Chapter 1 of 14

CBSE Class 12 Physics★ Unit I · Electrostatics, CBSE states no chapter-wise split

Electric Charges and Fields

Charge and its three basic properties, Coulomb's law and superposition, the electric field and its field lines, electric flux, the dipole, and Gauss's law with its standard applications. The Additional Exercises section has been removed from this chapter and from every other chapter of the current Physics book. CBSE Class 12 Physics Chapter 1.

⏱ 12 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1State and apply the three basic properties of charge: additivity, conservation and quantisation
2Apply Coulomb's law and the principle of superposition to find the net force on a charge
3Compute the electric field of point charges and describe fields using field lines
4Calculate electric flux through flat and closed surfaces, measuring the angle from the normal
5Find the dipole moment of a charge pair and the torque on a dipole in a uniform field
6Apply Gauss's law to derive the field of a line charge, a plane sheet and a charged shell

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Why this chapter matters

This is the foundation of the entire Electrostatics unit and of much of the rest of the Class 12 syllabus. Coulomb's law, superposition and Gauss's law reappear in capacitance, current electricity and magnetism, and the field concept introduced here is the one every later chapter builds on.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Vectors (Class 11): superposition requires vector addition of forces and fields, and the dipole torque is a cross product

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Newton's laws (Class 11): Coulomb forces obey the third law, which answers Exercise 1.2(b) without any calculation

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Scalar and vector products (Class 11): flux is a dot product and torque on a dipole is a cross product

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Surface areas and volumes (Class 9-10): Gaussian surfaces require the areas of spheres and cylinders, and Exercise 1.21 needs 4 pi r squared

Electric Charges and Fields

1. Check this before you revise anything

The "Additional Exercises" section has been removed — from this chapter and from every one of the 14 chapters in the Class 12 Physics book. A full-text search of all fourteen chapter PDFs returns zero occurrences of the phrase, and every chapter's questions now run contiguously from 1 upwards with no gaps in the numbering. The exercises were cut and the survivors renumbered cleanly, so nothing is missing from your copy.

For this chapter that leaves Exercises 1.1 to 1.23 — 23 questions, which is the entire problem set. Older editions carried a further block of harder problems beyond this point.

The old stub had no solutions file at all, and its structure duplicated sections that belong in the page's own data — a "Common Mistakes" heading, a "CBSE Exam Focus" heading and a "Self-Test" — all of which are now driven from meta.json instead. It also claimed a 22-minute read for a body of about 1200 words.

Textbook section	Topic
1.2 to 1.4	Electric charge, conductors and insulators, and the three basic properties
1.5 to 1.6	Coulomb's law and forces between multiple charges
1.7 to 1.8	Electric field and electric field lines
1.9	Electric flux
1.10 to 1.11	Electric dipole, and a dipole in a uniform external field
1.12 to 1.14	Continuous charge distribution, Gauss's law, and its applications

A transcription warning if you work from a PDF copy. Throughout the exercises the micro symbol renders as a plain "m" under most text extraction, so " $0.4 μ$ C" can appear as "0.4 mC". The difference is a factor of a thousand. In Exercise 1.2 the correct reading of $μ$ C gives a separation of 12 cm; reading it as mC would give 120 m for two "small spheres". Every value on this page has been checked against the printed page image.

2. Electric Charge and its Basic Properties (Textbook 1.2 to 1.4)

Charge is the property of matter responsible for electric force. There are two kinds, named positive and negative by Franklin's convention, and the rule is that like charges repel while unlike charges attract.

Rubbing two bodies together does not create charge. It merely transfers electrons from one to the other, leaving one with a deficit and the other with a surplus.

Three properties govern everything that follows.

Additivity. The total charge of a system is the algebraic sum of the individual charges, signs included. A system holding $+ 3 μ$ C and $- 5 μ$ C has total charge $- 2 μ$ C. Charges add as scalars, not as vectors — this is what distinguishes charge from the forces it produces.

Conservation. The total charge of an isolated system is constant. Charge can be moved from place to place, but it can be neither created nor destroyed.

This is why rubbing glass with silk yields equal and opposite charges: the pair began neutral and must remain so overall.

Quantisation. Charge exists only in integral multiples of the elementary charge:

$q = n e, e = 1.6 \times 1 0^{- 19} C, n = 0, \pm 1, \pm 2, \dots$

No body carries a fraction of $e$ , because charge moves only in whole electrons. At laboratory scale this is invisible — a charge of $1 μ$ C is already about $1 0^{13}$ electrons, so the steps are far too fine to detect and charge appears continuous.

Conductors and insulators (1.3). Conductors such as metals contain free electrons that move readily, so charge given to a conductor spreads over its surface and can be earthed away. Insulators such as glass hold charge where it is placed.

3. Coulomb's Law and Superposition (Textbook 1.5 to 1.6)

Coulomb's law gives the force between two stationary point charges:

$F = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r ^{2}} = k \frac{q _{1} q _{2}}{r ^{2}}, k = 9 \times 1 0^{9} N m^{2} C^{- 2}$

The force acts along the line joining the charges, is repulsive for like charges and attractive for unlike ones, and obeys Newton's third law — the two charges push or pull each other equally hard.

The inverse square is the part that does the work. Halving the separation quadruples the force. Exercise 1.12 exploits exactly this: doubling both charges multiplies the force by 4, and halving the distance multiplies it by another 4, for a net factor of 16.

The principle of superposition (1.6). When several charges are present, the force on any one of them is the vector sum of the forces each other charge would exert on it acting alone.

$F_{1} = F_{12} + F_{13} + F_{14} + \dots$

Each pairwise force is unaffected by the presence of the others. This is what makes the problem tractable, and it is why symmetry arguments are so powerful — in Exercise 1.6, four charges at the corners of a square produce zero net force at the centre because the diagonal pairs cancel, with no arithmetic required.

4. The Electric Field and Field Lines (Textbook 1.7 to 1.8)

Rather than describing charges as acting on each other across empty space, we say a charge sets up a field everywhere around it, and any other charge responds to the field at its own location.

$E = \frac{F}{q _{0}}, E = \frac{k q}{r ^{2}} for a point charge$

The field is force per unit charge, measured in N/C, and is a vector. The test charge $q_{0}$ must be small enough not to disturb the source charges it is being used to probe.

The force on a charge placed in a field is $F = q E$ . For a positive charge this is along the field; for a negative charge it is opposite to it. Exercise 1.8 turns on precisely that sign.

Field lines (1.8) are a way of picturing the field. The tangent at any point gives the field's direction, and the density of lines indicates its strength.

Two rules follow from the definition and are regularly examined:

Field lines are continuous and cannot break, because the field exists at every point along the path.
Field lines never cross. A crossing would allow two tangents, and hence two directions for the field, at a single point — impossible, since the force on a test charge there is one definite vector.

Lines begin on positive charges and end on negative ones, and they never form closed loops in electrostatics.

5. Electric Flux (Textbook 1.9)

Electric flux measures how much field passes through a surface:

$ϕ = E \cdot Δ S = E Δ S cos θ$

The angle is measured from the normal, not the plane. $θ$ is the angle between the field and the area vector $Δ S$ , which points perpendicular to the surface. Using the angle to the plane instead swaps $cos$ for $sin$ and is the standard error, which Exercise 1.14 tests directly.

Flux is a scalar despite being built from two vectors, and its unit is N m $^{2}$ /C.

When the surface is not flat or the field not uniform, divide the surface into elements small enough to treat as flat and sum the contributions.

For a closed surface the convention is that the area vector points outward, so flux leaving the surface counts positive and flux entering counts negative.

6. The Electric Dipole (Textbook 1.10 to 1.11)

An electric dipole is a pair of equal and opposite charges $+ q$ and $- q$ separated by a small distance $2 a$ . Its strength and orientation are captured by the dipole moment:

$p = q \times 2 a, directed from - q towards + q$

The direction convention matters and is frequently got backwards. Note also that $2 a$ is the full separation between the charges, not the distance from the centre — Exercise 1.9 places charges at $\pm 15$ cm, giving $2 a = 30$ cm.

Since the two charges are equal and opposite, the total charge of a dipole is zero, yet it still produces a field, because the two charges are at different places.

The field of a dipole falls off as $1/ r^{3}$ , faster than the $1/ r^{2}$ of a single point charge, because the two opposite contributions increasingly cancel at large distances.

A dipole in a uniform external field (1.11). The forces $+ q E$ and $- q E$ on the two charges are equal and opposite, so the net force is zero and the dipole does not translate.

They do not act along the same line, however, so they form a couple producing a torque:

$τ = p \times E, τ = pE sin θ$

The torque is zero when the dipole is aligned with the field and greatest when it is perpendicular to it, which is why $sin θ$ appears. The effect is to rotate the dipole into alignment with the field.

7. Gauss's Law and its Applications (Textbook 1.12 to 1.14)

Continuous charge distributions (1.12). For charge spread smoothly rather than sitting at points, we use densities: linear $λ$ (C/m), surface $σ$ (C/m $^{2}$ ) and volume $ρ$ (C/m $^{3}$ ).

Gauss's law. The total electric flux through any closed surface equals the net charge enclosed divided by $ε_{0}$ :

$ϕ = \oint E \cdot d S = \frac{q _{enclosed}}{ε _{0}}$

Three consequences are examined repeatedly.

The flux depends only on the enclosed charge — not on the size or shape of the surface, and not on where inside the charge sits. Exercises 1.18 and 1.19 make this point by giving surface dimensions that turn out to be irrelevant.

Charges outside the surface contribute nothing to the net flux, since their field lines enter and leave again.

Zero net flux means zero net enclosed charge, not the absence of charge. Equal positive and negative charges inside would give zero flux while plenty of charge is present, which is the whole point of Exercise 1.16(b).

Applications (1.14). Choosing a Gaussian surface that matches the symmetry of the charge distribution turns an intractable integral into simple arithmetic.

Distribution	Field	Note
Infinite line charge	$E = \frac{λ}{2 π ε _{0} r}$	Falls off as $1/ r$ , not $1/ r^{2}$
Infinite plane sheet	$E = \frac{σ}{2 ε _{0}}$	Independent of distance from the sheet
Uniformly charged shell, outside	$E = \frac{k q}{r ^{2}}$	Behaves as a point charge at the centre
Uniformly charged shell, inside	$E = 0$	No enclosed charge

Two parallel plates with opposite charges superpose to give $E = σ / ε_{0}$ between the plates and zero outside, which is Exercise 1.23 and the reason a capacitor's field is confined to its interior.

Summary

Charge comes in two kinds; like charges repel, unlike attract. Rubbing transfers electrons rather than creating charge.
Charge is additive (algebraic sum, signs included), conserved in an isolated system, and quantised as $q = n e$ with $e = 1.6 \times 1 0^{- 19}$ C.
Quantisation is invisible at macroscopic scale because $1 μ$ C is already about $1 0^{13}$ electrons.
Coulomb's law: $F = k q_{1} q_{2} / r^{2}$ with $k = 9 \times 1 0^{9}$ , acting along the line joining the charges and obeying Newton's third law.
Superposition: the force on one charge is the vector sum of the individual pairwise forces, each unaffected by the others.
$E = F / q_{0}$ , with $E = k q / r^{2}$ for a point charge; the force on a charge is $F = q E$ , reversed in direction for a negative charge.
Field lines are continuous and never cross, since a crossing would give the field two directions at one point.
Flux is $ϕ = E A cos θ$ with $θ$ measured from the normal to the surface, not from its plane.
Dipole moment $p = q \times 2 a$ points from $- q$ to $+ q$ , where $2 a$ is the full separation; the dipole's field falls off as $1/ r^{3}$ .
A dipole in a uniform field feels zero net force but a torque $τ = p \times E$ of magnitude $pE sin θ$ .
Gauss's law: $ϕ = q_{enclosed} / ε_{0}$ , independent of the surface's size, shape and the charge's position within it.
Zero net flux means zero net charge enclosed, not zero charge.
Standard results: line charge $λ /2 π ε_{0} r$ ; plane sheet $σ /2 ε_{0}$ ; between oppositely charged parallel plates $σ / ε_{0}$ , with zero field outside.
The Additional Exercises block has been removed from this and every other chapter of the current Physics book, leaving Exercises 1.1 to 1.23 here.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Quantisation of charge

q = n e, with e = 1.6 x 10^-19 C and n an integer

No body carries a fraction of e; a charge of 1 microcoulomb is already about 10^13 electrons

Coulomb's law

F = k q1 q2 / r^2, with k = 1/(4 pi epsilon_0) = 9 x 10^9 N m^2 C^-2

Acts along the line joining the charges and obeys Newton's third law

Principle of superposition

Net force on a charge is the VECTOR sum of the individual pairwise Coulomb forces

Each pairwise force is unaffected by the presence of the other charges

Electric field

E = F/q0, and E = k q / r^2 for a point charge

A vector measured in N/C; the test charge must be small enough not to disturb the source charges

Force on a charge in a field

F = qE, along E for a positive charge and opposite to E for a negative charge

The sign reversal for negative charges is what Exercise 1.8(b) tests

Electric flux

phi = E . dS = E dS cos(theta), where theta is measured from the NORMAL to the surface

A scalar despite being built from two vectors; using the angle to the plane gives sin instead of cos

Electric dipole moment

p = q x 2a, directed from -q towards +q

2a is the FULL separation between the charges, not the distance from the centre

Field of a dipole

Falls off as 1/r^3, faster than the 1/r^2 of a point charge

The two opposite contributions increasingly cancel at large distances

Torque on a dipole in a uniform field

tau = p x E, with magnitude p E sin(theta)

The net FORCE is zero, so the dipole rotates into alignment but does not translate

Gauss's law

phi = closed integral of E . dS = q_enclosed / epsilon_0

Independent of the size and shape of the surface and of where the charge sits inside it

Charges outside a Gaussian surface

Contribute zero net flux

Their field lines enter the surface and leave it again, cancelling exactly

Field of an infinite line charge

E = lambda / (2 pi epsilon_0 r)

Falls off as 1/r, NOT as 1/r^2 — the commonest error in Exercise 1.22

Field of an infinite plane sheet

E = sigma / (2 epsilon_0)

Independent of the distance from the sheet

Field of a uniformly charged shell

Outside: E = k q / r^2, as if a point charge sat at the centre. Inside: E = 0

Inside the shell no charge is enclosed, so Gauss's law gives zero directly

Field between oppositely charged parallel plates

E = sigma / epsilon_0 between the plates, and zero in both outer regions

The contributions add between the plates and cancel outside, which confines a capacitor's field

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Adding charges as vectors instead of algebraically

✓ Charge is a scalar and adds with signs: +3 microcoulomb and -5 microcoulomb give -2 microcoulomb. Only the forces and fields they produce are vectors.

WATCH OUT

✗ Measuring the flux angle from the plane of the surface rather than its normal

✓ Flux is E A cos(theta) with theta between the field and the area vector, which is perpendicular to the surface. Using the plane gives sin instead of cos.

WATCH OUT

✗ Taking the dipole separation as the distance from the centre to one charge

✓ p = q x 2a uses the FULL separation. Charges at plus and minus 15 cm are 30 cm apart, so 2a = 0.30 m.

WATCH OUT

✗ Pointing the dipole moment from the positive charge to the negative charge

✓ The convention is from -q towards +q. Reversing it flips the sign of every subsequent torque and potential energy result.

WATCH OUT

✗ Concluding that zero net flux means there is no charge inside the surface

✓ It means the NET enclosed charge is zero. Equal positive and negative charges inside would give zero flux with plenty of charge present.

WATCH OUT

✗ Using an inverse square law for the field of an infinite line charge

✓ A line charge gives E proportional to 1/r, not 1/r^2. Only point charges and spherically symmetric distributions fall off as 1/r^2.

WATCH OUT

✗ Assuming the size of a Gaussian surface affects the flux through it

✓ Gauss's law depends only on the enclosed charge. Doubling the radius of a sphere around a point charge leaves the flux unchanged, since E falls as 1/r^2 while area grows as r^2.

WATCH OUT

✗ Computing four separate forces in a symmetric arrangement before checking for cancellation

✓ Look at the geometry first. In Exercise 1.6 the diagonally opposite equal charges cancel in pairs, so the answer is zero with no arithmetic at all.

WATCH OUT

✗ Reading the micro symbol as milli when working from an extracted PDF

✓ The micro symbol often extracts as a plain 'm', a factor of a thousand. Check against the page image: Exercise 1.2 gives 12 cm with microcoulomb, but an absurd 120 m with millicoulomb.

WATCH OUT

✗ Forgetting to halve a given diameter before using the radius

✓ Exercise 1.21 specifies a 2.4 m diameter, so r = 1.2 m. Substituting 2.4 into 4 pi r squared quadruples the area and the charge.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex s

Exercises

Coulomb's law, superposition, electric field and field lines, flux, dipoles, and Gauss's law with its standard applications

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Electric Charges and Fields?

8 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

8 questions~6 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Like charges repel, unlike attract; rubbing transfers electrons rather than creating charge
•Charge is additive as an algebraic sum, with signs included
•Charge is conserved in an isolated system, which is why rubbing gives equal and opposite charges
•Charge is quantised: q = ne with e = 1.6 x 10^-19 C
•Quantisation is undetectable at macroscopic scale because 1 microcoulomb is about 10^13 electrons
•Coulomb's law: F = k q1 q2 / r^2 with k = 9 x 10^9, acting along the joining line
•Coulomb forces obey Newton's third law, so the two charges feel equal and opposite forces
•Superposition: net force is the vector sum of the pairwise forces, each unaffected by the others
•E = F/q0 and E = kq/r^2; the force on a charge is F = qE, reversed for a negative charge
•Field lines are continuous, never cross, and run from positive to negative charges
•Flux phi = E A cos(theta), with theta measured from the normal to the surface
•Dipole moment p = q x 2a points from -q to +q, and 2a is the full separation
•A dipole's field falls off as 1/r^3 because the opposite contributions partly cancel
•A dipole in a uniform field feels zero net force but a torque p E sin(theta)
•Gauss's law: phi = q_enclosed/epsilon_0, independent of the surface's size and shape
•Zero net flux means zero net enclosed charge, not the absence of charge
•Line charge: E = lambda/(2 pi epsilon_0 r), falling off as 1/r
•Plane sheet: E = sigma/(2 epsilon_0), independent of distance
•Charged shell: point-charge field outside, zero field inside
•Oppositely charged parallel plates: sigma/epsilon_0 between them, zero outside

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Unit I: Electrostatics, no chapter-wise split published by CBSE

Question type	Marks each	Typical count	What it tests
Charge Properties, Coulomb's Law and Superposition	2-3	1	Additivity, conservation and quantisation of charge; the inverse square law; scaling arguments; and vector addition of forces
Electric Field and Electric Dipole	3-4	1	Field of point charges, field line reasoning, dipole moment, and torque in a uniform field
Electric Flux and Gauss's Law	4-5	1	Flux through flat and closed surfaces, and Gauss's law applied to line, sheet and shell distributions

Prep strategy

Check the geometry for symmetry before computing individual forces — several questions collapse to zero by cancellation
Always measure the flux angle from the normal to the surface, never from its plane
Memorise the three standard Gauss's law results and note that the line charge falls off as 1/r, not 1/r^2
State directions with every force and field answer; a bare magnitude is an incomplete answer in this chapter
Watch for redundant data such as a Gaussian surface's dimensions, which signals that Gauss's law alone is needed

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Lightning conductors

Charge concentrates at sharp points, producing a very strong local field that ionises the air and guides a discharge safely to earth.

Electrostatic precipitators

Industrial chimneys charge smoke particles and collect them on oppositely charged plates, using the Coulomb force to remove pollutants from exhaust gases.

Photocopiers and laser printers

A charged drum attracts oppositely charged toner particles only where an image has been written, which is Coulomb attraction used for imaging.

Electrostatic shielding

The zero field inside a charged conducting shell, a direct consequence of Gauss's law, is what makes a metal enclosure protect sensitive electronics.

Inkjet printing

Ink droplets are given a controlled charge and then steered by deflecting plates, the same parallel-plate field geometry as Exercise 1.23.

Van de Graaff generators

Charge sprayed onto a moving belt accumulates on a hollow sphere's outer surface, illustrating that charge on a conductor resides on its surface.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Draw a diagram with all charges and the directions of each force before writing any equation

State the direction of every force and field answer, not merely its magnitude

Convert all lengths to metres at the start; centimetres left unconverted are the single largest source of lost marks

For Gauss's law questions, name the Gaussian surface you are choosing and say why its symmetry suits the distribution

Check whether the geometry allows cancellation before computing individual forces

Quote the flux angle explicitly as being measured from the normal, which earns the method mark even if arithmetic slips

Watch for data that is deliberately not needed, such as a Gaussian surface's dimensions or a sphere's radius when the field is measured outside it

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

Deriving the 1/r^3 dependence of the dipole field on the axial and equatorial lines by binomial expansion of the two point-charge contributions

STRETCH

The method of images, which replaces a grounded conducting plane with a fictitious mirror charge to solve otherwise intractable geometries

STRETCH

Proving that the field inside any closed conducting shell is zero regardless of the shell's shape, not merely for a sphere

STRETCH

Electrostatic potential energy of a continuous charge distribution, obtained by integrating the assembly work over the whole body

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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainGauss's law and surface independenceRecognising that the surface dimensions are irrelevant

A point charge of $2.0 μ$ C sits at the centre of a cubic Gaussian surface of edge 9.0 cm. Find the net electric flux through the surface.

Stuck? Show the approach

Apply Gauss's law directly and notice which given quantity is not needed.

Show the full solution

Gauss's law gives $ϕ = q_{enclosed} / ε_{0}$ for any closed surface. Substituting, $ϕ = \frac{2.0 \times 1 0 ^{- 6}}{8.854 \times 1 0 ^{- 12}} \approx 2.26 \times 1 0^{5}$ N m $^{2}$ /C. The edge length plays no part, since every field line leaving the charge must cross whatever closed surface surrounds it.

Answer: 2.26 x 10^5 N m^2/C

The trap

Attempting to compute the flux face by face, or assuming a larger cube gives more flux. The 9.0 cm is deliberately redundant information.

JEE MainScaling in Coulomb's lawTracking how two simultaneous changes combine

Two spheres carrying $6.5 \times 1 0^{- 7}$ C each are 50 cm apart. What happens to the force if each charge is doubled and the separation is halved?

Stuck? Show the approach

Work with scaling factors rather than recomputing the whole expression.

Show the full solution

The original force is $F = \frac{9 \times 1 0 ^{9} ( 6.5 \times 1 0 ^{- 7} ) ^{2}}{( 0.50 ) ^{2}} = 1.52 \times 1 0^{- 2}$ N. Doubling each charge multiplies the product $q_{1} q_{2}$ by $4$ ; halving $r$ divides $r^{2}$ by $4$ and so multiplies the force by a further $4$ . The net factor is $16$ , giving $F^{'} = 0.243$ N.

Answer: 0.243 N, a factor of 16 larger

The trap

Applying the factor of 2 only once, or forgetting that BOTH charges double, which gives 8 rather than 16.

JEE AdvancedSuperposition of two charged sheetsAdding sheet fields region by region

Two large parallel plates carry surface charge densities of equal magnitude $σ$ and opposite sign on their inner faces. Find the field in the two outer regions and between the plates.

Stuck? Show the approach

Each sheet gives $σ /2 ε_{0}$ on both sides; superpose in each of the three regions, tracking directions.

Show the full solution

A positive sheet's field points away from it and a negative sheet's towards it, each of magnitude $\frac{σ}{2 ε _{0}}$ . In either outer region the two contributions point oppositely and cancel, giving $E = 0$ . Between the plates both point from the positive plate to the negative one, so they add: $E = \frac{σ}{2 ε _{0}} + \frac{σ}{2 ε _{0}} = \frac{σ}{ε _{0}}$ .

Answer: Zero in both outer regions; sigma/epsilon_0 between the plates

The trap

Using sigma/epsilon_0 for a single sheet. One sheet gives sigma/(2 epsilon_0); the factor of 2 appears only after superposing both.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

It has been removed from the current edition, and not only here — a full-text search of all fourteen Class 12 Physics chapter PDFs finds zero occurrences of the phrase. Every chapter's questions now run contiguously from 1 upwards with no gaps, so the exercises were cut and the survivors renumbered. Nothing is missing from your copy of the book; this chapter's complete problem set is Exercises 1.1 to 1.23.

From the normal. Flux is the dot product of the field with the area vector, and the area vector points perpendicular to the surface. If a square's plane is parallel to the field, its normal is perpendicular to the field, so theta is 90 degrees and the flux is zero. Measuring from the plane instead swaps cos for sin and is the commonest error in Exercise 1.14.

No. Gauss's law gives the flux as the enclosed charge divided by epsilon_0, with no reference to the surface at all. Doubling the radius of a sphere around a point charge leaves the flux unchanged, because the field falls off as 1/r^2 while the area grows as r^2 and the two effects cancel exactly. Exercises 1.18 and 1.19 both make this point by giving dimensions that are not needed.

Not necessarily. Zero flux means the NET enclosed charge is zero. A box containing +5 microcoulomb and -5 microcoulomb has zero net charge and therefore zero net flux, while containing a great deal of charge. Gauss's law responds only to the algebraic sum.

Because the appropriate Gaussian surface is a cylinder, whose curved area grows as r rather than r^2. Applying Gauss's law then gives E = lambda/(2 pi epsilon_0 r). The 1/r^2 dependence belongs to point charges and spherically symmetric distributions, where the Gaussian sphere's area grows as r^2.

The forces on the two charges are +qE and -qE, equal in magnitude and opposite in direction, so they sum to zero and there is no translation. But they act at different points and so do not share a line of action, forming a couple. That couple produces a torque of magnitude p E sin(theta) which rotates the dipole into alignment with the field.

The tangent to a field line gives the direction of the field at that point. If two lines crossed, two different tangents could be drawn at the intersection, implying the field points in two directions at once. That is impossible, because the force on a test charge placed there is a single well-defined vector with one direction.

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Students who read Electric Charges and Fields usually read these next.

Electrostatic Potential and Capacitance

Potential as a scalar, potential due to point charges and dipoles, equipotential surfaces, potential energy, conductors and dielectrics, and capacitors in series and parallel with their stored energy. The Additional Exercises block has been removed, leaving 11 questions. CBSE Class 12 Physics Chapter 2.

Current Electricity

Drift velocity, Ohm's law and resistivity, the temperature dependence of resistance, internal resistance and terminal voltage, and Kirchhoff's rules applied to networks including the Wheatstone bridge. The Additional Exercises block has been removed, leaving 9 questions. CBSE Class 12 Physics Chapter 3.

Moving Charges and Magnetism

The Lorentz force, circular and helical motion in a magnetic field, the Biot-Savart law, Ampere's circuital law, the solenoid, the force between parallel currents that defines the ampere, and the torque on a current loop. Two topics the chapter's own text still promises — the cyclotron and the toroid — have been removed from it. CBSE Class 12 Physics Chapter 4.

Magnetism and Matter

The bar magnet and its field lines, the magnet as an equivalent solenoid, torque and potential energy of a magnetic dipole, Gauss's law for magnetism, magnetisation and magnetic intensity, and the classification of materials as diamagnetic, paramagnetic or ferromagnetic. The Earth's magnetism, permanent magnets and the hysteresis loop have all been removed. CBSE Class 12 Physics Chapter 5.

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