Electric Charges and FieldsNEET Physics · Class 12 · NCERT Chapter 1

Introduction

Class 12 Physics opens with electrostatics. This first chapter is about charges at rest, the forces between them, and the field they create around them. The next two chapters add potential, capacitance and currents. Almost half of the NEET Physics syllabus comes from electromagnetism, so the foundations you build here matter for the whole second year.

For NEET 2027 you can expect 1 to 2 questions from this chapter. The repeated favourites are: Coulomb's law for two or three charges, electric field at a point, dipole field on the axis vs equator, electric flux through a closed surface, and Gauss's law applications (line, plane, spherical shell). Lock these and you have most of the marks.

Electric charge and its properties

Electric charge is a basic property of matter, like mass. There are two kinds: positive and negative. SI unit: coulomb (C). Three properties to remember:

• Quantization: charge always comes in integer multiples of e = 1.6 × 10⁻¹⁹ C.
• Conservation: total charge of an isolated system stays fixed.
• Additivity: total charge of a body equals the sum of all individual charges on it.

Coulomb's law

The force between two stationary point charges $q_1$ and $q_2$ separated by distance r:

The force acts along the line joining the two charges. Like charges repel; unlike charges attract. The constant${ϵ}_0=8.854\times 10^{-12}{\text{C}}^2/({\text{N m}}^2)$ is called the permittivity of free space.

Force between two point charges. Positive charges in microcoulombs (μC); negative values give attraction.

Coulomb's law in a medium

Inside a medium with relative permittivity (dielectric constant) ${ϵ}_r$, the force is reduced by a factor of ${ϵ}_r$:

Principle of superposition

With many charges around, the force on a chosen charge is the vector sum of the forces from each of the others, computed independently:

Coulomb's law plus this rule is enough to handle any electrostatic problem with point charges, even though the algebra can get long.

Electric field

Electric field $\vec{E}$ at a point is the force per unit positive test charge placed at that point:

For a point charge q at distance r:

Direction: radially outward for positive q, radially inward for negative q. Once you know $\vec{E}$ at a point, the force on any charge q placed there is simply $\vec{F}=q\vec{E}$.

Field from a point charge falls off as 1/r². Direction is radially outward for positive charge, radially inward for negative charge.

Field of a system of charges

By superposition:

Always use vectors; the field can have components in any direction.

Two charges fixed at x = -0.5 m and x = +0.5 m. Move the test point and watch the resultant field vector.

Electric field lines

Field lines are imaginary curves drawn so that the tangent at every point gives the direction of the field there. Useful properties:

• Field lines start on positive charges and end on negative charges.
• The number of lines per unit area (perpendicular) is proportional to the field strength.
• Two field lines never cross (otherwise the field would have two directions at a point).
• For a positive point charge, lines go out radially. For negative, lines come in radially.
• Inside a conductor in electrostatic equilibrium, the field is zero, so no lines go through it.

Electric dipole

Two equal and opposite charges $+q$ and $-q$ separated by a small distance $2a$ form an electric dipole. The dipole moment vector:

Magnitude is $p=q\cdot 2a$. Unit: C·m. Many real molecules (HCl, water) have permanent dipole moments.

Field on the axis and on the equator of a dipole

For a point at distance r from the centre of the dipole (with r much greater than the dipole length):

• Axial (along the line of the dipole): $E=\frac{2kp}{r^3}$, along p.
• Equatorial (perpendicular bisector): $E=\frac{kp}{r^3}$, opposite to p.

Two takeaways: axial is twiceequatorial for the same r, and both fall off as 1/r³ (faster than a single point charge's 1/r²) because the two opposite charges partially cancel.

Field of an electric dipole at far distances (r much greater than dipole length 2 a). Two famous results: axial is twice as strong as equatorial, and both fall off as 1/r³.

Torque and PE of a dipole in a uniform field

Place a dipole in a uniform field $\vec{E}$ with angle theta between p and E. The two charges feel equal and opposite forces; the net force is zero, but the forces produce a couple:

Torque tries to rotate the dipole until p is parallel to E. The associated potential energy:

U is minimum at θ = 0 (stable), zero at 90°, and maximum at 180° (unstable). Work done in rotating from θ₁ to θ₂ in a uniform field:

Torque on a dipole in a uniform field tries to align p with E. Maximum torque at theta = 90°, zero at 0° and 180°.

Electric flux

Electric flux through a small flat patch of area dA is $d{\varphi }_E=\vec{E}\cdot d\vec{A}=EdA\cos \theta$, where theta is the angle between E and the normal to the surface. For a finite surface:

For a uniform field through a flat area A, this reduces to ${\varphi }_E=EA\cos \theta$. Flux is a scalar (signed). It measures the number of field lines piercing the surface.

Gauss's law

The total electric flux through any closed surface equals the total charge enclosed divided by ${ϵ}_0$:

Gauss's law is exact and always true. It is most useful when the charge distribution has enough symmetry (spherical, cylindrical or planar) that you can pull $E$ out of the integral.

Total electric flux through any closed surface depends only on the charge ENCLOSED, not on the shape or size of the surface.

Applications of Gauss's law

1) Infinite line of charge (linear density λ)

Use a coaxial cylindrical Gaussian surface. By symmetry, E is radial. Flux through the curved side is $E\cdot 2\pi rL$. Enclosed charge is $\lambda L$. So:

Field falls as 1/r (slower than a point charge).

2) Infinite plane sheet (surface density σ)

Use a cylinder pierced symmetrically through the plane. By symmetry, E is perpendicular to the plane and equal on both sides:

The field is uniform; it does not depend on the distance from the plane.

3) Spherical shell of charge Q

Use a concentric spherical Gaussian surface.

• Outside (r > R): $E=\frac{kQ}{r^2}$, the same as if all charge were concentrated at the centre.
• Inside (r < R): $E=0$. The shell provides perfect electrostatic shielding.

Three classic Gauss's law geometries plus the inside of a charged shell. Pick a shape, set the source density and distance.

Quick check: solid uniformly charged sphere

Outside: same as point charge. Inside (r < R): $E=(kQr)/R^3$, growing linearly with r from 0 at the centre to kQ/R² at the surface.

Worked NEET problems

Summary cheat sheet

• Coulomb: $F=k{q}_1{q}_2/r^2$, $k=9\times 10^9$.
• Field of point charge: $E=k|q|/r^2$.
• Dipole moment: $\vec{p}=q\cdot 2\vec{a}$, from -q to +q.
• Dipole, axial: $E=2kp/r^3$ along p.
• Dipole, equatorial: $E=kp/r^3$ opposite to p.
• Torque: $\vec{\tau }=\vec{p}\times \vec{E}$, $\tau =pE\sin \theta$.
• Dipole U: $U=-\vec{p}\cdot \vec{E}=-pE\cos \theta$.
• Flux: ${\varphi }_E=\int \vec{E}\cdot d\vec{A}$.
• Gauss: $\oint \vec{E}\cdot d\vec{A}=q_{\text{enc}}/{ϵ}_0$.
• Infinite line: $E=\lambda /(2\pi {ϵ}_{0}r)$.
• Infinite plane: $E=\sigma /(2{ϵ}_0)$.
• Outside shell: $E=kQ/r^2$. Inside: 0.

Next: try the interactive widgets for Coulomb's law, dipole field and Gauss applications, or work through the 32 NEET PYQs with full solutions. To time yourself, take the free 10-question mock test.