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

  • 1State the basic properties of electric charge
  • 2Apply Coulomb's law and the superposition principle
  • 3Calculate electric fields due to point charges and dipoles
  • 4Find torque and potential energy of a dipole in a field
  • 5Apply Gauss's law to symmetric charge distributions
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Why this chapter matters
Electrostatics is the foundation of all electrical phenomena. Coulomb's law, the electric field, the dipole, and Gauss's law explain the forces that hold matter together and let us calculate fields for symmetric charge distributions.

Before you start — revise these

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

Electric Charges and Fields

'Electric charge is the fundamental property of matter that gives rise to electric forces — the forces that hold atoms and molecules together.'

1. Chapter Overview

Electrostatics is the study of electric charges at REST. This chapter begins with the concept of electric charge, its properties (additivity, quantization, conservation), and the fundamental LAW governing the force between two point charges — COULOMB'S LAW. The concept of the ELECTRIC FIELD (the region around a charge where its influence is felt) is introduced, followed by the ELECTRIC DIPOLE (two equal and opposite charges separated by a small distance). Finally, GAUSS'S LAW — one of the most powerful tools in electrostatics — is presented and applied to calculate fields for symmetric charge distributions.


2. Electric Charge — Basic Properties

  • Charge: Quantized (q = ±ne, e = 1.6×10⁻¹⁹ C), CONSERVED (cannot be created or destroyed, only transferred), ADDITIVE (net charge = algebraic sum).
  • Conductors: Charges move freely. Insulators: Charges are bound.
  • Charging: By FRICTION (one body to another), by CONDUCTION (direct contact), by INDUCTION (without contact).

3. Coulomb's Law

  • Force between two point charges: F = k(q₁q₂)/r² where k = 1/(4πε₀) = 9×10⁹ N·m²/C²
  • Direction: ALONG the line joining the charges. LIKE charges REPEL, UNLIKE charges ATTRACT.
  • Coulomb's law in vector form: F₁₂ = k(q₁q₂/r²)·r̂₁₂ — 'pointing from the charge causing the force towards the charge experiencing the force.'
  • Principle of Superposition: Net force on a charge = VECTOR SUM of individual forces from all other charges.

Worked Example 1

Problem: Two point charges q₁ = +10 μC and q₂ = −5 μC are placed 20 cm apart. Find the magnitude and direction of the force on q₂. Solution: F = (9×10⁹)(10×10⁻⁶)(5×10⁻⁶)/(0.2)² = 9×10⁹×50×10⁻¹²/0.04 = 0.45/0.04 = 11.25 N. Force on q₂ is ATTRACTIVE towards q₁ (since q₂ is negative).


4. Electric Field (E)

  • Definition: E = F/q₀ — force per unit positive test charge.
  • Electric field due to a point charge: E = kq/r² (radially OUTWARD if q > 0, INWARD if q < 0).
  • Superposition: Net E = VECTOR SUM of fields from individual charges.

Electric Lines of Force

  • Imaginary lines showing the PATH a positive test charge would follow.
  • Properties: (1) Start at + charges, end at − charges. (2) NEVER CROSS. (3) Tangent at any point gives direction of E. (4) Closer lines → stronger field. (5) Perpendicular to surface of a conductor.

5. Electric Dipole

Dipole Moment

  • p⃗ = q × 2a⃗ (from −q to +q). SI unit: C·m.

Electric Field Due to a Dipole

  • Axial line (end-on): E = 2kp/r³ (along the dipole axis, in the SAME direction as p⃗).
  • Equatorial line (broadside-on): E = kp/r³ (OPPOSITE to p⃗).

Torque on a Dipole in a Uniform Field

  • τ⃗ = p⃗ × E⃗. Magnitude: τ = pE sin θ. 'The dipole tries to ALIGN itself with the field.'
  • Potential energy: U = −p⃗ · E⃗ = −pE cos θ.

6. Gauss's Law

  • Total electric flux through a closed surface: Φ_E = ∮ E⃗ · dA⃗ = q_enc/ε₀.
  • 'The flux through a closed surface depends ONLY on the charge enclosed — not on the shape of the surface or the distribution of charges outside.'

Applications of Gauss's Law

Charge DistributionElectric FieldKey
Infinite line chargeE = λ/(2πε₀r)Cylindrical Gaussian surface
Infinite plane sheetE = σ/(2ε₀)Cylindrical Gaussian surface (pillbox)
Spherical shell (outside)E = kQ/r²Spherical Gaussian surface
Spherical shell (inside)E = 0
Solid sphere (outside)E = kQ/r²Spherical Gaussian surface
Solid sphere (inside)E = kQr/R³Spherical Gaussian surface

7. Comparison Table: Coulomb's Law vs Gauss's Law

AspectCoulomb's LawGauss's Law
What it givesForce between chargesElectric flux and field
FormF = kq₁q₂/r²Φ = q_enc/ε₀
Best forPoint chargesSymmetric charge distributions
Vector or scalar?Vector equationScalar equation (but yields vector E with symmetry)
Fundamental natureExperimental lawDerived from Coulomb's law + superposition

8. Common Mistakes

  1. Using k = 1/(4πε₀) without its value: MEMORISE: k = 9×10⁹ N·m²/C². ε₀ = 8.85×10⁻¹² C²/N·m².
  2. Direction of dipole moment: p⃗ points from −q to +q, NOT from +q to −q.
  3. Gaussian surface choice: Must pass through the point where E is desired and have the SAME symmetry as the charge distribution.
  4. Flux depends only on enclosed charge: Charge OUTSIDE the Gaussian surface contributes ZERO net flux — a very common confusion.

9. CBSE Exam Focus

  1. Coulomb's law — numerical problems with superposition principle
  2. Electric field due to point charges — superposition
  3. Electric dipole — field at axial and equatorial points, torque
  4. Gauss's law — flux calculation, field due to symmetric distributions
  5. Electric field due to infinite line charge, plane sheet, spherical shell

10. Self-Test

Q1: Two charges 2 μC and −2 μC are placed 1 cm apart. Find the electric field at a point on the perpendicular bisector at a distance of 50 cm from the centre. A1: This is a dipole with q=2×10⁻⁶, 2a=0.01, r=0.5. Since r >> a, use equatorial formula: E = kp/r³ = (9×10⁹)(2×10⁻⁶×0.01)/(0.5)³ = (9×10⁹)(2×10⁻⁸)/0.125 = 180/0.125 = 1440 N/C (opposite to p).

Q2: An infinite line charge has linear charge density 5 μC/m. Find the electric field at a distance of 10 cm from it. A2: E = λ/(2πε₀r) = (5×10⁻⁶)/(2π×8.85×10⁻¹²×0.1) = (5×10⁻⁶)/(5.56×10⁻¹²) ≈ 9×10⁵ N/C.

Q3: Find the flux through a cube of side 0.1 m if a charge of 10 μC is placed at its centre. A3: By Gauss, Φ = q_enc/ε₀ = (10×10⁻⁶)/(8.85×10⁻¹²) = 1.13×10⁶ N·m²/C.

Q4: A uniformly charged conducting sphere of radius 0.1 m has total charge 5 μC. Find E at r=0.05 m and r=0.3 m. A4: Inside conductor: E=0. Outside: E=kQ/r²=(9×10⁹)(5×10⁻⁶)/(0.3)²=45000/0.09=5×10⁵ N/C.

Q5: An electric dipole is placed at an angle of 30° with a uniform electric field of 2×10⁴ N/C. If the dipole moment is 10⁻⁷ C·m, find the torque and potential energy. A5: τ = pE sin θ = (10⁻⁷)(2×10⁴)(sin 30°) = 2×10⁻³×0.5 = 10⁻³ N·m. U = −pE cos θ = −(10⁻⁷)(2×10⁴)(cos 30°) = −2×10⁻³×0.866 = −1.732×10⁻³ J.


11. Conclusion

Electrostatics is the FOUNDATION of all electrical phenomena:

  • COULOMB'S LAW: 'The force between charges decreases as the SQUARE of the distance.'
  • ELECTRIC FIELD: 'A region of influence — charges alter the space around them.'
  • DIPOLE: 'Two equal and opposite charges — the simplest neutral system with interesting properties.'
  • GAUSS'S LAW: 'A powerful symmetry tool — flux depends only on enclosed charge.'

'Electrostatics teaches us that charges are the source of electric fields, and fields transmit forces across empty space.'

Key formulas & results

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

Coulomb's law
F = k q1 q2 / r^2, k = 9e9 N m^2/C^2
Force along the line joining the charges.
Electric field of a point charge
E = k q / r^2
Radially outward for positive charge.
Dipole fields
Axial E = 2kp/r^3; equatorial E = kp/r^3
Dipole moment p points from -q to +q.
Gauss's law
Flux = q_enclosed / epsilon0
Depends only on enclosed charge.
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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
Pointing the dipole moment from +q to -q
The dipole moment vector points from the negative to the positive charge.
WATCH OUT
Thinking external charges contribute to Gaussian flux
Only the charge enclosed by the Gaussian surface affects the net flux; outside charges give zero net flux.
WATCH OUT
Adding electric fields as scalars
Electric field is a vector; use vector addition (superposition) for multiple charges.
WATCH OUT
Choosing a Gaussian surface without the right symmetry
Pick a surface matching the charge symmetry (spherical, cylindrical, planar) and passing through the field point.

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1MEDIUM· Dipole
Charges 2 uC and -2 uC are 1 cm apart. Find the field at 50 cm on the perpendicular bisector.
Show solution
This is the equatorial point with p = q(2a) = 2e-6 x 0.01 = 2e-8 C m. E = kp/r^3 = 9e9 x 2e-8 / 0.125 = 1440 N/C, opposite to p.
Q2MEDIUM· Line Charge
An infinite line charge has linear density 5 uC/m. Find the field at 10 cm.
Show solution
E = lambda/(2 pi epsilon0 r) = 5e-6/(2 pi x 8.85e-12 x 0.1) approximately 9e5 N/C.
Q3EASY· Gauss
Find the flux through a cube of side 0.1 m with 10 uC at its centre.
Show solution
Flux = q/epsilon0 = 10e-6/8.85e-12 = 1.13e6 N m^2/C, independent of the cube's size.
Q4MEDIUM· Sphere
A conducting sphere of radius 0.1 m carries 5 uC. Find E at r = 0.05 m and r = 0.3 m.
Show solution
Inside the conductor E = 0. Outside, E = kQ/r^2 = 9e9 x 5e-6 / 0.09 = 5e5 N/C.
Q5MEDIUM· Dipole Torque
A dipole (p = 1e-7 C m) is at 30 degrees to a field of 2e4 N/C. Find torque and potential energy.
Show solution
Torque = pE sin(theta) = 1e-7 x 2e4 x 0.5 = 1e-3 N m. U = -pE cos(theta) = -1e-7 x 2e4 x 0.866 = -1.73e-3 J.

5-minute revision

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

  • Charge is quantised, conserved, and additive; e = 1.6e-19 C.
  • Coulomb's law: F = k q1 q2 / r^2; like charges repel.
  • Electric field E = F/q0 = kq/r^2; use superposition.
  • Field lines start on + and end on - charges and never cross.
  • Dipole: axial E = 2kp/r^3, equatorial E = kp/r^3; torque = p x E.
  • Gauss's law: flux = q_enclosed/epsilon0.
  • Field is zero inside a conductor and inside a charged spherical shell.

CBSE marks blueprint

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

Typical chapter weightage: 6-8 marks across the chapter

Question typeMarks eachTypical countWhat it tests
Gauss's law applications3-51Flux and fields of symmetric distributions
Coulomb's law / field31Force and field with superposition
Dipole2-31Dipole field, torque, energy
Prep strategy
  • Memorise k and epsilon0 values
  • Use vector superposition for multiple charges
  • Choose Gaussian surfaces matching symmetry
  • Learn axial vs equatorial dipole formulas

Where this shows up in the real world

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

Electronics

Electrostatic forces and fields underlie the operation of every electronic device.

Photocopiers and printers

Charging and electric fields control toner placement in xerography and laser printing.

Shielding

Conductors with zero internal field provide electrostatic shielding (Faraday cages).

Exam strategy

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

  1. Draw a free-body-style diagram of charges and forces
  2. Add fields as vectors using components
  3. State the Gaussian surface and symmetry used
  4. Use axial/equatorial formulas correctly for dipoles

Going beyond the textbook

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

  • Derive the field of a uniformly charged solid sphere inside and outside.
  • Analyse the motion of a dipole in a non-uniform field (net force).

Where else this chapter is tested

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

CBSE Class 12 Physics examHigh
JEE Main and Advanced (Electrostatics)High
NEET PhysicsHigh

Questions students ask

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

Gauss's law states that the total flux equals the enclosed charge divided by epsilon0. Field lines from a charge inside the surface must all pass through it, contributing to the net flux. A charge outside the surface sends as many field lines into the surface as out of it, so its net contribution cancels to zero. Therefore the shape of the surface and any external charges do not change the net flux -- only the charge enclosed matters.

In a conductor the charges are free to move. If any field existed inside, it would push these free charges until they redistributed on the surface in a way that exactly cancels the internal field. At electrostatic equilibrium this rearrangement is complete, so the net field inside the conducting material is zero and all the excess charge resides on the outer surface.
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Last reviewed on 30 May 2026. Written and reviewed by subject-matter experts — read about our process.
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