Electric Charges and Fields Class 12 Notes | CBSE Physics Chapter 1

Think of scuffing your feet on a carpet and then getting a tiny shock off a doorknob. That spark is electrostatics in action. In this chapter you will learn exactly why that happens, how to put a number on the force between charges, how to picture the invisible field around them, and how one elegant idea, Gauss’s law, lets you crack fields that would otherwise need heavy calculus. Work through the derivations and diagrams below and you will be able to answer any board question on Electric Charges and Fields with confidence.

Exam Weightage: how much does this chapter matter?

Electric Charges and Fields sits inside Unit I (Electrostatics) of the CBSE Class 12 Physics syllabus. Electrostatics as a whole carries a healthy chunk of the 70-mark theory paper, and this chapter is the foundation that Chapter 2 (Potential and Capacitance) builds on. Use the split below to decide where to spend your revision time.

High-yield tip: the 5-mark derivations from Gauss’s law are asked almost every year. Learn all three (line, sheet, shell) cold.

Key Concepts

1. Electric Charge

Electric charge is a fundamental property of matter that makes it experience a force in an electric field. Rub a glass rod with silk and it can attract tiny bits of paper: that attraction is the charge you just created by transferring electrons. There are two kinds:

• Positive charge: what a proton carries.
• Negative charge: what an electron carries.

Like charges repel and unlike charges attract. The SI unit of charge is the coulomb (C). The charge on a single electron is e = 1.6 × 10⁻¹⁹ C.

Three properties you must state correctly

• Quantisation: charge always comes in whole-number multiples of e. Any charge q = ne, where n is an integer (positive or negative). You can never have half an electron’s worth of charge.
• Conservation: the total charge of an isolated system stays constant. Charge is never created or destroyed, only transferred from one body to another. When you charge a rod by friction, the silk gains exactly the charge the rod loses.
• Additivity: the total charge of a body is the algebraic sum (with signs) of all the charges on it. A body with +5 C and −3 C carries a net +2 C.

Methods of charging

2. Coulomb’s Law

Coulomb measured how two point charges push or pull on each other. His result: the force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

F = k q₁q₂ / r²

• k = 1/(4πε₀) = 9 × 10⁹ N·m²/C² (Coulomb’s constant in vacuum)
• ε₀ = 8.854 × 10⁻¹² C²/(N·m²), the permittivity of free space
• Like charges repel, unlike charges attract, and the force acts along the line joining the two charges.

Inside a medium: F = k q₁q₂ / (κ r²), where κ (kappa) is the dielectric constant of the medium. Since κ is greater than 1, the force is always weaker in a medium than in vacuum.

Deriving the vector form (step by step)

Marks are often lost because students give only the magnitude. The vector form carries the direction automatically. Let charge q₁ sit at position r₁ and q₂ at position r₂.

Step 1. The vector pointing from q₂ to q₁ is r₁₂ = r₁ − r₂, and its length is the separation r = |r₁₂|.

Step 2. The unit vector along that direction is r̂₁₂ = r₁₂ / r.

Step 3. The force on q₁ due to q₂ points along this unit vector when the charges are alike (repulsion) and opposite to it when they are unlike (attraction). Both cases are captured in one line by keeping the signs of the charges:

F₁₂ = (1/4πε₀) · (q₁q₂ / r²) r̂₁₂

Step 4 (check the signs). If q₁q₂ is positive (like charges), F₁₂ points along r̂₁₂, that is away from q₂: repulsion. If q₁q₂ is negative (unlike charges), the force reverses: attraction. By Newton’s third law, F₂₁ = −F₁₂.

Superposition principle

When more than two charges are present, deal with them in pairs. The net force on any one charge is the vector sum of the separate forces from every other charge, each calculated by Coulomb’s law as if the others were absent:

F₁ = F₁₂ + F₁₃ + F₁₄ + …

This is why a charge sitting exactly between two equal like charges feels zero net force: the two pulls are equal and opposite.

3. Electric Field

Rather than talk about action at a distance, physics uses a field. The electric field at a point is the force experienced by a small unit positive test charge placed there.

E = F / q₀ = k Q / r²

The second form is the field a distance r away from a single point charge Q. The unit of electric field is N/C, which is the same as V/m. Electric field is a vector: it points away from a positive charge and towards a negative charge. We keep the test charge q₀ vanishingly small so it does not disturb the charge whose field we are measuring.

Electric field due to common configurations

4. Electric Field Lines

Field lines are imaginary curves drawn so that the tangent at any point gives the direction of the field there, and the crowding of the lines shows its strength. They turn an invisible vector field into a picture you can read at a glance.

Properties (state these exactly in exams)

• Field lines start on positive charges and end on negative charges.
• Two field lines never cross. If they did, the field would have two directions at one point, which is impossible.
• Where lines are closer together the field is stronger.
• They meet the surface of a conductor at right angles.
• They do not form closed loops (unlike magnetic field lines) and have no breaks in a charge-free region.

5. Electric Dipole

An electric dipole is a pair of equal and opposite charges, +q and −q, held a small distance 2a apart. Molecules like HCl and water behave as tiny dipoles, which is what makes water such a good solvent.

The dipole moment measures how strong the dipole is and which way it faces:

p = q × 2a

It is a vector of unit C·m that points from the negative charge to the positive charge.

Field on the axial line (derivation)

Take a point P on the axis of the dipole, a distance r from its centre. The +q is nearer (distance r − a) and the −q is farther (distance r + a).

Step 1. Field due to +q, pointing away from it (along p): E₊ = kq/(r − a)².

Step 2. Field due to −q, pointing towards it (opposite to p): E₋ = kq/(r + a)².

Step 3. The net axial field is their difference (they point opposite ways):

E = kq [ 1/(r − a)² − 1/(r + a)² ]

Step 4. Put over a common denominator: (r + a)² − (r − a)² = 4ar, so

E = kq · 4ar / (r² − a²)²

Step 5. Since p = 2aq, substitute 2aq for p:

E_axial = (1/4πε₀) · 2pr / (r² − a²)²

Step 6 (short dipole). For a point far away, r is much greater than a, so a² is negligible next to r²:

E_axial = 2kp / r³, directed along p.

Field on the equatorial line (derivation)

Now take a point P on the perpendicular bisector, a distance r from the centre. Each charge is the same distance √(r² + a²) from P, so the two fields have equal magnitude kq/(r² + a²).

Step 1. Resolve each field into a component along the axis and one perpendicular to it. By symmetry the perpendicular components are equal and opposite, so they cancel.

Step 2. The components along the axis (pointing from +q towards −q, that is opposite to p) add up. Each contributes a factor cosθ, where cosθ = a/√(r² + a²):

E = 2 · [ kq/(r² + a²) ] · a/√(r² + a²) = 2kqa / (r² + a²)^3/2

Step 3. Again p = 2aq, so

E_equat = kp / (r² + a²)^3/2, directed opposite to p.

Step 4 (short dipole). For r much greater than a: E_equat = kp/r³. Notice the axial field is exactly twice the equatorial field at the same distance.

Torque on a dipole in a uniform field (derivation)

Place the dipole in a uniform field E at an angle θ to the field. The +q feels a force qE one way and the −q feels qE the opposite way.

Step 1. The two forces are equal, opposite and not along the same line, so they form a couple. The net force is zero, which means the dipole does not move off, it only turns.

Step 2. The torque of a couple is force times the perpendicular distance between the two forces. That perpendicular distance is 2a sinθ:

τ = qE × (2a sinθ) = (q · 2a) E sinθ

Step 3. Since p = q · 2a,

τ = pE sinθ, or in vector form τ = p × E

The torque is maximum (pE) when θ = 90° and zero when θ = 0° or 180°. The dipole turns until it lines up with the field.

6. Electric Flux and Gauss’s Law

Electric flux through a surface tells you how many field lines pierce it. If the field E crosses an area A whose normal makes an angle θ with the field, then

Φ = E · A = EA cosθ

Its unit is N·m²/C (the same as V·m). Flux is largest when the surface faces the field square on (θ = 0) and zero when the surface lies along the field (θ = 90°).

Gauss’s Law

Gauss’s law is the shortcut that makes hard field problems easy whenever there is symmetry. It states that the total electric flux through any closed surface equals 1/ε₀ times the net charge enclosed by that surface:

∮ E · dA = q_enclosed / ε₀

The clever part: charges outside the surface contribute zero net flux, and the shape of the surface does not matter, only the charge inside. Pick a closed surface (a Gaussian surface) that matches the symmetry of the problem so that E is constant and either parallel or perpendicular to it everywhere. Here are the three classic applications, each derived in full.

(a) Field due to an infinite line charge

Step 1. A wire with linear charge density λ has cylindrical symmetry, so choose a coaxial cylinder of radius r and length l as the Gaussian surface.

Step 2. E is radial, so it is parallel to the flat end caps (no flux through them) and perpendicular to the curved surface. Flux = E × (curved area) = E · 2πrl.

Step 3. Charge enclosed by length l of wire = λl. Apply Gauss’s law:

E · 2πrl = λl / ε₀

Step 4. Cancel l and solve:

E = λ / (2πε₀r)

The field of a line charge falls off as 1/r, more slowly than a point charge.

(b) Field due to an infinite plane sheet

Step 1. A sheet with surface charge density σ has planar symmetry. The field points straight out from both faces. Choose a pill-box (a small cylinder) that pierces the sheet, with each flat face of area A.

Step 2. E is perpendicular to both flat faces and parallel to the curved side (no flux there). Flux through the two faces = EA + EA = 2EA.

Step 3. Charge enclosed = σA. Apply Gauss’s law:

2EA = σA / ε₀

Step 4. Cancel A and solve:

E = σ / (2ε₀)

Remarkably, this does not depend on distance: the field of an infinite sheet is uniform on each side.

(c) Field due to a uniformly charged thin spherical shell

Step 1. A shell of radius R carrying charge Q has spherical symmetry. Choose a concentric sphere of radius r as the Gaussian surface. E is radial and constant on it, so flux = E · 4πr².

Step 2 (outside, r > R). The whole charge Q is enclosed:

E · 4πr² = Q / ε₀  ⇒  E = kQ/r²

So from outside, the shell behaves as if all its charge sat at the centre.

Step 3 (on the surface, r = R). E = kQ/R².

Step 4 (inside, r < R). No charge is enclosed by the inner sphere, so:

E · 4πr² = 0 / ε₀  ⇒  E = 0

The field is zero everywhere inside a hollow charged shell or conductor. (For a uniformly charged solid sphere the inside field instead grows linearly, E = kQr/R³.)

7. Continuous Charge Distributions

Real charged objects are not single points; the charge is spread out. We describe that spread with a charge density and add up (integrate) the contributions of every tiny element dq.

• Linear charge density λ = charge per unit length (C/m). A charged wire uses dq = λ dl.
• Surface charge density σ = charge per unit area (C/m²). A charged sheet or plate uses dq = σ dA.
• Volume charge density ρ = charge per unit volume (C/m³). A charged solid uses dq = ρ dV.

The field of the whole body is the vector sum of the fields of all its elements:

E = (1/4πε₀) ∫ (dq / r²) r̂

Doing that integral directly is hard, which is exactly why Gauss’s law is such a gift whenever the distribution is symmetric.

8. Conductors in an Electrostatic Field

Metals are full of free electrons, and that gives a charged conductor at rest four properties you should be ready to prove or state.

• The field inside a conductor is zero. If any field remained, the free electrons would keep moving; they only stop once the internal field is cancelled everywhere.
• Any excess charge sits on the outer surface. Put a Gaussian surface just inside the metal: the field there is zero, so the enclosed charge must be zero, which forces all the charge to the surface.
• The field just outside is perpendicular to the surface and has magnitude σ/ε₀.
• The whole conductor is at one potential (you will use this in Chapter 2).

Field just outside a charged conductor (derivation)

Step 1. Take a small pill-box Gaussian surface with one flat face just outside the conductor and the other just inside, each of area A.

Step 2. Inside the conductor E = 0, so that face passes no flux. Outside, E is perpendicular to the surface, so only the outer face carries flux = EA.

Step 3. Charge enclosed = σA. Apply Gauss’s law: EA = σA/ε₀.

E = σ / ε₀

Note this is twice the field of an isolated sheet, because for a conductor all the field is pushed out to one side.

9. Two Useful Comparisons and an Application

Electrostatic shielding. Because the field inside a conductor is zero, the space inside a hollow conductor is completely protected from any outside electric field. This is called electrostatic shielding, and it is why you are safe inside a car during a lightning strike and why sensitive electronics sit inside metal cans. The hollow conductor acts as a Faraday cage.

Coulomb’s force compared with gravity. The two forces look alike on paper, but their strengths and behaviour differ sharply. Knowing the contrast is a common 1-mark or 2-mark question.

Both obey the inverse-square law and both act along the line joining the two bodies, which is why the mathematics of the two chapters rhymes so closely.

Important Definitions