Electrostatics for JEE Main

Electrostatics opens the electricity half of JEE Main Physics, and almost everything after it leans on it. Capacitors, current and even magnetism reuse the same ideas: a field is force per unit charge, potential is energy per unit charge, and fields from several sources simply add. If you can add vectors cleanly and choose a good Gaussian surface, most questions here are short.

The official 2026 syllabus covers charge and its conservation, Coulomb's law and superposition (including continuous distributions), field lines, the electric dipole and its field and torque, flux and Gauss's law for an infinite wire, an infinite sheet and a thin spherical shell, potential of a point charge, a dipole and a system of charges, equipotentials, potential energy of two charges and of a dipole in a field, and conductors, dielectrics and polarisation. Capacitors, the last part of the unit, have their own guide.

Charge and Coulomb's law

Charge is conserved and quantised: any charge is q = ne, with e = 1.6 × 10⁻¹⁹ C. Two point charges in vacuum exert

F = kq₁q₂/r², with k = 1/(4πε₀) ≈ 9 × 10⁹ N m² C⁻²

along the line joining them. In a medium of dielectric constant K the force becomes F/K.

When several charges act on one, find each force separately and add them as vectors. This is the superposition principle, and it is where most marks are lost: students add magnitudes instead of components.

Electric field and field lines

The field is the force on a unit positive test charge, E = F/q₀. For a point charge, E = kq/r², pointing away from a positive charge and towards a negative one.

Field lines start on positive charges and end on negative ones, never cross, and are closer together where the field is stronger. Just outside a conductor they meet the surface at right angles.

Continuous charge: a ring. For a ring of charge Q and radius R, the field on its axis at distance x from the centre is

E = kQx / [(R² + x²)√(R² + x²)]

It is zero at the centre (by symmetry), largest at x = R/√2, and behaves like kQ/x² far away.

The electric dipole

Two charges +q and −q separated by 2a form a dipole of moment p = q × 2a, directed from −q to +q. For a distance r much larger than a:

In general, V = kp cos θ/r². Dipole fields fall as 1/r³, faster than a point charge's 1/r².

In a uniform field the net force on a dipole is zero, but there is a torque:

• τ = pE sin θ, which turns p towards E
• potential energy U = −pE cos θ
• work to turn it from θ₁ to θ₂: W = pE(cos θ₁ − cos θ₂)

θ = 0 is stable equilibrium and θ = 180° is unstable. Turning a dipole from 0 to 180° takes 2pE of work.

Gauss's law and its three standard uses

The flux through a closed surface is Φ = q_enclosed/ε₀, whatever the shape of the surface and wherever the charge sits inside it. Charges outside contribute zero net flux.

Gauss's law gives the field quickly only when symmetry makes E constant over parts of the surface. The syllabus names three cases:

Derivation for the line. Take a cylinder of radius r and length L around the wire. The flat ends have no flux, because E is radial. The curved surface gives E × 2πrL = λL/ε₀, so E = λ/(2πε₀r).

A uniformly charged solid sphere is not named in the syllabus, but the same method gives kQ/r² outside and kQr/R³ inside, rising linearly from zero at the centre.

Flux trick. A point charge q at the centre of a cube sends q/(6ε₀) through each face. At a corner of the cube, only one-eighth of its flux passes through the cube, q/(8ε₀) in total.

Potential and potential energy

Potential is work per unit charge in bringing a test charge from infinity. For a point charge, V = kq/r. Potential is a scalar, so for several charges you add the values with their signs: no components needed.

Field and potential are linked by E = −dV/dr. The field points towards falling potential.

• Equipotential surfaces are always perpendicular to field lines, and no work is done moving a charge along one.
• Thin shell: V = kQ/r outside and kQ/R everywhere inside, constant, even though E = 0 there.

The potential energy of two charges is U = kq₁q₂/r. For a group of charges, add kqᵢqⱼ/r over every pair once. The work an external agent does to assemble the group slowly equals U.

Conductors and dielectrics

In electrostatic equilibrium a conductor has:

• zero field inside its material
• all excess charge on its outer surface
• the same potential everywhere (its surface is an equipotential)
• a field just outside of σ/ε₀, normal to the surface

When two spheres of radii R₁ and R₂ are joined by a wire, their potentials become equal, so charge divides in proportion to radius and σ ∝ 1/R. Charge crowds at sharp points, which is why lightning conductors are pointed.

A dielectric has no free charges, but an applied field polarises it: tiny dipoles line up and reduce the field inside by the factor K. This is what raises a capacitor's capacitance.

Worked problems

Problem 1: where the force vanishes. Charges +4q and +q are fixed 3 m apart. Where must a third charge be placed so that the net force on it is zero? What charge must it have for all three to be in equilibrium?

Between the charges, at distance x from +4q: k(4q)/x² = kq/(3 − x)². Taking square roots, 2/x = 1/(3 − x), so 6 − 2x = x and x = 2 m (1 m from +q). This does not depend on the third charge.

For +q to be in equilibrium as well: k(4q)q/9 + kQq/1² = 0, so Q = −4q/9.

Problem 2: field of a line charge. A long straight wire carries 2 μC/m. Find the field 10 cm from it.

E = 2kλ/r = (2 × 9 × 10⁹ × 2 × 10⁻⁶)/0.1 = 36,000/0.1 = 3.6 × 10⁵ N/C, radially outward.

Problem 3: dipole in a field. A dipole of moment 2 × 10⁻⁸ C m sits in a uniform field of 5 × 10⁴ N/C. Find the maximum torque and the work needed to turn it from the stable position through 90° and through 180°.

τ_max = pE = 2 × 10⁻⁸ × 5 × 10⁴ = 10⁻³ N m.
To 90°: W = pE(1 − 0) = 10⁻³ J. To 180°: W = pE(1 − (−1)) = 2 × 10⁻³ J.

Problem 4 (numerical answer): assembling charges. Three charges of 1 μC each are placed at the corners of an equilateral triangle of side 10 cm. Find the work needed to assemble them from far apart, in mJ.

There are three pairs, each at 0.1 m: U = 3 × (9 × 10⁹ × 10⁻¹²)/0.1 = 3 × 0.09 = 0.27 J = 270 mJ.

Practice set

1. What is the electric field at the centre of a uniformly charged ring, and at what distance on its axis is the field largest?
2. A point charge q sits at the centre of a cube. Find the flux through one face.
3. How much work is needed to bring a 2 μC charge from infinity to 0.3 m from a fixed 3 μC charge?
4. Two conducting spheres of radii 1 cm and 3 cm are joined by a thin wire and charged. Find the ratio of their surface charge densities (small : large).
5. A dipole's field at distance r on its axis is E. What is the field at the same distance on its equatorial line?
6. An infinite sheet has charge density σ. Compare the field at 1 cm and at 1 m from it.
7. Charges +q, +q, −q, −q sit at the corners of a square of side a, the two positive charges on adjacent corners. Find the potential and the field at the centre.
8. A thin shell of radius R carries charge Q. Find the field and the potential at its centre.

Answers

1. Zero at the centre; largest at x = R/√2.
2. By symmetry, q/(6ε₀).
3. W = kq₁q₂/r = 9 × 10⁹ × 6 × 10⁻¹²/0.3 = 0.18 J.
4. Equal potentials give Q ∝ R and σ ∝ 1/R, so 3 : 1.
5. E/2, pointing opposite to the dipole moment.
6. The same, σ/(2ε₀), because the sheet is infinite.
7. V = 0 (equal distances, charges sum to zero). E is not zero: each charge gives 2kq/a², and they add to 4√2 kq/a², from the positive side towards the negative side.
8. E = 0; V = kQ/R.

What to do next

• Derive the fields of a line, a sheet and a shell from Gauss's law without looking.
• Solve 15 superposition problems with charges on triangles and squares, resolving into components each time.
• Practise dipole torque and energy questions until the sign of U = −pE cos θ is automatic.
• Move on to capacitors, then current electricity.