⚡Physics • Chapter Revision

Electrostatics Revision Notes

Quick facts, formula checklists, and concept reminders for JEE Main preparation.

8–12%
Weightage
~3 Q
Avg Qs
Hard
Difficulty

Chapter Overview

Electrostatics is the study of electromagnetic phenomena that occur when there are no moving charges. It introduces key concepts like electric force, fields, potentials, and capacitance, accounting for a high mark volume in JEE and NEET.

Theory & Concepts

1Coulomb's Law & Superposition

The electrostatic force between two point charges $q_1$ and $q_2$ separated by distance $r$ in vacuum is: $$F = k\frac{|q_1 q_2|}{r^2}$$ where $k = \frac{1}{4\pi\varepsilon_0} \approx 9 \times 10^9 \text{ N m}^2/\text{C}^2$. * **Superposition Principle**: The net force acting on any single charge due to a system of surrounding charges is the vector sum of all individual forces acting on it: $\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + \dots + \vec{F}_n$.

2Electric Field & Potential

* **Electric Field ($\vec{E}$)**: Force per unit charge: $\vec{E} = \frac{\vec{F}}{q_0}$. For a point charge, $E = \frac{kq}{r^2}$. * **Electric Potential ($V$)**: Work done per unit charge in bringing a test charge from infinity: $V = \frac{kq}{r}$. Potential is a scalar quantity. * **Relation between $E$ and $V$**: $\vec{E} = -\vec{\nabla}V$. In 1D: $E = -\frac{dV}{dx}$.

3Gauss's Law and Flux

Electric flux $\Phi$ through any closed surface is proportional to the enclosed net charge $Q_{enclosed}$: $$\Phi = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\varepsilon_0}$$ Gauss's Law is highly useful to compute electric fields for symmetric charge systems (spherical shell, infinite wire, infinite sheet).

Core Terminology

Permittivity of Free Space ($\varepsilon_0$)

A constant representing the capability of vacuum to permit electric fields, valued at $8.854 \times 10^{-12}\text{ C}^2/\text{N m}^2$.

Equipotential Surface

A surface where all points are at the same electric potential. No work is done in moving a charge along an equipotential surface, and electric field lines are always perpendicular to it.

Electric Dipole Moment

A vector pointing from the negative charge to the positive charge of a dipole, defined as $\vec{p} = q \cdot \vec{d}$.

Concept Application (Solved Examples)

Example 1: Question

Find the electric field at a distance of $3\text{ m}$ from a point charge of $+9\text{ }\mu\text{C}$ in vacuum.

Step-by-Step Solution:

1. **Identify Given Parameters**: - Charge, $q = 9 \times 10^{-6}\text{ C}$ - Distance, $r = 3\text{ m}$ - Constant, $k = 9 \times 10^9\text{ N m}^2/\text{C}^2$ 2. **Apply Electric Field Formula**: $$E = \frac{kq}{r^2} = \frac{(9 \times 10^9) \times (9 \times 10^{-6})}{3^2} = \frac{8.1 \times 10^4}{9} = 9 \times 10^3\text{ N/C}$$ 3. **Direction**: Points radially outward away from the positive charge.
Exam Tip: Always convert micro-coulombs ($\mu\text{C}$) to standard coulombs ($\text{C}$) before multiplying.

Common Mistakes & Pitfalls to Avoid

  • Treating Potential as a Vector: Potential is a scalar. When calculating potential from multiple charges, sum them algebraically (including signs), not vectorially.
  • Ignoring dielectric medium: If a medium with dielectric constant $K$ is introduced, the force and electric field decrease by a factor of $K$ ($F_{medium} = \frac{F_{vacuum}}{K}$), while capacitance increases ($C_{medium} = K \cdot C_{vacuum}$).

Syllabus Guidelines

Electric charges: conservation, Coulomb's law
Electric field due to point charge, electric dipole
Torque on dipole in uniform electric field
Electric flux, Gauss's law and its applications
Field due to infinitely long wire, infinite plane, thin spherical shell
Electric potential for point charge, dipole, system of charges
Potential difference, equipotential surfaces
Conductors and insulators, dielectrics, polarization
Capacitors, combination of capacitors, energy stored in capacitor