PhysicsISC Class 12 · Chapter 1

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

Q: Do I have to derive the dipole field for ISC 2027, or only state it?

Derive it. The ISC 2027 Physics syllabus asks for the derivation of E on the axis (end-on) and on the perpendicular bisector (broadside-on) of a dipole, and the short-dipole forms. The torque \tau = \vec{p} \times \vec{E} must also be derived. (The 'qualitative only' note applies to the magnetic dipole in Chapter 5, not to the electric dipole.)

Q: Is the proof of Gauss's theorem required?

No. You state it, explain flux and the Gaussian surface, and apply it to an infinite line charge, an infinite thin plane sheet and a thin spherical shell (inside, on the surface and outside).

Q: Do I need position vectors $\vec{r}_1$ and $\vec{r}_2$ in Coulomb's law?

No. The syllabus says Coulomb's law in vector form is required, but the position coordinates \vec{r}_1, \vec{r}_2 are not necessary. Use the unit vector along the line joining the charges.

Q: How many marks does this chapter carry?

Electrostatics and Current Electricity together carry 14 of the 70 theory marks. Expect a mix of one-mark items, a 2-mark reasoning or numerical item and often a 3-mark derivation from this chapter.

Q: What is the difference between a thin charged sheet and a charged conducting plate?

For a thin sheet with density σ, Gauss's theorem gives \sigma/2\varepsilon_0 on each side. Two oppositely charged sheets (as in a capacitor) give \sigma/\varepsilon_0 between them. Keep to the thin-sheet result unless the question describes a pair of plates.

Q: Which sign convention is used for the dipole field direction?

On the axis the field points along \vec{p} (from −q to +q); on the equatorial line it points opposite to \vec{p}. State the direction with the magnitude in every answer.

Updated 14 October 2026 · Fact-checked

Electric Charges and Fields is the first chapter of ISC Class 12 Physics. It covers quantisation and conservation of charge, Coulomb's law in vector form, superposition, field lines, the electric dipole (axial and equatorial field, torque) and Gauss's theorem, applied to an infinite line charge, an infinite plane sheet and a thin spherical shell.

What Electric Charges and Fields is about

Electrostatics is the study of charges at rest. This chapter starts with what charge is: two kinds, positive and negative; charge is conserved and quantised ( $q = n e$ ). Coulomb's law gives the force between two point charges, and the superposition principle lets you add the forces from several charges as vectors. You are expected to compare Coulomb's law with Newton's law of gravitation and to know how a dielectric medium reduces the force (relative permittivity $ε_{r} = K$ ).

The electric field $E = F / q_{0}$ describes what a charge does to the space around it, and lines of force picture it. The ISC syllabus asks for the field lines of an isolated positive and negative charge, a dipole, two like charges close together and the uniform field between oppositely charged parallel plates.

An electric dipole is a pair of equal and opposite charges a small distance apart. You must derive the field on the axis (end-on position) and on the perpendicular bisector (broadside-on position), reduce both to the short-dipole forms for $r ≫ 2 l$ , and derive the torque $τ = p \times E$ in a uniform field.

The last part is electric flux and Gauss's theorem, $\oint E \cdot d A = q / ε_{0}$ . You state it, list the properties of a good Gaussian surface, and use it for three shapes: an infinite line of charge, an infinite thin plane sheet and a thin hollow spherical shell (inside, on the surface and outside), including the graph of E against r for the shell.

This chapter and the next two (potential and capacitance; current electricity) together carry 14 of the 70 theory marks. Derivations from this chapter are regular 3-mark items in Section C, and Gauss's-theorem reasoning turns up in short answers.

Weightage: Electric Charges and Fields is part of the unit Electrostatics; Current Electricity, which carries 14 of the 70 marks in the ISC Physics Paper 1 (Theory), shared with 2 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Physics syllabus, examination year 2027, distribution of marks

Key concepts

Charge: two kinds, conservation and quantisation: Rubbing two bodies transfers electrons from one to the other: the body that gains electrons becomes negative, the one that loses them becomes positive. No charge is created, so the total charge of an isolated system stays constant (conservation). Charge also comes only in whole-number multiples of the electronic charge $e = 1.6 \times 1 0^{- 19} C$ (quantisation): $q = \pm n e$ . Because e is tiny, everyday charges look continuous. Conductors (metals) have free electrons that can move through them; insulators hold their electrons bound to atoms. The SI unit of charge, the coulomb, is defined from the ampere: $1 C = 1 A \times 1 s$ .; Worked example
A polythene strip rubbed with wool acquires a charge of $- 3.2 \times 1 0^{- 7} C$ . (a) How many electrons were transferred, and in which direction? (b) By how much does the mass of the strip change? ( $m_{e} = 9.1 \times 1 0^{- 31} k g$ )
Try it first, then show the solutionSolution
The strip is negative, so electrons moved from the wool to the strip.
$n = \frac{q}{e} = \frac{3.2 \times 1 0 ^{- 7}}{1.6 \times 1 0 ^{- 19}} = 2 \times 1 0^{12}$
Gain in mass $= n m_{e} = 2 \times 1 0^{12} \times 9.1 \times 1 0^{- 31} = 1.82 \times 1 0^{- 18} k g$
Answer: (a) $2 \times 1 0^{12}$ electrons, from the wool to the strip (b) mass increases by $1.82 \times 1 0^{- 18} k g$
Coulomb's law in vector form, and the effect of a medium: The force between two point charges is along the line joining them, proportional to the product of the charges and inversely proportional to the square of the distance: $F_{12} = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r ^{2}} \overset{r}{^}_{21}$ , where $\overset{r}{^}_{21}$ points from charge 1 to charge 2 and $F_{12}$ is the force on 2 due to 1. Like charges repel, unlike attract, and $F_{12} = - F_{21}$ . In air $\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} N m^{2} C^{- 2}$ and $ε_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2}$ . In a medium of permittivity $ε = ε_{r} ε_{0}$ the force falls to $F / ε_{r}$ . So relative permittivity (dielectric constant) $K = ε_{r} = F_{ai r} / F_{m e d i u m}$ : 'relative permittivity of water is 81' means the force between two charges in water is $\frac{1}{81}$ of the force between the same charges at the same distance in vacuum. Compared with gravitation: both are inverse-square and central, but the electric force can attract or repel, depends on the medium and is about $1 0^{36}$ times stronger between two protons.; Worked example
Charges $+ 2 μ C$ and $- 3 μ C$ are 30 cm apart in air. Find the force between them. What is the force if the same pair is placed, at the same distance, in a liquid of dielectric constant 4?
Try it first, then show the solutionSolution
$F = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r ^{2}} = \frac{9 \times 1 0 ^{9} \times 2 \times 1 0 ^{- 6} \times 3 \times 1 0 ^{- 6}}{( 0.30 ) ^{2}}$
$= \frac{54 \times 1 0 ^{- 3}}{0.09} = 0.6 N$ , attractive (unlike charges)
In the liquid: $F^{'} = \frac{F}{K} = \frac{0.6}{4} = 0.15 N$ , still attractive
Answer: 0.6 N attractive in air; 0.15 N attractive in the liquid
Superposition principle and continuous charge distributions: The force on one charge due to several others is the vector sum of the forces each would exert alone: $F_{1} = F_{12} + F_{13} + F_{14} + \dots$ . The same holds for fields. For charge spread continuously, define densities: linear $λ = d q / d l$ (C m⁻¹), surface $σ = d q / d A$ (C m⁻²) and volume $ρ = d q / d V$ (C m⁻³). Then the field is a sum (integral) of the fields of small elements $d q$ . Always draw each force with its direction before adding; use components or the parallelogram rule.; Worked example
Three equal charges of $+ 1 μ C$ sit at the corners of an equilateral triangle of side 10 cm in air. Find the net force on any one charge.
Try it first, then show the solutionSolution
Each pair: $F = \frac{9 \times 1 0 ^{9} \times ( 1 0 ^{- 6} ) ^{2}}{( 0.10 ) ^{2}} = 0.9 N$ , repulsive.
The two forces on one charge are at $6 0^{\circ}$ to each other. Resultant $= 2 F cos 3 0^{\circ} = 3 F$ .
$3 \times 0.9 = 1.56 N$ , along the bisector of the angle, pointing away from the triangle.
Answer: 1.56 N, directed outward along the angle bisector at that corner
Electric field, field of a point charge and lines of force: The electric field at a point is the force per unit positive test charge, $E = F / q_{0}$ (unit N C⁻¹ or V m⁻¹), measured with $q_{0}$ so small that it does not disturb the source. For a point charge q: $E = \frac{1}{4 π ε _{0}} \frac{q}{r ^{2}}$ , radially outward if q is positive and inward if negative. A charge q in a field experiences $F = q E$ . Lines of force are drawn so that the tangent at any point gives the direction of $E$ and their crowding shows its strength. Properties: they start on positive and end on negative charges; two lines never cross (the field has one direction at a point); they do not form closed loops; they meet a conductor's surface at right angles; between oppositely charged parallel plates they are parallel and equally spaced (uniform field). At a neutral point the fields of two charges cancel.; Worked example
Charges $+ 4 μ C$ and $+ 9 μ C$ are 50 cm apart. Where, on the line joining them, is the electric field zero?
Try it first, then show the solutionSolution
Like charges: the neutral point lies between them, nearer the smaller charge. Let it be x from the $4 μ C$ charge.
$\frac{k ( 4 )}{x ^{2}} = \frac{k ( 9 )}{( 0.5 - x ) ^{2}} \Rightarrow \frac{2}{x} = \frac{3}{0.5 - x}$
$1 - 2 x = 3 x \Rightarrow x = 0.2 m$
Answer: 20 cm from the $+ 4 μ C$ charge (30 cm from the $+ 9 μ C$ charge)
Electric dipole: dipole moment and field on the axis and equator: A dipole is $+ q$ and $- q$ separated by $2 l$ . Its dipole moment $p = q (2 l)$ points from $- q$ to $+ q$ (unit C m). Its net charge is zero, but its field is not. Axial (end-on) point at distance r from the centre: $E = \frac{1}{4 π ε _{0}} \frac{2 p r}{( r ^{2} - l ^{2} ) ^{2}}$ , along $p$ ; for a short dipole ( $r ≫ 2 l$ ) $E_{a x ia l} = \frac{1}{4 π ε _{0}} \frac{2 p}{r ^{3}}$ . Equatorial (broadside-on) point on the perpendicular bisector: $E = \frac{1}{4 π ε _{0}} \frac{p}{( r ^{2} + l ^{2} ) ^{3/2}}$ , opposite to $p$ ; for a short dipole $E_{e q} = \frac{1}{4 π ε _{0}} \frac{p}{r ^{3}}$ . So at equal distances $E_{a x ia l} : E_{e q} = 2 : 1$ , and a dipole field falls as $1/ r^{3}$ , faster than the $1/ r^{2}$ of a single charge.; Worked example
A dipole is made of $\pm 2 n C$ charges 4 mm apart. Find (a) its dipole moment, (b) the field at 20 cm from its centre on the axis and (c) on the equatorial line.
Try it first, then show the solutionSolution
$p = q \times 2 l = 2 \times 1 0^{- 9} \times 4 \times 1 0^{- 3} = 8 \times 1 0^{- 12} C m$
20 cm is much larger than 4 mm, so use the short-dipole forms.
Axial: $E = \frac{2 k p}{r ^{3}} = \frac{2 \times 9 \times 1 0 ^{9} \times 8 \times 1 0 ^{- 12}}{( 0.2 ) ^{3}} = \frac{0.144}{0.008} = 18 N C^{- 1}$ , along $p$
Equatorial: $E = \frac{k p}{r ^{3}} = 9 N C^{- 1}$ , opposite to $p$
Answer: (a) $8 \times 1 0^{- 12} C m$ (b) $18 N C^{- 1}$ along $p$ (c) $9 N C^{- 1}$ opposite to $p$
Dipole in a uniform field: zero net force, torque p × E: In a uniform field $E$ the charges feel $+ q E$ and $- q E$ : equal, opposite and along parallel lines, so the net force is zero but they form a couple. The perpendicular distance between the lines of action is $2 l sin θ$ , so the torque is $τ = q E \times 2 l sin θ = pE sin θ$ , or $τ = p \times E$ . It turns the dipole to line up with the field. $τ$ is maximum ( $pE$ ) at $θ = 9 0^{\circ}$ and zero at $0^{\circ}$ (stable) and $18 0^{\circ}$ (unstable). In a non-uniform field the two forces are unequal, so there is also a net force: this is why a charged comb attracts neutral bits of paper (induced dipoles in a non-uniform field).; Worked example
A dipole of moment $8 \times 1 0^{- 12} C m$ makes $3 0^{\circ}$ with a uniform field of $5 \times 1 0^{4} N C^{- 1}$ . Find the torque and the net force on it.
Try it first, then show the solutionSolution
$τ = pE sin θ = 8 \times 1 0^{- 12} \times 5 \times 1 0^{4} \times sin 3 0^{\circ}$
$= 4 \times 1 0^{- 7} \times 0.5 = 2 \times 1 0^{- 7} N m$
Net force is zero because the field is uniform.
Answer: Torque $2 \times 1 0^{- 7} N m$ (turning $p$ towards $E$ ); net force zero
Electric flux and Gauss's theorem: Flux measures how much field passes through a surface, just as $Q = v \cdot A$ measures the volume of liquid flowing per second through an area. For a uniform field, $ϕ_{E} = E \cdot A = E A cos θ$ , where θ is the angle between $E$ and the area vector (the normal). Special cases: $θ = 0^{\circ}$ gives EA, $9 0^{\circ}$ gives 0, $18 0^{\circ}$ gives $- E A$ . For a non-uniform field, $ϕ_{E} = \int E \cdot d A$ . Unit: N m² C⁻¹ (or V m). Gauss's theorem: the total flux through any closed surface is $\frac{1}{ε _{0}}$ times the net charge enclosed, $\oint E \cdot d A = \frac{q}{ε _{0}}$ . Charges outside contribute zero net flux; the size and shape of the surface do not matter. A good Gaussian surface is closed, passes through the point where E is wanted, and follows the symmetry so that E is constant in magnitude and either along or perpendicular to $d A$ on each part.; Worked example
A closed surface encloses a net charge of $8.85 n C$ . (a) Find the flux through it. (b) What happens to the flux if the surface is made twice as large? (c) What if the charge is replaced by a dipole? ( $ε_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2}$ )
Try it first, then show the solutionSolution
$ϕ = \frac{q}{ε _{0}} = \frac{8.85 \times 1 0 ^{- 9}}{8.85 \times 1 0 ^{- 12}} = 1000 N m^{2} C^{- 1}$
Flux depends only on the enclosed charge, so enlarging the surface leaves it at 1000 N m² C⁻¹.
A dipole has net charge $+ q - q = 0$ , so the flux becomes zero (though the field at points on the surface is not zero).
Answer: (a) $1.0 \times 1 0^{3} N m^{2} C^{- 1}$ (b) unchanged (c) zero
Applications of Gauss's theorem: line, sheet and thin shell: Infinite line charge (λ): Gaussian surface a coaxial cylinder of radius r and length l. Flux only through the curved surface: $E (2 π r l) = λ l / ε_{0}$ , so $E = \frac{λ}{2 π ε _{0} r}$ (falls as $1/ r$ , not $1/ r^{2}$ ). Infinite thin plane sheet (σ): a cylinder (pillbox) piercing the sheet, end faces of area A on each side. Flux only through the two ends: $2 E A = σ A / ε_{0}$ , so $E = \frac{σ}{2 ε _{0}}$ , independent of distance. Thin spherical shell (charge q, radius R): concentric spherical Gaussian surface of radius r. Outside ( $r > R$ ): $E = \frac{1}{4 π ε _{0}} \frac{q}{r ^{2}}$ , as if all the charge were at the centre. On the surface: $E = \frac{1}{4 π ε _{0}} \frac{q}{R ^{2}} = \frac{σ}{ε _{0}}$ . Inside ( $r < R$ ): no charge is enclosed, so $E = 0$ . The E-r graph is zero up to R, jumps to its maximum at R, then falls as $1/ r^{2}$ .; Worked example
A thin spherical shell of radius 10 cm carries $2 μ C$ . Find the field at (a) 5 cm, (b) 10 cm and (c) 20 cm from its centre.
Try it first, then show the solutionSolution
(a) 5 cm is inside: the Gaussian surface encloses no charge, so $E = 0$ .
(b) On the surface: $E = \frac{9 \times 1 0 ^{9} \times 2 \times 1 0 ^{- 6}}{( 0.10 ) ^{2}} = 1.8 \times 1 0^{6} N C^{- 1}$
(c) Outside: $E = \frac{9 \times 1 0 ^{9} \times 2 \times 1 0 ^{- 6}}{( 0.20 ) ^{2}} = 4.5 \times 1 0^{5} N C^{- 1}$
Answer: (a) 0 (b) $1.8 \times 1 0^{6} N C^{- 1}$ (c) $4.5 \times 1 0^{5} N C^{- 1}$ , all radially outward where non-zero

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.

Quantisation of charge

Suppose you shuffle across a nylon carpet on a dry day and feel a spark when you touch a steel door handle.

Suppose your body picked up $- 8 n C$ from the carpet. How many extra electrons is that, and could the charge ever be $- 8.08 \times 1 0^{- 19} C$ ?

Try it first, then show the solutionSolution

$n = \frac{8 \times 1 0 ^{- 9}}{1.6 \times 1 0 ^{- 19}} = 5 \times 1 0^{10}$ electrons
$\frac{8.08 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}} = 5.05$ , not a whole number.

Answer: $5 \times 1 0^{10}$ extra electrons; $- 8.08 \times 1 0^{- 19} C$ is impossible because charge comes only in whole multiples of e.

Memory hook: Billions of electrons on you, but never a fraction of one.

Force on a charge in a field (F = qE)

Imagine an electrostatic precipitator in a factory chimney: smoke particles pick up charge and a strong field pulls them onto collecting plates.

Suppose a soot particle of mass $1.6 \times 1 0^{- 13} k g$ carries $3.2 \times 1 0^{- 15} C$ in a field of $5 \times 1 0^{5} N C^{- 1}$ . How many times larger than its weight is the electric force? ( $g = 10 m s^{- 2}$ )

Try it first, then show the solutionSolution

Electric force $F = q E = 3.2 \times 1 0^{- 15} \times 5 \times 1 0^{5} = 1.6 \times 1 0^{- 9} N$
Weight $= m g = 1.6 \times 1 0^{- 13} \times 10 = 1.6 \times 1 0^{- 12} N$
Ratio $= \frac{1.6 \times 1 0 ^{- 9}}{1.6 \times 1 0 ^{- 12}} = 1000$

Answer: The electric force is 1000 times the weight, so gravity hardly matters: the particle is driven onto the plate.

Memory hook: For tiny charged dust, qE beats mg by miles.

Field of a thin spherical shell (zero inside)

Suppose a science museum has a Van de Graaff generator with a hollow metal dome of radius 15 cm. Air near the dome breaks down (sparks) when the field reaches about $3 \times 1 0^{6} N C^{- 1}$ .

What is the largest charge the dome can hold before sparking, and what is the field at its centre?

Try it first, then show the solutionSolution

The field is largest at the surface: $E = \frac{k Q}{R ^{2}}$ , so $Q = \frac{E R ^{2}}{k}$ .
$Q = \frac{3 \times 1 0 ^{6} \times ( 0.15 ) ^{2}}{9 \times 1 0 ^{9}} = 7.5 \times 1 0^{- 6} C$
Inside a charged thin shell no charge is enclosed by a Gaussian sphere, so the field at the centre is zero.

Answer: About $7.5 μ C$ ; the field at the centre is zero.

Memory hook: Shell rule: all the field outside, none inside.

Topics in this chapter

Electric Charges and Fields formula sheet

Quantisation of charge: $q = \pm n e, e = 1.6 \times 1 0^{- 19} C$; n is a whole number. A charge that is not a whole multiple of e (such as $2.2 \times 1 0^{- 19} C$ ) is impossible.
Coulomb's law (vector form): $F_{12} = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r ^{2}} \overset{r}{^}_{21}$; $\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} N m^{2} C^{- 2}$ ; $F_{12} = - F_{21}$ .
Force in a medium: $F_{m e d i u m} = \frac{1}{4 π ε _{0} ε _{r}} \frac{q _{1} q _{2}}{r ^{2}}, ε_{r} = K = \frac{ε}{ε _{0}}$; Dielectric constant K has no unit; K = 1 for vacuum (about 1 for air).
Superposition: $F_{1} = F_{12} + F_{13} + \dots + F_{1 n}$; Vector sum. The same rule adds fields.
Charge densities: $λ = \frac{d q}{d l}, σ = \frac{d q}{d A}, ρ = \frac{d q}{d V}$; Units C m⁻¹, C m⁻², C m⁻³.
Electric field: $E = \frac{F}{q _{0}}, F = q E$; Unit N C⁻¹ = V m⁻¹. A negative charge feels a force opposite to E.
Field of a point charge: $E = \frac{1}{4 π ε _{0}} \frac{q}{r ^{2}}$; Radially outward for +q, inward for −q.
Dipole moment: $p = q (2 l)$; Directed from −q to +q. Unit C m.
Field on the axis of a dipole: $E_{a x ia l} = \frac{1}{4 π ε _{0}} \frac{2 p r}{( r ^{2} - l ^{2} ) ^{2}} \approx \frac{1}{4 π ε _{0}} \frac{2 p}{r ^{3}}$; Approximation for $r ≫ 2 l$ . Direction: along $p$ .
Field on the equatorial line of a dipole: $E_{e q} = \frac{1}{4 π ε _{0}} \frac{p}{( r ^{2} + l ^{2} ) ^{3/2}} \approx \frac{1}{4 π ε _{0}} \frac{p}{r ^{3}}$; Direction: opposite to $p$ . At equal distance $E_{a x ia l} = 2 E_{e q}$ .
Torque on a dipole: $τ = p \times E, τ = pE sin θ$; Net force in a uniform field is zero.
Electric flux: $ϕ_{E} = E \cdot A = E A cos θ, ϕ_{E} = \int E \cdot d A$; θ is measured from the normal (area vector). Unit N m² C⁻¹.
Gauss's theorem: $\oint E \cdot d A = \frac{q _{e n c l ose d}}{ε _{0}}$; Closed surfaces only; charges outside give zero net flux.
Infinite line of charge: $E = \frac{λ}{2 π ε _{0} r} = \frac{2 k λ}{r}$; $E \propto λ / r$ .
Infinite thin plane sheet: $E = \frac{σ}{2 ε _{0}}$; Independent of distance. Between two oppositely charged parallel sheets: $σ / ε_{0}$ ; outside them: 0.
Thin spherical shell: $E_{o u t} = \frac{1}{4 π ε _{0}} \frac{q}{r ^{2}}; E_{s u r f a ce} = \frac{1}{4 π ε _{0}} \frac{q}{R ^{2}} = \frac{σ}{ε _{0}}; E_{in} = 0$; Outside, the shell acts like a point charge at its centre.

Electric Charges and Fields: diagrams to practise

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

Lines of force: isolated charges, a dipole, two like charges and parallel plates

The syllabus names these four patterns; asked as 'draw the lines of force' in 1- or 2-mark items or inside a reasoning question.

Draw

Isolated $+ q$ : straight radial lines pointing outward, equally spaced. Isolated $- q$ : the same with arrows pointing inward.
Dipole: lines leave $+ q$ , curve round and end on $- q$ ; the line along the axis between the charges is straight.
Two equal positive charges close together: lines leave both and bend away from each other; leave an empty region (neutral point) midway.
Parallel plates: straight, parallel, equally spaced lines from the + plate to the − plate, with slight bulging at the edges.

Label

+q, −q
Arrowheads on every line
Neutral point N
Uniform field region between plates

Watch out: Lines must never cross, never start or stop in empty space, and must meet a conductor at 90°. Crowd them near the charges.

Dipole geometry for the axial and equatorial derivations

Needed with every 3-mark dipole derivation; examiners award a mark for a correct labelled figure.

Draw

Draw a horizontal line; mark −q at A and +q at B, 2l apart, centre O. Draw $p$ from A to B.
Axial: mark P on the extended axis beyond +q at distance r from O; show $E_{+ q}$ (away from B) longer than $E_{- q}$ (towards A).
Equatorial: mark Q on the perpendicular bisector at distance r from O; join AQ and BQ (each $r^{2} + l^{2}$ ).
At Q draw $E_{+ q}$ along BQ produced and $E_{- q}$ along QA; resolve both, showing the vertical parts cancelling and the horizontal parts adding, opposite to $p$ .

Label

−q, +q, O
2l
r
$p$
$E_{+ q}$ , $E_{- q}$ , resultant E
angle θ

Watch out: Label the angle θ between QB and the bisector; $cos θ = l / r^{2} + l^{2}$ is the step most often skipped.

Gaussian surfaces for a line, a sheet and a spherical shell, and the E-r graph

Gauss's-theorem applications (3 or 5 marks) expect the surface to be drawn; the E-r graph for a shell is named in the syllabus.

Draw

Line: a long straight line with a coaxial cylinder of radius r and length l round it; show E radially outward on the curved surface and $d A$ along the axis on the end faces.
Sheet: a vertical plane with a short cylinder piercing it, end faces of area A on both sides; show E perpendicular to the sheet and outward through each end.
Shell: a circle of radius R with a dotted concentric circle of radius r inside (and another outside).
Graph: E on the y-axis, r on the x-axis; E = 0 from 0 to R, a vertical jump at R to $\frac{k q}{R ^{2}}$ , then a falling $1/ r^{2}$ curve.

Label

Gaussian surface (dotted)
λ, σ or q
$E$ and $d A$
r, R
E = 0 inside

Watch out: For the sheet, write why the curved surface gives zero flux (E is perpendicular to $d A$ there).

How the board asks Electric Charges and Fields

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

Question type	How it is asked	Method
One-mark recall and MCQ (Section A)	Quantisation (is a given charge possible?), properties of lines of force, units of flux or permittivity, the field inside a shell, odd-one-out and match-the-column items.	Test each option against the exact property. For 'is this charge possible', divide by $1.6 \times 1 0^{- 19}$ and check for a whole number.
How will it change? (Section B, 2 marks)	'How does the flux or field change if the radius is doubled, the charge is doubled, the distance is halved, a dipole replaces the charge?'	Write the working formula first ( $ϕ = q / ε_{0}$ , $E = λ /2 π ε_{0} r$ , $E = σ /2 ε_{0}$ ), then read off the dependence.
Derivation (Section C, 3 marks)	Field on the axis or the equatorial line of a dipole; torque on a dipole; field of a line, a sheet or a shell by Gauss's theorem.	Labelled figure, the two field contributions with directions, the algebra, the final expression with direction, then the short-dipole form if asked.
Numericals on Coulomb's law and superposition	Force between charges in air or a medium, net force at a corner of a triangle or square, neutral point between two charges.	Convert µC and cm to SI, find each force separately with direction, then add as vectors.
Gauss's theorem reasoning and numericals	Flux through a closed surface or one face of a cube, charge from flux, field between two charged sheets.	Count only enclosed charge; use symmetry for one face of a cube (one-sixth of the total).
Long answer (Section D, 5 marks)	State Gauss's theorem, derive one application, then a numerical on it; or a dipole derivation plus a torque numerical.	Split your answer by the sub-part marks; keep the statement exact and the units on every answer.

Tricks to remember Electric Charges and Fields

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

Axis gets two
Short dipole at the same distance: axial $\frac{2 k p}{r ^{3}}$ , equatorial $\frac{k p}{r ^{3}}$ . 'Axis gets two, equator gets one, and equator points back.'
Use it when: Ratio questions ( $E_{1} : E_{2} = 2 : 1$ ) and checking derivations.
Line-sheet-shell powers
Line: $1/ r$ . Sheet: $r^{0}$ (constant). Shell outside: $1/ r^{2}$ . Dipole: $1/ r^{3}$ . 'One, none, two, three.'
Use it when: How-will-it-change questions on distance.
Flux counts only who is inside
Gauss is a doorman: he counts only the charges inside the room, not the size of the room or who stands outside.
Use it when: Flux when the surface grows, shrinks, changes shape or has a charge outside.
Torque is sin, energy is cos
$τ = pE sin θ$ is zero when aligned; the dipole 'sins' when it is crooked.
Use it when: Picking the right trig function for torque.
Neutral point near the smaller charge
Like charges: neutral point between them, nearer the smaller. Unlike charges: outside, beyond the smaller.
Use it when: Setting up the neutral-point equation correctly.
Sheet half, pair whole
One sheet gives $σ /2 ε_{0}$ ; two opposite sheets give the whole $σ / ε_{0}$ between them and nothing outside.
Use it when: Parallel-plate 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.

Example 1: Which charge is possible? · 1 mark · MCQBoard level

Which of the following cannot be the charge on a body?
(a) $3.2 \times 1 0^{- 19} C$
(b) $4.8 \times 1 0^{- 19} C$
(c) $2.4 \times 1 0^{- 19} C$
(d) $8.0 \times 1 0^{- 19} C$

Try it first, then show the solutionSolution

Divide each by $e = 1.6 \times 1 0^{- 19} C$ : 2, 3, 1.5, 5.
Only (c) gives a non-integer.

Answer: (c) $2.4 \times 1 0^{- 19} C$

Example 2: Assertion-reason on lines of force · 1 mark · Assertion-reasonBoard level

Assertion: Two electric lines of force can never intersect.
Reason: At any point the electric field has only one direction.
(a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Try it first, then show the solutionSolution

If two lines crossed, the tangents at the crossing would give two directions of E at one point.
The field has a single direction, so this cannot happen; the reason explains the assertion.

Answer: (a)

Example 3: Keeping the force the same in oil · 2 marks · Short answerHarder board level

Two point charges in air exert a force of 0.6 N on each other. They are immersed in oil of dielectric constant 3. To what fraction of the original distance must they be brought so that the force is again 0.6 N? [2]

Try it first, then show the solutionSolution

In air: $F = \frac{k q _{1} q _{2}}{r ^{2}}$ . In oil at distance r': $F^{'} = \frac{k q _{1} q _{2}}{K r ^{'2}}$ .
Set $F^{'} = F$ : $K r^{'2} = r^{2} \Rightarrow r^{'} = \frac{r}{K} = \frac{r}{3} = 0.577 r$

Answer: $r^{'} = r / 3 \approx 0.58$ of the original distance

Example 4: Derivation: field on the axis of a dipole · 3 marks · Short answerHarder board level

Derive an expression for the electric field intensity at a point on the axial line of an electric dipole of length 2l at distance r from its centre. Reduce it for a short dipole. [3]

Try it first, then show the solutionSolution

Let −q be at A and +q at B, centre O, and P on the axis beyond B with OP = r. Then AP = r + l and BP = r − l.
$E_{+ q} = \frac{1}{4 π ε _{0}} \frac{q}{( r - l ) ^{2}}$ along OP (away from +q); $E_{- q} = \frac{1}{4 π ε _{0}} \frac{q}{( r + l ) ^{2}}$ towards A. $E_{+ q}$ is larger.
Net $E = \frac{q}{4 π ε _{0}} [\frac{1}{( r - l ) ^{2}} - \frac{1}{( r + l ) ^{2}}] = \frac{q}{4 π ε _{0}} \frac{4 r l}{( r ^{2} - l ^{2} ) ^{2}}$
With $p = 2 q l$ : $E = \frac{1}{4 π ε _{0}} \frac{2 p r}{( r ^{2} - l ^{2} ) ^{2}}$ , along $p$ .
Short dipole ( $r ≫ l$ ): $(r^{2} - l^{2})^{2} \approx r^{4}$ , so $E = \frac{1}{4 π ε _{0}} \frac{2 p}{r ^{3}}$ .

Answer: $E = \frac{1}{4 π ε _{0}} \frac{2 p r}{( r ^{2} - l ^{2} ) ^{2}}$ along $p$ ; for a short dipole $E = \frac{1}{4 π ε _{0}} \frac{2 p}{r ^{3}}$

Example 5: Net force at the corner of a triangle · 3 marks · Short answerHarder board level

Three charges of $+ 1 μ C$ each are at the vertices of an equilateral triangle of side 10 cm in air. Calculate the magnitude and direction of the resultant force on one of them. [3]

Try it first, then show the solutionSolution

Each force: $F = \frac{9 \times 1 0 ^{9} \times 1 0 ^{- 6} \times 1 0 ^{- 6}}{0.01} = 0.9 N$ , repulsive, along the two sides meeting at the corner.
Angle between them is $6 0^{\circ}$ : $R = F^{2} + F^{2} + 2 F^{2} cos 6 0^{\circ} = 3 F$
$R = 1.732 \times 0.9 = 1.56 N$ , along the bisector of the angle at that vertex, produced outwards (away from the centre).

Answer: 1.56 N, directed away from the centroid of the triangle

Example 6: Gauss's theorem: plane sheet and two parallel sheets · 5 marks · Long answerHarder board level

(a) State Gauss's theorem. [1]
(b) Use it to obtain the field due to a uniformly charged infinite thin plane sheet of surface charge density σ. [2]
(c) Two such large parallel sheets carry $+ 8.85 \times 1 0^{- 7} C m^{- 2}$ and $- 8.85 \times 1 0^{- 7} C m^{- 2}$ . Find the field between them and outside them. [2]

Try it first, then show the solutionSolution

(a) The total electric flux through any closed surface equals $\frac{1}{ε _{0}}$ times the net charge enclosed by it: $\oint E \cdot d A = q / ε_{0}$ .
(b) Take a cylinder piercing the sheet, end faces of area A parallel to the sheet. By symmetry E is perpendicular to the sheet, outward on both sides. Curved surface: $E ⊥ d A$ , flux 0.
Flux through the two ends $= E A + E A = 2 E A$ . Charge enclosed $= σ A$ . So $2 E A = σ A / ε_{0} \Rightarrow E = \frac{σ}{2 ε _{0}}$ .
(c) Between the sheets both fields point from + to −, so they add: $E = \frac{σ}{ε _{0}} = \frac{8.85 \times 1 0 ^{- 7}}{8.85 \times 1 0 ^{- 12}} = 1 \times 1 0^{5} N C^{- 1}$ .
Outside, the two fields are equal and opposite, so $E = 0$ .

Answer: (b) $E = σ /2 ε_{0}$ , perpendicular to the sheet (c) $1 \times 1 0^{5} N C^{- 1}$ from the positive to the negative sheet between them; zero outside

Example 7: Shell field and its graph · 3 marks · Short answerHOTS / competency

A thin hollow sphere of radius R carries charge q spread uniformly. Using Gauss's theorem, explain why the field is zero inside it, and sketch E against r from r = 0 to r = 3R. If the field just outside the surface is $E_{0}$ , what is it at r = 2R and r = 3R? [3]

Try it first, then show the solutionSolution

Inside, draw a concentric Gaussian sphere of radius r < R: it encloses no charge, so $\oint E \cdot d A = 0$ ; by symmetry E is the same at every point of the sphere, hence E = 0.
Outside, $E = \frac{k q}{r ^{2}}$ , so $E \propto 1/ r^{2}$ : at 2R it is $E_{0} /4$ and at 3R it is $E_{0} /9$ .
Graph: flat at zero up to R, jumps to $E_{0}$ at R, then falls along a $1/ r^{2}$ curve through $E_{0} /4$ and $E_{0} /9$ .

Answer: E = 0 inside; $E_{0} /4$ at 2R; $E_{0} /9$ at 3R

Example 8: Comparing axial and equatorial fields at different distances · 2 marks · Short answerHOTS / competency

The field of a short dipole at distance r on its axis is E. What is the field at distance 2r on its equatorial line, and in which direction? [2]

Try it first, then show the solutionSolution

$E = \frac{2 k p}{r ^{3}}$ on the axis, so $k p = \frac{E r ^{3}}{2}$ .
Equatorial at 2r: $E^{'} = \frac{k p}{( 2 r ) ^{3}} = \frac{E r ^{3} /2}{8 r ^{3}} = \frac{E}{16}$ , opposite to $p$ .

Answer: $E /16$ , directed opposite to the dipole moment

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 2 + 1 marks.

Steering ink drops in a printer

Suppose an inkjet printer gives each ink drop a charge and then sends it between two deflecting plates to place it on the paper. A drop of mass $1.0 \times 1 0^{- 10} k g$ carries $- 2.0 \times 1 0^{- 13} C$ and enters the region between the plates horizontally at $20 m s^{- 1}$ . The plates are 1.0 cm long and produce a uniform vertical field of $8 \times 1 0^{5} N C^{- 1}$ . Ignore gravity.

(i) · 1 mark
Calculate the electric force on the drop.
Answer
Answer: $1.6 \times 1 0^{- 7} N$
$F = q E = 2.0 \times 1 0^{- 13} \times 8 \times 1 0^{5} = 1.6 \times 1 0^{- 7} N$
(ii) · 1 mark
Find the vertical acceleration of the drop.
Answer
Answer: $1.6 \times 1 0^{3} m s^{- 2}$
$a = F / m = \frac{1.6 \times 1 0 ^{- 7}}{1.0 \times 1 0 ^{- 10}} = 1600 m s^{- 2}$
(iii) · 2 marks
How long does the drop spend between the plates, and how far is it deflected sideways in that time?
Answer
Answer: $5 \times 1 0^{- 4} s$ ; 0.2 mm
Time $t = \frac{0.01}{20} = 5 \times 1 0^{- 4} s$ . Deflection $y = \frac{1}{2} a t^{2} = 0.5 \times 1600 \times (5 \times 1 0^{- 4})^{2} = 2 \times 1 0^{- 4} m = 0.2 mm$ .
(iv) · 1 mark
Is the drop deflected towards the positive or the negative plate? Give a reason.
Answer
Answer: Towards the positive plate.
The drop is negative, so the force $q E$ is opposite to $E$ , which points from the positive plate to the negative plate.

A charge inside a cube

Suppose a student places a small charge of $+ 6 μ C$ at the exact centre of a hollow plastic cube of side 20 cm and wants to find the flux through its faces. Take $ε_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2}$ .

(i) · 1 mark
What is the total flux through the cube?
Answer
Answer: $6.78 \times 1 0^{5} N m^{2} C^{- 1}$
$ϕ = \frac{q}{ε _{0}} = \frac{6 \times 1 0 ^{- 6}}{8.85 \times 1 0 ^{- 12}} = 6.78 \times 1 0^{5} N m^{2} C^{- 1}$
(ii) · 1 mark
What is the flux through one face?
Answer
Answer: $1.13 \times 1 0^{5} N m^{2} C^{- 1}$
By symmetry all six faces share it equally: $\frac{6.78 \times 1 0 ^{5}}{6} = 1.13 \times 1 0^{5}$ .
(iii) · 1 mark
The cube is replaced by one of side 40 cm with the same charge at its centre. How does the total flux change?
Answer
Answer: It does not change.
Flux depends only on the enclosed charge, not on the size of the surface.
(iv) · 2 marks
A second charge of $- 6 μ C$ is now placed just outside the cube. What is the new total flux through the cube? Is the field on the faces still the same as before?
Answer
Answer: Total flux unchanged at $6.78 \times 1 0^{5} N m^{2} C^{- 1}$ ; the field on the faces changes.
The outside charge sends as much flux in as out, so the net flux still equals $q_{in} / ε_{0}$ . But the field at each point is the vector sum of both charges' fields, so E on the faces is different (and the face-by-face share is no longer equal).

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.

Charges $+ 4 μ C$ and $+ 9 μ C$ are 50 cm apart. Find the position of the neutral point. (2 marks)

Step	Marks
States the point is between the charges and writes $\frac{k ( 4 μ C )}{x ^{2}} = \frac{k ( 9 μ C )}{( 0.5 - x ) ^{2}}$	1
Solves to x = 0.2 m, i.e. 20 cm from the $4 μ C$ charge	1

Derive the electric field at a point on the equatorial line of a dipole and write it for a short dipole. (3 marks)

Step	Marks
Labelled figure; magnitudes $E_{+ q} = E_{- q} = \frac{k q}{r ^{2} + l ^{2}}$ with directions shown	1
Resolves: perpendicular components cancel; parallel components add, $E = 2 E_{+ q} cos θ$ with $cos θ = \frac{l}{r ^{2} + l ^{2}}$	1
$E = \frac{1}{4 π ε _{0}} \frac{p}{( r ^{2} + l ^{2} ) ^{3/2}}$ , opposite to $p$ ; short dipole $E = \frac{1}{4 π ε _{0}} \frac{p}{r ^{3}}$	1

State Gauss's theorem. Use it to find the field due to an infinitely long line of charge of linear density λ. Find the field at 6 cm from a line with $λ = 0.2 μ C m^{- 1}$ and the force on an electron there. (5 marks)

Step	Marks
Correct statement: total flux through a closed surface $= q_{e n c l ose d} / ε_{0}$	1
Cylindrical Gaussian surface (radius r, length l) drawn; flux through end faces zero, through curved surface $E \times 2 π r l$	1
$E \cdot 2 π r l = λ l / ε_{0} \Rightarrow E = \frac{λ}{2 π ε _{0} r}$	1
$E = \frac{2 \times 9 \times 1 0 ^{9} \times 0.2 \times 1 0 ^{- 6}}{0.06} = 6 \times 1 0^{4} N C^{- 1}$	1
$F = e E = 1.6 \times 1 0^{- 19} \times 6 \times 1 0^{4} = 9.6 \times 1 0^{- 15} N$ , towards the line	1

Where students lose marks in Electric Charges and Fields

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

Mistake: Thinking flux changes when the Gaussian surface changes size
CISCE examiners' comments, 2024: almost all candidates thought the flux from a charge at the centre of a hollow sphere changes when the radius is doubled, and some confused flux with field.
Wrong working
Radius doubled, so area is 4 times: flux becomes 4X.
Right working
$ϕ = q / ε_{0}$ depends only on q enclosed: flux stays X (the field falls to a quarter, the area grows four times).
Do this: Quote Gauss's theorem before answering any 'how will the flux change' question.
Usually costs: 1 mark per part.
Mistake: Treating a line charge as a point charge
CISCE examiners' comments, 2024: when asked how the field of an infinite line charge changes if the distance is halved, many used the inverse-square law and answered 'four times'.
Wrong working
$E \propto 1/ r^{2}$ , so halving r makes E four times.
Right working
$E = \frac{λ}{2 π ε _{0} r}$ , so halving r doubles E; doubling λ doubles E.
Do this: Write the formula for the right shape: line $1/ r$ , sheet constant, point or shell outside $1/ r^{2}$ .
Usually costs: 1 mark per part.
Mistake: Defining relative permittivity loosely
CISCE examiners' comments, 2024: candidates explaining 'relative permittivity of water is 81' linked it to refractive index or permeability.
Wrong working
Water slows light 81 times.
Right working
The force (or field) between two charges in water is 1/81 of that between the same charges, at the same distance, in vacuum.
Do this: Use the ratio definition $K = F_{v a c uu m} / F_{m e d i u m}$ ; keep permittivity (electric) apart from permeability (magnetic).
Usually costs: 1 mark.
Mistake: Answering 'yes' or giving no reason for an impossible charge
CISCE examiners' comments, 2024: on whether a body can carry $2.2 \times 1 0^{- 19} C$ , many said yes or gave no reason.
Wrong working
Yes, any small charge is possible.
Right working
$\frac{2.2 \times 1 0 ^{- 19}}{1.6 \times 1 0 ^{- 19}} = 1.375$ , not a whole number, so no: charge is quantised.
Do this: Always show the division by e; the reason carries the mark.
Usually costs: 1 mark.
Mistake: Naming the zero-potential locus or the field ratio wrongly for a dipole
CISCE examiners' comments, 2025: candidates wrote 'equipotential surface' instead of the equatorial plane, mixed up broadside-on and equatorial terms, and did not use the short-dipole formulas for the axial-to-equatorial field ratio.
Wrong working
$E_{1} / E_{2} = 1$ because the distances are equal.
Right working
Short dipole: $E_{1} = \frac{2 k p}{r ^{3}}$ , $E_{2} = \frac{k p}{r ^{3}}$ , so $E_{1} / E_{2} = 2$ .
Do this: Learn the two short-dipole forms side by side; 'broadside-on' and 'equatorial' mean the same position.
Usually costs: 1 mark each.
Mistake: Wrong direction or subtraction when adding two sheet fields
CISCE examiners' comments, 2025: for two oppositely charged sheets, candidates confused a thin sheet with a thick conducting plate, subtracted fields acting in the same direction and could not give the direction of the resultant.
Wrong working
$E = \frac{σ}{2 ε _{0}} - \frac{σ}{2 ε _{0}} = 0$ between the sheets.
Right working
Between them both fields point from + to −: $E = \frac{σ}{2 ε _{0}} + \frac{σ}{2 ε _{0}} = \frac{σ}{ε _{0}}$ , towards the negative sheet.
Do this: Draw arrows for each sheet's field in each region before adding.
Usually costs: 1-2 marks in a 3-mark question.

Practice questions on Electric Charges and Fields

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The SI unit of the permittivity of free space is:
(a) $N m^{2} C^{- 2}$
(b) $C^{2} N^{- 1} m^{- 2}$
(c) $N C^{- 1}$
(d) $N m^{2} C^{- 1}$
Hint
Rearrange $F = \frac{q _{1} q _{2}}{4 π ε _{0} r ^{2}}$ for $ε_{0}$ .
Answer
(b) $C^{2} N^{- 1} m^{- 2}$
Q2 · 1 mark · MCQBoard level
The electric field at a point inside a uniformly charged thin spherical shell is:
(a) maximum at the centre
(b) the same as on the surface
(c) zero
(d) proportional to the distance from the centre
Hint
How much charge does a Gaussian sphere inside the shell enclose?
Answer
(c) zero
Q3 · 1 mark · MCQBoard level
An electric dipole is placed in a uniform electric field. Which statement is correct?
(a) Net force and torque are both always zero.
(b) Net force is zero; torque may be non-zero.
(c) Net force is non-zero; torque is zero.
(d) Both are always non-zero.
Hint
The forces on the two charges are equal and opposite but along different lines.
Answer
(b)
Q4 · 1 mark · Very short answerBoard level
Why can electrostatic lines of force not form closed loops?
Hint
Where do they start and end?
Answer
They start on positive charges and end on negative charges; a closed loop would need a line with no start or end, which would mean work done round a closed path, impossible for an electrostatic field.
Q5 · 2 marks · Short answerBoard level
A metal sphere has a charge of $+ 6.4 μ C$ . How many electrons has it lost? Has its mass increased or decreased?
Hint
$n = q / e$ .
Answer
$n = \frac{6.4 \times 1 0 ^{- 6}}{1.6 \times 1 0 ^{- 19}} = 4 \times 1 0^{13}$ electrons lost; its mass has decreased (by $4 \times 1 0^{13} \times 9.1 \times 1 0^{- 31} = 3.64 \times 1 0^{- 17} k g$ ).
Q6 · 2 marks · Short answerHOTS / competency
Two identical small metal spheres carry $+ 6 μ C$ and $- 2 μ C$ and attract each other with force F. They are touched together and put back at the same separation. Find the new force in terms of F and state whether it attracts or repels.
Hint
After contact, identical spheres share the total charge equally.
Answer
Total charge $= + 4 μ C$ , so each gets $+ 2 μ C$ . $F \propto 6 \times 2 = 12$ before and $2 \times 2 = 4$ after, so $F^{'} = F /3$ , now repulsive.
Q7 · 2 marks · Short answerHarder board level
An infinite line charge produces a field of $6 \times 1 0^{4} N C^{- 1}$ at 6 cm. What is the field at 12 cm if the linear charge density is also doubled?
Hint
$E \propto λ / r$ .
Answer
Both λ and r double, so $E = \frac{2 λ}{2 π ε _{0} ( 2 r )}$ is unchanged: $6 \times 1 0^{4} N C^{- 1}$ .
Q8 · 2 marks · Short answerHarder board level
Find the ratio of the electric force to the gravitational force between two protons. ( $m_{p} = 1.67 \times 1 0^{- 27} k g$ , $G = 6.67 \times 1 0^{- 11} N m^{2} k g^{- 2}$ , $e = 1.6 \times 1 0^{- 19} C$ )
Hint
The distance cancels in the ratio.
Answer
$\frac{F _{e}}{F _{g}} = \frac{k e ^{2}}{G m _{p}^{2}} = \frac{9 \times 1 0 ^{9} \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{6.67 \times 1 0 ^{- 11} \times ( 1.67 \times 1 0 ^{- 27} ) ^{2}} \approx 1.24 \times 1 0^{36}$
Q9 · 3 marks · Short answerHarder board level
Two point charges of $\pm 2 n C$ are 4 mm apart. Calculate the field at a point 20 cm from the centre (a) on the axis and (b) on the equatorial line. Why may you use the short-dipole formulas here?
Hint
Find p first.
Answer
$p = 8 \times 1 0^{- 12} C m$ . (a) $\frac{2 k p}{r ^{3}} = 18 N C^{- 1}$ along $p$ (b) $\frac{k p}{r ^{3}} = 9 N C^{- 1}$ opposite to $p$ . Short-dipole forms apply because $r = 20 c m ≫ 2 l = 4 mm$ .
Q10 · 3 marks · Short answerHarder board level
A dipole of moment $4 \times 1 0^{- 9} C m$ is in a uniform field of $2 \times 1 0^{4} N C^{- 1}$ . (a) Find the torque when it makes $6 0^{\circ}$ with the field. (b) At what angles is the torque half its maximum value? (c) In which position is it in stable equilibrium?
Hint
$τ = pE sin θ$ .
Answer
(a) $τ = 4 \times 1 0^{- 9} \times 2 \times 1 0^{4} \times sin 6 0^{\circ} = 6.93 \times 1 0^{- 5} N m$ (b) $sin θ = 0.5$ : $3 0^{\circ}$ or $15 0^{\circ}$ (c) $θ = 0^{\circ}$ , $p$ along $E$ .
Q11 · 3 marks · Short answerHarder board level
The net flux through a closed surface is found to be $2.26 \times 1 0^{5} N m^{2} C^{- 1}$ . (a) Find the net charge inside. (b) If a charge of $- 1 μ C$ is added outside the surface, what is the new flux? (c) If the enclosed charge were a dipole, what would the flux be?
Hint
$q = ε_{0} ϕ$ .
Answer
(a) $q = 8.85 \times 1 0^{- 12} \times 2.26 \times 1 0^{5} = 2.0 \times 1 0^{- 6} C = 2 μ C$ (b) unchanged, $2.26 \times 1 0^{5}$ , because outside charges contribute no net flux (c) zero, as a dipole's net charge is zero.
Q12 · 5 marks · Long answerHarder board level
(a) State Gauss's theorem and list two properties of a Gaussian surface. [2]
(b) Use the theorem to derive the field at a point outside, on and inside a thin spherical shell of radius R carrying charge q. [2]
(c) A shell of radius 10 cm carries $2 μ C$ . Find the field at 20 cm from its centre. [1]
Hint
Use a concentric spherical Gaussian surface each time.
Answer
(a) $\oint E \cdot d A = q / ε_{0}$ ; the surface must be closed and pass through the point, and should follow the symmetry so E is constant and normal (or parallel) on it. (b) Outside: $E \cdot 4 π r^{2} = q / ε_{0}$ , $E = \frac{k q}{r ^{2}}$ ; on the surface $E = \frac{k q}{R ^{2}}$ ; inside no charge is enclosed so $E = 0$ . (c) $E = \frac{9 \times 1 0 ^{9} \times 2 \times 1 0 ^{- 6}}{0.04} = 4.5 \times 1 0^{5} N C^{- 1}$ .
Q13 · 2 marks · Short answerHarder board level
A charged comb attracts small neutral bits of paper, but a dipole in a uniform field feels no net force. Explain the difference.
Hint
Think about the field of a comb and induced dipoles.
Answer
The comb induces a dipole in each bit of paper (unlike charge on the near side). The comb's field is non-uniform: stronger on the near side, so the attraction on the near charge exceeds the repulsion on the far charge, giving a net pull. In a uniform field the two forces on a dipole are equal and opposite, so the net force is zero.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the principle of conservation of charge.Show answer
The total charge of an isolated system remains constant; charge can be transferred but not created or destroyed.
2.What is meant by quantisation of charge?Show answer
Every charge is a whole-number multiple of the electronic charge: $q = \pm n e$ , $e = 1.6 \times 1 0^{- 19} C$ .
3.Write Coulomb's law in vector form.Show answer
$F_{12} = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r ^{2}} \overset{r}{^}_{21}$ , the force on $q_{2}$ due to $q_{1}$ .
4.Define electric field intensity and give its SI unit.Show answer
Force per unit positive test charge, $E = F / q_{0}$ ; unit N C⁻¹ (or V m⁻¹).
5.Define electric dipole moment. What is its direction?Show answer
$p = q \times 2 l$ ; directed from the negative charge to the positive charge. Unit C m.
6.What is the torque on a dipole in a uniform field?Show answer
$τ = p \times E$ , magnitude $pE sin θ$ ; the net force is zero.
7.State Gauss's theorem.Show answer
The total electric flux through any closed surface equals the net charge enclosed divided by $ε_{0}$ .
8.Field due to an infinite plane sheet?Show answer
$E = σ /2 ε_{0}$ , perpendicular to the sheet, independent of distance.
9.Field inside a uniformly charged thin spherical shell?Show answer
Zero, because a Gaussian surface inside encloses no charge.

Best order to study it

Revise charge, conservation and quantisation, and practise $n = q / e$ .
Learn Coulomb's law in vector form and the effect of a medium; compare with gravitation.
Practise superposition numericals (triangle, square, neutral point) with vector diagrams.
Learn the field of a point charge and draw the four line-of-force patterns.
Derive the dipole field on the axis and the equatorial line, then the torque; write the short-dipole forms.
Learn flux, Gauss's theorem and the Gaussian-surface rules; derive the line, sheet and shell results and draw the E-r graph.
Finish with mixed Section B and C questions and one 5-mark Gauss's-theorem answer under time.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you say whether a given charge is possible by dividing by e?
Can you write Coulomb's law in vector form and explain the meaning of $ε_{r} = 81$ for water?
Can you find the neutral point for like and for unlike charges?
Can you draw lines of force for $+ q$ , $- q$ , a dipole, two like charges and parallel plates?
Can you derive the axial and equatorial fields of a dipole and reduce them for a short dipole?
Can you derive $τ = pE sin θ$ and state when it is zero and maximum?
Can you state Gauss's theorem exactly and list properties of a Gaussian surface?
Can you derive the field of a line, a sheet and a shell, and sketch E against r for the shell?
Do you know that flux depends only on enclosed charge, not on the surface's size or shape?

Electric Charges and Fields: frequently asked questions

Do I have to derive the dipole field for ISC 2027, or only state it?

Derive it. The ISC 2027 Physics syllabus asks for the derivation of E on the axis (end-on) and on the perpendicular bisector (broadside-on) of a dipole, and the short-dipole forms. The torque \tau = \vec{p} \times \vec{E} must also be derived. (The 'qualitative only' note applies to the magnetic dipole in Chapter 5, not to the electric dipole.)

Is the proof of Gauss's theorem required?

No. You state it, explain flux and the Gaussian surface, and apply it to an infinite line charge, an infinite thin plane sheet and a thin spherical shell (inside, on the surface and outside).

Do I need position vectors $\vec{r}_1$ and $\vec{r}_2$ in Coulomb's law?

No. The syllabus says Coulomb's law in vector form is required, but the position coordinates \vec{r}_1, \vec{r}_2 are not necessary. Use the unit vector along the line joining the charges.

How many marks does this chapter carry?

Electrostatics and Current Electricity together carry 14 of the 70 theory marks. Expect a mix of one-mark items, a 2-mark reasoning or numerical item and often a 3-mark derivation from this chapter.

What is the difference between a thin charged sheet and a charged conducting plate?

For a thin sheet with density σ, Gauss's theorem gives \sigma/2\varepsilon_0 on each side. Two oppositely charged sheets (as in a capacitor) give \sigma/\varepsilon_0 between them. Keep to the thin-sheet result unless the question describes a pair of plates.

Which sign convention is used for the dipole field direction?

On the axis the field points along \vec{p} (from −q to +q); on the equatorial line it points opposite to \vec{p}. State the direction with the magnitude in every answer.

Based on: CISCE ISC 2027 Physics syllabus (revised, uploaded February 2026); CISCE ISC 2027 Physics specimen Paper 1 (Theory) question paper and answer key; CISCE Analysis of Pupil Performance, ISC Physics, 2025; CISCE Analysis of Pupil Performance, ISC Physics, 2024