JEE Mains Physics · Electrostatics

Coulomb's Law and Equilibrium of Charges

Two point charges push or pull along the line joining them with F = kq₁q₂/r²; with several charges, add the forces as vectors, and a charge at rest has the electric force balanced by the others.

All Electrostatics notes JEE Mains Physics strategy

Why this matters

Twenty-two PYQs, fourteen of them multiple choice, and two from 2026. Nine use the force law itself: in a medium, in vector form, from a charged wire or ring, or set against gravity. Six touch identical spheres together or split a charge into two parts. Seven hold charges still: balls hanging from threads, a charge on an incline or a rough table, and a third charge placed where the forces cancel.

Concept 1 of 3: Coulomb's law and adding forces

Two point charges act along the line joining them: like charges repel, unlike charges attract. The force falls as the square of the distance. In a medium the force is smaller by the dielectric constant K, because the medium's own charges partly screen each charge. With many charges, each pair acts as if the others were not there, and the forces add as vectors.

Definition

$F = \dfrac{kq_1q_2}{r^{2}}$ , with $k = \dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}\ \text{N m}^{2}\,\text{C}^{-2}$ .
In a medium of dielectric constant K: $F_m = F/K$ . A separation r in the medium gives the same force as $r\sqrt{K}$ in vacuum.
Vector form, force on $q_2$ due to $q_1$ : $\vec F = \dfrac{kq_1q_2}{|\vec r_2 - \vec r_1|^{3}}(\vec r_2 - \vec r_1)$ . Keep the signs of the charges; they set the direction.
Several charges: add the forces as vectors. Two forces at right angles combine by Pythagoras.
A rod of length L and charge Q, with q on its line at distance a from the near end: $F = \dfrac{kqQ}{a(a + L)}$ .
A ring of charge Q and radius R with q at its centre: the ring is stretched, with tension $T = \dfrac{kqQ}{2\pi R^{2}}$ .
Electric and gravitational forces both fall as $1/r^{2}$ , so their ratio for two particles does not depend on the distance.

Coulomb's law in vector form

\vec F_{21} = \frac{kq_1q_2}{|\vec r_2 - \vec r_1|^{3}}\,(\vec r_2 - \vec r_1), \qquad F_{\text{medium}} = \frac{F}{K}

Worked example

A charge of

2\ \mu\text{C}

sits at

(1, 0, 2)

m and a charge of

5\ \mu\text{C}

(3, 2, 3)

m. Find the force on the

5\ \mu\text{C}

charge.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 8 Apr 2026 Shift 2 · Q16Moderate

Example 1 · Electrostatics · Coulomb's Law and Equilibrium of Charges

Two point charges

q_{1}= 3\mu C

and

q_{2}= - 4\mu C

are placed at points

(2\widehat{i}+ 3\widehat{j}+ 3\widehat{k})

and

(\widehat{i}+\widehat{j}+\widehat{k})

respectively. Force on charge

q_{2}

\_\_\_\_

N. (Take

\frac{1}{4\pi\epsilon_{0}}= 9 \times10^{9}

SI Units)

The vector points from the source to the target

The force on q₂ uses r₂ − r₁. Using r₁ − r₂ gives the force on q₁, which is the same size and points the other way. That reversed vector is usually among the options.

K divides the force; √K scales the distance

A medium cuts the force by K. To get the same force in vacuum, the charges must be √K times farther apart, not K times.

Forces add as vectors

Two forces at right angles of 3 N and 4 N give 5 N, not 7 N. Draw each force on the charge first, then add.

Drill 8 more on coulomb's law and adding forces

Concept 2 of 3: Sharing charge by contact

Touch two identical conductors and they end at the same potential, so they hold equal charges. The total charge, with signs, does not change. If the charges are unlike, they first cancel in part, then the rest is shared. For a fixed total Q split into two parts, the force is greatest when the parts are equal.

Definition

Two identical spheres touched: each ends with $\dfrac{q_1 + q_2}{2}$ , signs included.
An uncharged identical sphere touched to one holding q takes $q/2$ .
A sphere touched to several others in turn: track its charge after every contact, then use Coulomb's law on the final pair.
Splitting a total Q into q and $Q - q$ at a fixed distance: $F \propto q(Q - q)$ , greatest at $q = Q/2$ .
Spheres of different sizes touched end at the same potential, so they share in proportion to their radii, not equally.

Sharing and the largest force

q' = \frac{q_1 + q_2}{2}, \qquad F \propto q(Q - q) \ \text{is largest at}\ q = \frac{Q}{2}

Worked example

Two identical small spheres carry

+6\ \mu\text{C}

and

-2\ \mu\text{C}

and attract with a force F. They are touched together and put back in place. Find the new force in terms of F.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 29 July 2022 · Q91Moderate

Example 2 · Electrostatics · Coulomb's Law and Equilibrium of Charges

Two identical metallic spheres

A

and

B

when placed at certain distance in air repel each other with a force of

F

. Another identical uncharged sphere

C

is first placed in contact with

A

and then in contact with

B

and finally placed at midpoint between spheres A and B. The force experienced by sphere

C

will be:

Add the signs before halving

Spheres with +6 and −2 share +4, so each gets +2. Halving 6 + 2 = 8 gives the wrong charge and the wrong direction of force.

Order of contact matters

A sphere carries what it picked up into the next contact. Touching A then B gives a different result from B then A whenever A and B differ.

Equal sharing needs identical spheres

Two spheres of different radii end at a common potential, so the larger one takes more charge, in proportion to its radius.

Drill 5 more on sharing charge by contact

Concept 3 of 3: Charged bodies at rest

A charge at rest has no net force on it. A charged ball hanging from a thread has three forces: weight down, tension along the thread, and the electric push sideways. They close into a triangle, so the thread's angle from the vertical tells you the ratio of the electric force to the weight. On a line between two fixed charges, a third charge rests only where the two forces point opposite ways and match in size.

Definition

Hanging ball with the thread at θ to the vertical: $\tan\theta = \dfrac{F_e}{mg}$ , and $T = \sqrt{(mg)^{2} + F_e^{2}}$ .
Two equal balls on threads of length l, each at θ from the vertical, are $2l\sin\theta$ apart. For small angles, $x^{3} = \dfrac{q^{2}l}{2\pi\varepsilon_0 mg}$ , so $x \propto q^{2/3}$ .
In a liquid of density σ, the force falls by K and the weight by buoyancy. The angle stays the same only if $K = \dfrac{\rho}{\rho - \sigma}$ , with ρ the density of the balls.
On a smooth incline: $\dfrac{kq^{2}}{r^{2}} = mg\sin\theta$ . On a rough table at the point of slipping: $\dfrac{kq^{2}}{r^{2}} = \mu mg$ .
Third charge on the line of two fixed charges: like charges, it rests between them, nearer the smaller, at $x = \dfrac{r}{1 + \sqrt{q_2/q_1}}$ from $q_1$ . Unlike charges, it rests outside, beyond the smaller one.

Hanging balls and the liquid condition

\tan\theta = \frac{F_e}{mg}, \qquad K = \frac{\rho}{\rho - \sigma}

Worked example

Two identical balls of mass 30 g hang from one point on silk threads 50 cm long. They carry equal charges q, and each thread makes an angle θ with the vertical, where

\sin\theta = 0.6

. Find q. (g = 10 m/s²)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 30 Jan 2024 · Q117Moderate

Example 3 · Electrostatics · Coulomb's Law and Equilibrium of Charges

Two identical charged spheres are suspended by string of equal lengths. The string make an angle of

37^{\circ}

with each other. When suspended in a liquid of density

0.7\text{ }g/cm^{3}

, the angle remains same. If density of material of the sphere is

1.4\text{ }g/cm^{3}

, the dielectric constant of the liquid is _____

\left( \tan37^{\circ}=\frac{3}{4} \right)

Angle from the vertical, not between the threads

If the threads make an angle 2θ with each other, each makes θ with the vertical. Use θ in tan θ = F/mg.

A liquid changes two forces

The electric force falls by K and the weight falls by the buoyancy. Changing only one of them gives the wrong K.

The net force on a ball at rest is zero

When a question asks for the force on a ball in equilibrium, it means the electric force. The net force is zero by definition.

Drill 6 more on charged bodies at rest

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Coulomb's law and adding forces
Coulomb's law in vector form
$\vec F_{21} = \frac{kq_1q_2}{|\vec r_2 - \vec r_1|^{3}}\,(\vec r_2 - \vec r_1), \qquad F_{\text{medium}} = \frac{F}{K}$
Sharing charge by contact
Sharing and the largest force
$q' = \frac{q_1 + q_2}{2}, \qquad F \propto q(Q - q) \ \text{is largest at}\ q = \frac{Q}{2}$
Charged bodies at rest
Hanging balls and the liquid condition
$\tan\theta = \frac{F_e}{mg}, \qquad K = \frac{\rho}{\rho - \sigma}$

Watch out for (9)

The vector points from the source to the target
→ Coulomb's law and adding forces
K divides the force; √K scales the distance
→ Coulomb's law and adding forces
Forces add as vectors
→ Coulomb's law and adding forces
Add the signs before halving
→ Sharing charge by contact
Order of contact matters
→ Sharing charge by contact
Equal sharing needs identical spheres
→ Sharing charge by contact
Angle from the vertical, not between the threads
→ Charged bodies at rest
A liquid changes two forces
→ Charged bodies at rest
The net force on a ball at rest is zero
→ Charged bodies at rest

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