Coulomb's Law Numericals for JEE Main: Full Guide

Coulomb's law is the foundation of electrostatics, and JEE Main tests it through force calculations, equilibrium positions, and the superposition of multiple charges. The numerical problems look varied, but they all reduce to applying F = kq₁q₂/r², resolving vector components, and setting net forces to zero. This guide drills every type systematically so you are never caught off-guard.

The Fundamental Formula and Constants

Coulomb's law: F = kq₁q₂/r², where k = 1/(4πε₀) = 9 × 10⁹ N·m²/C². Like charges repel, unlike charges attract. The force is along the line joining the charges. Always express charges in coulombs (1 μC = 10⁻⁶ C, 1 nC = 10⁻⁹ C) and distances in metres. The magnitude formula is straightforward; the challenge is always the vector direction in multi-charge problems. A clean free-body diagram drawn before writing any equation prevents 90% of sign errors. For the full electrostatics treatment, see our electrostatics complete guide.

Superposition Principle: Three-Charge Problems

The superposition principle states that the force on any charge due to multiple other charges is the vector sum of the individual forces. For three charges in a line, find F₁₂ and F₁₃ separately, noting their signs, then add algebraically. For charges not in a line, resolve each force into x and y components, sum the components, and find the resultant magnitude and direction. The symmetry of equilateral triangle and square configurations often zeroes out components, reducing the calculation dramatically — spot the symmetry before computing.

Equilibrium Position of a Third Charge

Place a third charge q₃ at position x between two charges q₁ (at origin) and q₂ (at distance d). For q₃ to be in equilibrium, the force from q₁ must equal and oppose the force from q₂: kq₁q₃/x² = kq₂q₃/(d−x)². This simplifies to q₁/x² = q₂/(d−x)², giving x = d√q₁/(√q₁ + √q₂). If q₁ and q₂ have the same sign, equilibrium lies between them; if opposite signs, it lies outside the smaller charge. Recognising which case you face before computing saves time.

Force in a Dielectric Medium

In a medium with dielectric constant K, Coulomb's law becomes F = kq₁q₂/(Kr²). The force reduces by factor K compared to free space. JEE problems may ask how deep in water (K = 80) a charge must be placed so that the force equals some fraction of the vacuum value — a direct application of this modified formula. Note that the medium reduces only the electrostatic force, not the charge magnitudes themselves.

Charged Spheres and Induced Charges

When two identical conducting spheres at different charges are brought into contact and then separated, charge redistributes equally: each sphere gets (q₁ + q₂)/2. Excess charge on a conductor resides on the outer surface, and on a sphere it distributes uniformly. These conducting-sphere problems appear frequently in JEE, often combined with the Coulomb force calculation before and after contact. Take a free mock test to work through these under exam conditions. For induced charge and field problems, connect to our electric field and conductors guide.