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Class 12 Physics Notes

Electric Charges and Fields Class 12 Notes

Complete, exam-ready notes on electric charge, Coulomb's law, electric field and field lines, the electric dipole, flux and Gauss's law — written for CBSE boards, JEE and NEET revision.

Class12SubjectPhysicsCoversCBSE · JEE · NEET

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is an electric field in one line?

The electric field at a point is the force per unit positive test charge experienced there, a vector quantity directed away from a positive source charge and towards a negative one, given by E = F/q₀.

Electric Charge and Its Properties

Electric charge

An intrinsic property of matter (protons/electrons) responsible for electric force. The SI unit is the coulomb (C). Charge is quantised — it always exists in integer multiples of the elementary charge e=1.6×1019Ce = 1.6\times10^{-19}\,\text{C}.

  • Additive:\text{Additive}:
  • the net charge of a system is the algebraic sum of individual charges.
  • Conserved:\text{Conserved}:
  • charge can be transferred but never created or destroyed (net charge of an isolated system is constant).
  • Quantised:\text{Quantised}:
  • q=neq = ne
  • , where
  • nn
  • is an integer.
  • Invariant:\text{Invariant}:
  • charge does not depend on the speed of the charge.

Like repels, unlike attracts

Two like charges repel and two unlike charges attract. The force acts along the line joining the two charges.

Coulomb's Law

Coulomb's law

F=kq1q2r2=14πε0q1q2r2F = k\,\frac{q_1q_2}{r^2} = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}

The electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. k=9×109N m2C2k = 9\times10^9\,\text{N m}^2\,\text{C}^{-2} and ε0=8.85×1012C2N1m2\varepsilon_0 = 8.85\times10^{-12}\,\text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}.

Coulomb's law is valid only for point charges and obeys the principle of superposition: the net force on a charge is the vector sum of the forces from each other charge taken one at a time.

Force in a medium

In a medium of relative permittivity εr\varepsilon_r, the force reduces by the factor εr\varepsilon_r: F=14πε0εrq1q2r2F = \frac{1}{4\pi\varepsilon_0 \varepsilon_r}\frac{q_1q_2}{r^2}.

Electric Field and Field Lines

Electric field

E=Fq0\vec{E} = \frac{\vec{F}}{q_0}

The electric field at a point is the force experienced by a unit positive test charge placed there. For a point charge QQ: E=14πε0Qr2E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2} along the radial direction.

  • Field lines start on positive charges and end on negative charges.
  • Two field lines never cross each other.
  • The tangent to a field line at a point gives the direction of the electric field there.
  • The density of lines (closer spacing) indicates a stronger field.
  • Field lines are perpendicular to the surface of a conductor.
  • In a uniform field the lines are parallel and equally spaced.

Field lines and work

A field line is imaginary and is not a path of a charged particle (unless it also happens to coincide with the force direction under special conditions).

Electric Dipole

Electric dipole

p=qd\vec{p} = q\,\vec{d}

A pair of equal and opposite charges +q+q and q-q separated by a small distance dd. The dipole moment points from the negative to the positive charge, unit C·m.

Axial field (on the axis, far from the dipole): Ea=2kpr3E_a = \frac{2kp}{r^3}. Equatorial field: Ee=kpr3E_e = \frac{kp}{r^3}. On an external uniform field E\vec{E}, the dipole experiences a torque τ=p×E\vec{\tau} = \vec{p}\times\vec{E} but no net force, so it rotates to align with the field.

U=pE=pEcosθU = -\vec{p}\,\cdot\,\vec{E} = -pE\cos\theta
Potential energy of a dipole in a uniform field

Electric Flux and Gauss's Law

Electric flux

ΦE=EdA\Phi_E = \oint \vec{E}\,\cdot\,d\vec{A}

A measure of the number of electric field lines passing through a surface. It is a scalar, equal to the surface integral of EcosθdAE\cos\theta\,dA, with units N·m²/C.

Gauss's law

ΦE=EdA=qencε0\Phi_E = \oint \vec{E}\,\cdot\,d\vec{A} = \frac{q_{\text{enc}}}{\varepsilon_0}

The total electric flux through any closed surface equals the net charge enclosed by the surface divided by ε0\varepsilon_0. Charges outside the surface contribute zero net flux.

  • Infinite sheet:\text{Infinite sheet}:
  • E=σ2ε0E = \frac{\sigma}{2\varepsilon_0}
  • , independent of distance.
  • Solid conducting sphere:\text{Solid conducting sphere}:
  • E=0E = 0
  • inside;
  • E=14πε0Qr2E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}
  • outside (r > R).
  • Thin infinite line:\text{Thin infinite line}:
  • E=λ2πε0rE = \frac{\lambda}{2\pi\varepsilon_0 r}
  • .

Key Formulas

F=kq1q2r2F = k\,\frac{q_1q_2}{r^2}
Coulomb's law
E=14πε0Qr2E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}
Electric field of a point charge
E=σ2ε0E = \frac{\sigma}{2\varepsilon_0}
Field of an infinite sheet
Ea=2kpr3,Ee=kpr3E_a = \frac{2kp}{r^3},\quad E_e = \frac{kp}{r^3}
Dipole axial vs equatorial field
τ=pEsinθ,U=pEcosθ\tau = pE\sin\theta,\quad U = -pE\cos\theta
Dipole torque and potential energy
ΦE=EdA=qencε0\Phi_E = \oint \vec{E}\,\cdot\,d\vec{A} = \frac{q_{\text{enc}}}{\varepsilon_0}
Gauss's law

Solved Examples

Example: Two charges q1=2μCq_1 = 2\,\mu\text{C} and q2=3μCq_2 = 3\,\mu\text{C} are 10 cm apart. Find the magnitude of the force between them.

Solution: Using F=kq1q2/r2F = k\,q_1q_2/r^2: F=9×109×(2×106)(3×106)(0.1)2=5.4NF = 9\times10^9 \times \frac{(2\times10^{-6})(3\times10^{-6})}{(0.1)^2} = 5.4\,\text{N}.

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Coulomb's law

F=kq1q2r2F = k\,\frac{q_1q_2}{r^2}

Electric field (point charge)

E=14πε0Qr2E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}

Field of an infinite sheet

E=σ2ε0E = \frac{\sigma}{2\varepsilon_0}

Dipole axial field

E=2kpr3E = \frac{2kp}{r^3}

Dipole equatorial field

E=kpr3E = \frac{kp}{r^3}

Dipole torque

τ=pEsinθ\tau = pE\sin\theta

Dipole potential energy

U=pEcosθU = -pE\cos\theta

Gauss's law

ΦE=qencε0\Phi_E = \frac{q_{\text{enc}}}{\varepsilon_0}

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • Inside a conductor in electrostatic equilibrium the electric field is zero.
  • Charge is quantised (q = ne) and conserved.
  • Gauss's law: Φ = q_enc/ε₀ — use for sheet, sphere and line geometries.
  • Dipole axial field = 2 × equatorial field, both ∝ 1/r³.
  • Field lines never cross and start/end only on charges.

FAQ

Common questions

Is Coulomb's law an inverse-square law?

Yes — the force varies as 1/r². This inverse-square dependence is what makes Gauss's law valid and is confirmed experimentally to high precision.

Why is the field zero inside a conductor?

In electrostatic equilibrium, free charges redistribute on the surface so that the internal field cancels exactly. Any excess charge resides on the surface.

What is the SI unit of electric flux?

N·m²/C (also written V·m). It is a scalar quantity measuring the total field lines through a surface.

What is a dipole moment?

It is the product of the charge magnitude and the separation distance (p = qd), directed from negative to positive charge. Its unit is the coulomb-metre (C·m).

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