delta S}$ ( $\text{C m}^{-2}$ ), $\rho = \dfrac{\Delta Q}{\Delta V}$ ( $\text{C m}^{-3}$ ).
The elements $\Delta l$ , $\Delta S$ , $\Delta V$ are small on the macroscopic scale but contain very many charges, so quantisation and microscopic discontinuity are averaged away.
The field is then found by superposing the fields of the elements: $\vec{E} = \dfrac{1}{4\pi\varepsilon_0}\displaystyle\sum\dfrac{\rho\,\Delta V}{r'^2}\hat{r}'$ , becoming an integral in the limit.

1.13 Gauss's law

Statement: the total electric flux through any closed surface is $\dfrac{1}{\varepsilon_0}$ times the net charge enclosed by it — $\displaystyle\oint_S\vec{E}\cdot d\vec{S} = \dfrac{q_{\text{enc}}}{\varepsilon_0}$ .
It holds for any closed surface, of any shape or size, and for any field, uniform or not.
$\vec{E}$ in the integral is the field of all charges, inside and outside; but only the charge inside appears on the right.
Zero net flux means zero net enclosed charge (a dipole inside a box gives zero flux), not necessarily zero field.
The chosen closed surface is called the Gaussian surface. It may be imaginary, but it must not pass through a discrete point charge, where the field is undefined.
Gauss's law follows from Coulomb's inverse-square law: for a sphere of radius $r$ around a point charge, $\phi = \dfrac{q}{4\pi\varepsilon_0 r^2}\times 4\pi r^2 = \dfrac{q}{\varepsilon_0}$ , independent of $r$ .

1.14 Applications of Gauss's law

Infinite straight wire, linear density $\lambda$ : coaxial cylindrical Gaussian surface; $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$ , radial, $E\propto\dfrac{1}{r}$ .
Infinite plane sheet, surface density $\sigma$ : pillbox Gaussian surface through the sheet; $E = \dfrac{\sigma}{2\varepsilon_0}$ , normal to the sheet and independent of distance from it.
Thin spherical shell, radius $R$ , total charge $q$ : concentric spherical Gaussian surface; $E = \dfrac{q}{4\pi\varepsilon_0 r^2}$ for $r\ge R$ , and $E = 0$ for $r<R$ . Outside, the shell acts as a point charge at its centre.
Solid sphere of uniform volume charge density: outside, the same as a point charge; inside, $E = \dfrac{\rho r}{3\varepsilon_0}$ , rising linearly with $r$ .
Conductors: the field inside the material of a conductor is zero, so all excess charge sits on its outer surface, and a hollow conductor shields its interior (Faraday cage — a car body is the everyday example).
Cavity with a charge: $q$ at the centre of a cavity in a conductor induces $-q$ on the inner surface and leaves $q$ (plus any charge already on the conductor) on the outer surface.

Comparisons CBSE asks for

Point of comparison	Coulomb (electrostatic) force	Gravitational force
Depends on	product of charges	product of masses
Sign	attractive or repulsive	always attractive
Medium	reduced by factor $K$	unaffected by medium
Relative strength	about $2.4\times10^{39}$ times stronger for a proton-electron pair	far weaker

Point of comparison	Axial point	Equatorial point
Exact field	$\dfrac{1}{4\pi\varepsilon_0}\dfrac{2pr}{(r^2-a^2)^2}$	$\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{(r^2+a^2)^{3/2}}$
Short dipole	$\dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3}$	$\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^3}$
Direction	along $\vec{p}$	antiparallel to $\vec{p}$
Ratio at equal $r$	$2$	$1$

Orientation	Stable equilibrium	Unstable equilibrium
Angle $\theta$ between $\vec{p}$ and $\vec{E}$	$0^\circ$ , $\vec{p}$ parallel to $\vec{E}$	$180^\circ$ , $\vec{p}$ antiparallel to $\vec{E}$
Torque	zero	zero
Potential energy	$-pE$ , minimum	$+pE$ , maximum
On small displacement	restoring torque returns it	torque turns it further away

Distribution	Field	Dependence on distance
Point charge	$\dfrac{q}{4\pi\varepsilon_0 r^2}$	$E\propto\dfrac{1}{r^2}$
Short dipole	$\dfrac{2p}{4\pi\varepsilon_0 r^3}$ on axis	$E\propto\dfrac{1}{r^3}$
Infinite line charge	$\dfrac{\lambda}{2\pi\varepsilon_0 r}$	$E\propto\dfrac{1}{r}$
Infinite plane sheet	$\dfrac{\sigma}{2\varepsilon_0}$	independent of distance
Spherical shell, outside	$\dfrac{q}{4\pi\varepsilon_0 r^2}$	$E\propto\dfrac{1}{r^2}$
Spherical shell, inside	$0$	zero everywhere

Behaviour	Uniform field	Non-uniform field
Net force on dipole	zero	non-zero
Torque on dipole	$pE\sin\theta$	generally non-zero
Resulting motion	rotation only	translation and rotation

Graphs to be able to draw

Field of an infinite line charge: $E$ against $r$ is a rectangular hyperbola, $E\propto1/r$ , falling steeply and tending to zero.
Field of a spherical shell: $E = 0$ from the centre out to $r = R$ , then a vertical jump to $\dfrac{\sigma}{\varepsilon_0}$ at $r = R$ , then a $1/r^2$ decay for $r>R$ .
Field of a uniformly charged solid sphere: a straight line rising from the origin, $E\propto r$ , up to $r = R$ , then a $1/r^2$ decay.
Field of a short dipole: $E$ against $r$ falls as $1/r^3$ — steeper than the point-charge curve drawn on the same axes.
Field of an infinite plane sheet: a horizontal line, $E$ constant on both sides.

Standard diagrams described in words

Field lines of a single $+q$ : straight lines radiating outward in all directions; for $-q$ , the same lines directed inward.
Two equal positive charges: lines curve away from each other, with a neutral point midway on the line joining them where no line passes.
Dipole ( $+q$ and $-q$ ): lines leave $+q$ , curve round and terminate on $-q$ , densest along the axis between the charges; the equatorial plane carries lines running from the $+q$ side to the $-q$ side.
Point charge $+q$ near an uncharged conducting plate: lines radiate from $+q$ and bend so as to strike the near face of the plate at right angles, that face carrying an induced negative charge. For $-q$ the lines run in the reverse sense, into the charge, leaving induced positive charge on the near face. Field lines always meet a conductor's surface normally.
Gaussian surface for a line charge: a cylinder of radius $r$ and length $l$ drawn coaxially with the wire — flux only through the curved side, area $2\pi rl$ .
Gaussian surface for a plane sheet: a cylinder or box pierced through the sheet, with its two flat faces of area $A$ parallel to the sheet, one on each side — flux $EA$ through each face, none through the curved side.
Gaussian surface for a shell: a sphere concentric with the shell, of radius $r$ greater than $R$ for the outside field and less than $R$ for the inside field.

Traps that cost marks

The angle in $\phi = ES\cos\theta$ is between $\vec{E}$ and the normal to the surface. If the question gives the angle the plane makes with the field, use $90^\circ$ minus that angle. A plane at $30^\circ$ to $\vec{E}$ means $\theta = 60^\circ$ .
Flux is a scalar. Never quote a direction for it, only a sign.
Only enclosed charge counts in Gauss's law; a charge just outside the surface contributes nothing to the flux even though it does contribute to $\vec{E}$ on the surface.
Increasing the size of a Gaussian surface leaves the flux unchanged, provided the enclosed charge is unchanged.
Zero flux means zero net enclosed charge — it does not mean the field inside is zero.
$\vec{p}$ points from $-q$ to $+q$ , the opposite sense to the field between the charges. Marking it the other way loses the direction mark.
Use the exact dipole formulas unless the question says short dipole or $r\gg a$ ; state the approximation explicitly when you drop $a^2$ .
The equatorial field is antiparallel to $\vec{p}$ . Writing it along $\vec{p}$ loses a mark even if the magnitude is right.
A dipole in a uniform field has zero net force. Writing "force $= pE$ " is a standard error.
After two identical conductors touch, each carries $\dfrac{q_1+q_2}{2}$ — with signs. $-Q$ and $+3Q$ become $+Q$ each, and the force becomes repulsive.
Charge is quantised: any option offering $3.5e$ is impossible, however well the total adds up.
$E = \dfrac{\sigma}{2\varepsilon_0}$ for a single sheet but $\dfrac{\sigma}{\varepsilon_0}$ between two oppositely charged sheets. Keep the factor of two straight.
For several sheets, add the contributions as signed vectors: each positive sheet pushes away from itself on both sides, each negative sheet pulls towards itself on both sides.
Inside a conductor $E = 0$ , but inside a cavity that contains a charge the field is not zero.
Attraction does not prove a body is charged — a neutral body is attracted by induction. Only repulsion is a sure test.
Field lines never cross and never close on themselves; they must also be drawn perpendicular to any conducting surface.
Convert centimetres to metres before substituting: an area of $10\,\text{cm}\times10\,\text{cm}$ is $10^{-2}\,\text{m}^2$ , not $100$ .

Solved Numerical ExamplesWorked numericals to attempt first, each with a hint before the full solution.Coming soon

These are being added chapter by chapter. They are solved numericals from the prescribed book and the reference guides — not previous board questions, which live under Practice — so that you can work through the method before meeting it in a real paper.

Important DerivationsEvery derive / obtain / prove question in this chapter.19

1
Derivation 1
A thin circular ring of radius $r$ is charged uniformly so that its linear charge density becomes $\lambda$ . Derive an expression for the electric field at a point $P$ at a distance $x$ from it along the axis of the ring. Hence, prove that at large distances $(x \gg r)$ , the ring behaves as a point charge.
Asked in: CBSE Class 12 Physics — Long Answer · 3/5, 2020, Ev · Section E, 5 marks
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Consider an elementary charge $dq = \lambda\,dl$ on the ring, where $dl$ is an elementary length. The distance from this element to the axial point $P$ is $\sqrt{r^2+x^2}$ . The field due to $dq$ at $P$ is $dE = \dfrac{dq}{4\pi\varepsilon_0(r^2+x^2)}$ , resolved into $dE_x$ (along the axis) and $dE_y$ (perpendicular to it). By symmetry, the perpendicular components cancel around the ring and only the axial components add: $dE_x = dE\cos\theta = \dfrac{\lambda\,x\,dl}{4\pi\varepsilon_0(r^2+x^2)^{3/2}}$ $\therefore\ E_x = \displaystyle\int dE\cos\theta = \int\dfrac{\lambda\,x\,dl}{4\pi\varepsilon_0(r^2+x^2)^{3/2}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{x\lambda l}{(r^2+x^2)^{3/2}}$ where $l = 2\pi r$ is the circumference (total charge $q=\lambda l$ ), so $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{xq}{(r^2+x^2)^{3/2}}$ When $x \gg r$ : $(r^2+x^2)^{3/2} \approx x^3$ , so $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{xq}{x^3} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{x^2}$ which is exactly the field of a point charge $q$ at distance $x$ — hence at large distances, the ring behaves as a point charge.
2
Derivation 2
Consider a system of $n$ charges $q_1, q_2, \ldots, q_n$ with position vectors $\vec{r}_1, \vec{r}_2, \vec{r}_3, \ldots, \vec{r}_n$ relative to some origin $O$ . Deduce the expression for the net electric field $\vec{E}$ at a point $P$ with position vector $\vec{r}_P$ due to this system of charges.
Asked in: CBSE Class 12 Physics — Long Answer · 3/5, Foreign 2015 · Section E, 5 marks
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According to Coulomb's law, the force on a test charge $q_0$ placed at $P$ due to charge $q_1$ is $\vec{F}_1 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_0}{r_{1P}^2}\hat{r}_{1P}$ where $\hat{r}_{1P}$ is the unit vector from $q_1$ to $P$ and $r_{1P}$ is the distance between them. The electric field at $P$ due to $q_1$ is $\vec{E}_1 = \dfrac{\vec{F}_1}{q_0} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1}{r_{1P}^2}\hat{r}_{1P}$ Similarly, the field due to $q_2$ is $\vec{E}_2 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_2}{r_{2P}^2}\hat{r}_{2P}$ , and so on. By the principle of superposition, the net field at $P$ is $\vec{E} = \vec{E}_1+\vec{E}_2+\ldots+\vec{E}_n = \dfrac{1}{4\pi\varepsilon_0}\displaystyle\sum_{i=1}^{n}\dfrac{q_i}{r_{iP}^2}\hat{r}_{iP}$
3
Derivation 3
Obtain the expression for the electric field at a point on the equatorial line of an electric dipole.
Asked in: CBSE Class 12 Physics — SA · 2019C · Section B, 2 marks
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The strength of an electric dipole is measured by its dipole moment, $\vec{p} = q(2a)$ , directed from the negative to the positive charge along the line joining them ( $2a$ = separation). Consider a point $P$ on the perpendicular bisector (equatorial line), at distance $r$ from the centre. The fields $E_{PA}$ (due to $-q$ ) and $E_{PB}$ (due to $+q$ ) are equal in magnitude ( $E_{PA}=E_{PB}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2+a^2}$ ); their components perpendicular to $\vec{p}$ cancel, and the components along $-\vec{p}$ add: $E_{net} = 2E_{PA}\cos\theta$ , where $\cos\theta = \dfrac{a}{(r^2+a^2)^{1/2}}$ $E_{net} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q\cdot2a}{(r^2+a^2)^{3/2}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{(r^2+a^2)^{3/2}}$ directed antiparallel to $\vec{p}$ . For a short dipole ( $r \gg a$ ): $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^3}$ , or in vector form, $\vec{E} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{-\vec{p}}{r^3}$
4
Derivation 4
Derive an expression for the electric field due to a dipole of dipole moment $\vec{p}$ at a point on its perpendicular bisector (equatorial line).
Asked in: CBSE Class 12 Physics — SA · 2/3, Delhi 2019, 2017 · Section C, 3 marks
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(Same derivation as Q26.) Electric field on the equatorial line at any point on the perpendicular bisector of the dipole, at distance $r$ from its centre: $E_{net} = E_{PA}\cos\theta+E_{PB}\cos\theta = 2E_{PA}\cos\theta$ ( $E_{PA}=E_{PB}$ , vertical components cancel) $E_{net} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q\cdot2a}{(r^2+a^2)^{3/2}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{(r^2+a^2)^{3/2}}$ directed antiparallel to $\vec{p}$ . For a short dipole $(r \gg a)$ : $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^3}$
5
Derivation 5
(a) Derive an expression for the electric field $E$ due to a dipole of length ' $2a$ ' at a point distant $r$ from the centre of the dipole on the axial line. (b) Draw a graph of $E$ versus $r$ for $r \gg a$ .
Asked in: CBSE Class 12 Physics — Long Answer · 3/5, AI 2017, An · Section E, 5 marks
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(a) Consider a dipole with charges $+q$ (at distance $r-a$ ) and $-q$ (at distance $r+a$ ) from an axial point $P$ , with dipole moment $\vec{p}$ pointing from $-q$ to $+q$ . Field due to $-q$ at $P$ : $\vec{E}_{-q} = \dfrac{-q}{4\pi\varepsilon_0(r+a)^2}\hat{p}$ (towards $-q$ , i.e. left) Field due to $+q$ at $P$ : $\vec{E}_{+q} = \dfrac{q}{4\pi\varepsilon_0(r-a)^2}\hat{p}$ (towards right) Resultant field: $\vec{E}_{axial} = \vec{E}_{+q}+\vec{E}_{-q} = \dfrac{q}{4\pi\varepsilon_0}\left[\dfrac{1}{(r-a)^2}-\dfrac{1}{(r+a)^2}\right]\hat{p}$ $= \dfrac{q}{4\pi\varepsilon_0}\cdot\dfrac{4ar}{(r^2-a^2)^2}\hat{p} = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2pr}{(r^2-a^2)^2}\hat{p}$ (using $p=q\cdot2a$ ) For $r \gg a$ , $a^2$ can be neglected compared to $r^2$ : $\vec{E}_{axial} = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2\vec{p}}{r^3}$ (directed along $+\hat{p}$ ) (b) $E$ falls off as $1/r^3$ : a steep monotonically decreasing curve of $E$ against $r$ , approaching zero as $r$ increases.
6
Derivation 6
An electric dipole of dipole moment $\vec{p}$ consists of point charges $+q$ and $-q$ separated by a distance $2a$ apart. Deduce the expression for the electric field $\vec{E}$ due to the dipole at a distance $x$ from the centre of the dipole on its axial line, in terms of the dipole moment $\vec{p}$ . Hence show that in the limit $x \gg a$ , $\vec{E} \to 2\vec{p}/(4\pi\varepsilon_0 x^3)$ .
Asked in: CBSE Class 12 Physics — Long Answer · 3/5, Delhi 2015, Ev · Section E, 5 marks
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(Same derivation as Q28, with $r$ replaced by $x$ .) Field at an axial point at distance $x$ from the centre: $\vec{E}_{axial} = \dfrac{q}{4\pi\varepsilon_0}\left[\dfrac{1}{(x-a)^2}-\dfrac{1}{(x+a)^2}\right]\hat{p} = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2px}{(x^2-a^2)^2}\hat{p}$ For $x \gg a$ , $a^2$ is negligible compared to $x^2$ : $\vec{E} \to \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2\vec{p}}{x^3}$ — as required.
7
Derivation 7
Derive the expression for the torque acting on an electric dipole, when it is held in a uniform electric field. Identify the orientation of the dipole in the electric field, in which it attains a stable equilibrium.
Asked in: CBSE Class 12 Physics — SA · 2020 · Section B, 2 marks
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When an electric dipole ( $+q$ at $A$ , $-q$ at $B$ , separation $2a$ ) is placed in a uniform field $\vec{E}$ making angle $\theta$ with $\vec{p}$ , the two charges experience equal and opposite forces $qE$ , which cancel — so the net force is zero. However, these forces are not collinear, so they constitute a couple. Torque $=$ magnitude of either force $\times$ perpendicular distance between the lines of action $= qE\times(2a\sin\theta) = (q\cdot2a)E\sin\theta$ $\therefore\ \tau = pE\sin\theta$ , or in vector form, $\vec{\tau} = \vec{p}\times\vec{E}$ Stable equilibrium: when $\theta = 0°$ (i.e., $\vec{p}$ parallel to $\vec{E}$ ), $\tau = 0$ and any small angular displacement produces a restoring torque bringing the dipole back — this is the position of stable equilibrium.
8
Derivation 8
(a) Obtain the expression for the torque $\vec{\tau}$ experienced by an electric dipole of dipole moment $\vec{p}$ in a uniform electric field $\vec{E}$ . (b) What will happen if the field were not uniform?
Asked in: CBSE Class 12 Physics — Long Answer · Delhi 2017 · Section E, 5 marks
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(a) (Same derivation as Q33.) When a dipole is placed in a uniform field, the two equal and opposite forces $qE$ on its charges cancel (net force zero), but being non-collinear they produce a torque: $\tau = Fr_\perp = qE\cdot2a\sin\theta = (q\cdot2a)E\sin\theta$ , i.e., $\tau = pE\sin\theta$ , or $\vec{\tau}=\vec{p}\times\vec{E}$ — this torque tends to align the dipole along the field direction. (b) If the field is not uniform, the forces on the two charges no longer cancel exactly, so the dipole experiences a net force in addition to the torque, and undergoes both translational and rotational motion.
9
Derivation 9
An electric dipole of dipole moment $\vec{p}$ is placed in a uniform electric field $\vec{E}$ . Obtain the expression for the torque $\vec{\tau}$ experienced by the dipole. Identify two pairs of perpendicular vectors in the expression.
Asked in: CBSE Class 12 Physics — Long Answer · Delhi 2015C, Cr · Section E, 5 marks
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(Same derivation as Q33/36.) $\tau = pE\sin\theta$ , or in vector form, $\vec{\tau} = \vec{p}\times\vec{E}$ . Since $\vec{\tau}$ is the cross product of $\vec{p}$ and $\vec{E}$ , it is perpendicular to both. The two pairs of perpendicular vectors are: (i) $(\vec{\tau},\vec{p})$ and (ii) $(\vec{\tau},\vec{E})$ .
10
Derivation 10
(a) Define electric flux. Is it a scalar or a vector quantity? A point charge $q$ is at a distance of $d/2$ directly above the centre of a square of side $d$ . Use Gauss's law to obtain the expression for the electric flux through the square. (b) If the point charge is now moved to a distance ' $d$ ' from the centre of the square and the side of the square is doubled, explain how the electric flux will be affected.
Asked in: CBSE Class 12 Physics — Long Answer · 2018, An · Section E, 5 marks
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(a) Electric flux linked with a surface is the number of electric lines of force cutting through it normally; it is a scalar quantity: $\phi = \displaystyle\int_S\vec{E}\cdot d\vec{S}$ . Imagine the given square as one face of a cube of edge $d$ . Since the charge $q$ is at distance $d/2$ directly above the centre of the square, it lies at the centre of this imaginary cube, so by Gauss's theorem, the electric flux linked with the whole cube is $\phi = \dfrac{q}{\varepsilon_0}$ . By symmetry, this flux is divided equally among the cube's 6 faces, so the flux through the given square is $\phi_{sq} = \dfrac{\phi}{6} = \dfrac{q}{6\varepsilon_0}$ . (b) Now the distance of the charge from the plane doubles ( $d$ ) and the side of the square also doubles ( $2d$ ), so the same symmetry holds with a cube of side $2d$ — the charge again sits at the centre of an (enlarged) imaginary cube. Hence the flux through the square is unaffected: $\phi_{sq} = \dfrac{q}{6\varepsilon_0}$ , unchanged.
11
Derivation 11
Consider two hollow concentric spheres, $S_1$ and $S_2$ , enclosing charges $2Q$ and $4Q$ respectively as shown in the figure. (i) Find the ratio of the electric flux through them. (ii) How will the electric flux through the sphere $S_1$ change if a medium of dielectric constant ' $\varepsilon_r$ ' is introduced in the space inside $S_1$ in place of air? Deduce the necessary expression.
Asked in: CBSE Class 12 Physics — Long Answer · AI 2014, Ap · Section E, 5 marks
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(i) Charge enclosed by $S_1$ is $2Q$ , so flux $\phi_1 = \dfrac{2Q}{\varepsilon_0}$ . Charge enclosed by $S_2$ is $2Q+4Q=6Q$ , so flux $\phi_2 = \dfrac{6Q}{\varepsilon_0}$ . Ratio, $\phi_1:\phi_2 = 2:6 = 1:3$ (ii) When a medium of dielectric constant $\varepsilon_r$ is introduced inside $S_1$ , the flux through $S_1$ becomes $\phi_1' = \dfrac{2Q}{\varepsilon_0\varepsilon_r}$ .
12
Derivation 12
Apply Gauss's law to show that for a charged spherical shell, the electric field outside the shell is as if the entire charge were concentrated at the centre.
Asked in: CBSE Class 12 Physics — SA · AI 2019, R · Section B, 2 marks
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Consider a spherical Gaussian surface of radius $r$ ( $r>R$ , the shell's radius), concentric with the shell. By symmetry, $\vec{E}$ is radial and has the same magnitude at every point of this surface. Using Gauss's theorem at point $P$ on this surface: $\displaystyle\oint\vec{E}\cdot d\vec{s} = \dfrac{Q}{\varepsilon_0}$ $\displaystyle\int E\,4\pi r^2\cos0° = \dfrac{Q}{\varepsilon_0}$ ( $E$ is constant over the surface) $E = \dfrac{Q}{4\pi\varepsilon_0 r^2}$ This is exactly the field of a point charge $Q$ placed at the centre — hence the field outside a uniformly charged spherical shell is as if the entire charge were concentrated at its centre.
13
Derivation 13
(a) An infinitely long thin straight wire has a uniform linear charge density $\lambda$ . Obtain the expression for the electric field $(E)$ at a point lying at a distance $x$ from the wire, using Gauss' law. (b) Show graphically the variation of this electric field $E$ as a function of distance $x$ from the wire.
Asked in: CBSE Class 12 Physics — Long Answer · 2020 · Section E, 5 marks
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(a) Consider a cylindrical Gaussian surface $S$ of radius $x$ and length $l$ , coaxial with the charged wire. The flux through the two flat circular ends is zero (field is perpendicular to their normal), and through the curved surface: $\phi = \displaystyle\oint\vec{E}\cdot d\vec{S} = \int E\,dS\cos90°+\int E\,dS\cos0°+\int E\,dS\cos90° = E\cdot2\pi xl$ By Gauss's theorem, $\phi = \dfrac{q}{\varepsilon_0} = \dfrac{\lambda l}{\varepsilon_0}$ (charge enclosed on length $l$ is $\lambda l$ ). $\therefore\ E\times2\pi xl = \dfrac{\lambda l}{\varepsilon_0} \ \Rightarrow\ E = \dfrac{\lambda}{2\pi\varepsilon_0 x}$ , directed radially outward from the wire. (b) Since $E \propto \dfrac{1}{x}$ , the graph of $E$ vs $x$ is a rectangular-hyperbola-like curve, steeply decreasing and approaching zero as $x$ increases.
14
Derivation 14
A hollow conducting sphere of inner radius $r_1$ and outer radius $r_2$ has a charge $Q$ on its surface. A point charge $-q$ is also placed at the centre of the sphere. (a) What is the surface charge density on the (i) inner and (ii) outer surface of the sphere? (b) Use Gauss' law of electrostatics to obtain the expression for the electric field at a point lying outside the sphere.
Asked in: CBSE Class 12 Physics — Long Answer · AI 2020 · Section E, 5 marks
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(a) By induction, charge $+q$ is induced on the inner surface (to neutralise the field inside the conductor due to $-q$ at the centre), so the inner surface charge is $+q$ : $\sigma_{inner} = \dfrac{q}{4\pi r_1^2}$ Charge conservation: total charge on the conductor is $Q$ , so the outer surface carries $(Q-q)$ : $\sigma_{outer} = \dfrac{Q-q}{4\pi r_2^2}$ (b) For a point outside the sphere at distance $x>r_2$ , using a spherical Gaussian surface, the total charge enclosed is $(Q-q)+q-q = Q-q$ ... more directly, the net charge enclosed is $(-q)+Q = Q-q$ : $\displaystyle\oint\vec{E}\cdot d\vec{S} = \dfrac{Q-q}{\varepsilon_0} \ \Rightarrow\ E\times4\pi x^2 = \dfrac{Q-q}{\varepsilon_0}$ $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q-q}{x^2}$
15
Derivation 15
State Gauss's law on electrostatics and derive an expression for the electric field due to a long straight thin uniformly charged wire (linear charge density $\lambda$ ) at a point lying at a distance $r$ from the wire.
Asked in: CBSE Class 12 Physics — Long Answer · 3/5, 2020 · Section E, 5 marks
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Gauss's law: the total flux over a closed surface $S$ in vacuum is $\dfrac{1}{\varepsilon_0}$ times the total charge enclosed by that surface: $\phi = \displaystyle\oint_S\vec{E}\cdot d\vec{S} = \dfrac{q_{enclosed}}{\varepsilon_0}$ (Derivation, same as Q57(a).) Using a coaxial cylindrical Gaussian surface of radius $r$ and length $l$ : flux through the flat ends is zero, and through the curved surface $\phi = E\times2\pi rl$ . By Gauss's theorem, $\phi = \dfrac{\lambda l}{\varepsilon_0}$ , so $E\times2\pi rl = \dfrac{\lambda l}{\varepsilon_0} \ \Rightarrow\ E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$ , directed radially outward (for $\lambda>0$ ).
16
Derivation 16
Using Gauss law, derive an expression for electric field due to a spherical shell of uniform charge distribution $\sigma$ and radius $R$ at a point lying at a distance $x$ from the centre of the shell, such that (i) $0<x<R$ , and (ii) $x>R$ . OR Use Gauss's law to show that due to a uniformly charged spherical shell of radius $R$ , the electric field at any point situated outside the shell at a distance $r$ from its centre is equal to the electric field at the same point when the entire charge on the shell were concentrated at its centre. Also plot the graph showing the variation of electric field with $r$ , for $r\le R$ and $r\ge R$ .
Asked in: CBSE Class 12 Physics — Long Answer · 3/5, 2020 · Section E, 5 marks
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(i) For a point inside the shell ( $x<R$ ), a spherical Gaussian surface of radius $x$ encloses no charge (all the charge $q=\sigma\times4\pi R^2$ resides on the shell of radius $R>x$ ). By Gauss's law, $E\times4\pi x^2 = 0 \ \Rightarrow\ E=0$ . (ii) For a point outside the shell ( $x>R$ ), take a concentric spherical Gaussian surface of radius $x$ . By symmetry $\vec{E}$ is radial and uniform over this surface: $\displaystyle\oint\vec{E}\cdot d\vec{S} = E\times4\pi x^2$ Charge enclosed $= \sigma\times4\pi R^2 = q$ (total charge on shell), so by Gauss's law: $E\times4\pi x^2 = \dfrac{\sigma\times4\pi R^2}{\varepsilon_0} = \dfrac{q}{\varepsilon_0}$ $E = \dfrac{q}{4\pi\varepsilon_0 x^2}$ , or vectorially, $\vec{E} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{x^2}\hat{r}$ — identical to the field of the entire charge $q$ concentrated at the centre, confirming the required result. The graph of $E$ vs $r$ is zero for $r<R$ , and falls off as $E\propto1/r^2$ for $r>R$ (with a discontinuous jump to $\sigma/\varepsilon_0$ exactly at $r=R$ ).
17
Derivation 17
(a) Use Gauss' law to derive the expression for the electric field $(\vec{E})$ due to a straight uniformly charged infinite line of charge density $\lambda\,\text{C m}^{-1}$ . (b) Draw a graph to show the variation of $E$ with perpendicular distance $r$ from the line of charge. (c) Find the work done in bringing a charge $q$ from perpendicular distance $r_1$ to $r_2\,(r_2>r_1)$ .
Asked in: CBSE Class 12 Physics — Long Answer · 2018 · Section E, 5 marks
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(a),(b) (Same derivation as Q57/60.) $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$ , directed radially outward; $E \propto 1/r$ , so the graph is a steadily decreasing curve. (c) Work done in moving charge $q$ from $r_2$ to $r_1$ (against/along the field): $W = \displaystyle\int_{r_2}^{r_1}\vec{F}\cdot d\vec{r} = \int_{r_2}^{r_1}\dfrac{q\lambda}{2\pi\varepsilon_0 r}dr\cos0° = \dfrac{q\lambda}{2\pi\varepsilon_0}\int_{r_2}^{r_1}\dfrac{dr}{r}$ $= \dfrac{q\lambda}{2\pi\varepsilon_0}[\ln r]_{r_2}^{r_1} = \dfrac{q\lambda}{2\pi\varepsilon_0}\ln\dfrac{r_1}{r_2}$
18
Derivation 18
Use Gauss's law to prove that the electric field inside a uniformly charged spherical shell is zero.
Asked in: CBSE Class 12 Physics — Long Answer · 3/5, AI 2015 · Section E, 5 marks
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(Same as the $r<R$ case of Q61.) For a point $P$ inside the shell, take a concentric spherical Gaussian surface of radius $r<R$ through $P$ . Since all the charge $q$ resides on the shell of radius $R$ (outside this Gaussian surface), the charge enclosed by this surface is zero. By Gauss's law, $E\times4\pi r^2 = \dfrac{0}{\varepsilon_0} = 0 \ \Rightarrow\ E=0$ Hence the electric field at every point inside a uniformly charged spherical shell is zero.
19
Derivation 19
A small conducting sphere of radius ' $r$ ' carrying a charge $+q$ is surrounded by a large concentric conducting shell of radius $R$ on which a charge $+Q$ is placed. Using Gauss's law, derive the expressions for the electric field at a point ' $x$ ' (i) between the sphere and the shell $(r<x<R)$ , and (ii) outside the spherical shell.
Asked in: CBSE Class 12 Physics — Long Answer · 3/5, Foreign 2015, An · Section E, 5 marks
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Consider a Gaussian spherical surface of radius $x$ , centred at $O$ . Total flux through this surface: $\phi = \displaystyle\oint E\,ds = E\oint ds = E\times4\pi x^2$ ...(i) (since $\oint ds = 4\pi x^2$ ) By Gauss's theorem, $\phi = \dfrac{q_{enclosed}}{\varepsilon_0}$ ...(ii) (i) For $r<x<R$ : the Gaussian surface encloses only the inner sphere's charge, $q_{enclosed}=q$ . From (i) and (ii): $E\times4\pi x^2 = \dfrac{q}{\varepsilon_0} \ \Rightarrow\ E = \dfrac{q}{4\pi\varepsilon_0 x^2}$ (ii) For $x>R$ (outside the shell): the Gaussian surface encloses both charges, $q_{enclosed}=q+Q$ . From (i) and (ii): $E\times4\pi x^2 = \dfrac{q+Q}{\varepsilon_0} \ \Rightarrow\ E = \dfrac{q+Q}{4\pi\varepsilon_0 x^2}$

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