Electrostatics

Master electric charges, Coulomb's law, electric fields, potential, and capacitors for JEE Physics.

Electrostatics deals with electric charges at rest. It forms the foundation for understanding electricity and magnetism.

Overview

graph TD
    A[Electrostatics] --> B[Electric Charges]
    A --> C[Electric Field]
    A --> D[Electric Potential]
    A --> E[Capacitance]
    B --> B1[Coulomb's Law]
    B --> B2[Superposition]
    C --> C1[Field Lines]
    C --> C2[Gauss's Law]
    D --> D1[Potential Energy]
    D --> D2[Equipotential Surfaces]
    E --> E1[Parallel Plate]
    E --> E2[Combinations]

Electric Charges

Fundamental Properties

  1. Conservation of Charge: Total charge in an isolated system remains constant
  2. Quantization: Charge exists in discrete packets: $q = ne$ where $e = 1.6 \times 10^{-19}$ C
  3. Additive: Total charge is algebraic sum of individual charges

Types of Charging

  • Friction: Transfer of electrons between materials
  • Conduction: Direct contact transfer
  • Induction: Redistribution of charges without contact

Coulomb’s Law

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

$$\boxed{\vec{F} = k\frac{q_1 q_2}{r^2}\hat{r} = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}\hat{r}}$$

where:

  • $k = 9 \times 10^9$ N·m²/C²
  • $\varepsilon_0 = 8.85 \times 10^{-12}$ C²/N·m²

Vector Form

$$\vec{F}_{12} = k\frac{q_1 q_2}{r_{12}^2}\hat{r}_{12}$$
JEE Tip
The force is repulsive for like charges (positive $F$) and attractive for unlike charges (negative $F$). The direction is along the line joining the charges.

Superposition Principle

For multiple charges, the net force on a charge is the vector sum:

$$\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + ...$$

Electric Field

The electric field at a point is the force per unit positive test charge:

$$\vec{E} = \frac{\vec{F}}{q_0} = k\frac{Q}{r^2}\hat{r}$$

Electric Field Due to Various Charge Distributions

ConfigurationElectric Field
Point charge$E = \frac{kQ}{r^2}$
Infinite line charge$E = \frac{\lambda}{2\pi\varepsilon_0 r}$
Infinite plane sheet$E = \frac{\sigma}{2\varepsilon_0}$
Uniformly charged sphere (outside)$E = \frac{kQ}{r^2}$
Uniformly charged sphere (inside)$E = \frac{kQr}{R^3}$

Electric Field Lines

Properties:

  • Start from positive charges, end at negative charges
  • Never cross each other
  • Perpendicular to equipotential surfaces
  • Density indicates field strength

Interactive Demo: Electric Field Lines

Click the buttons to change the charges and see how electric field lines are affected:

Electric Dipole

An electric dipole consists of two equal and opposite charges $±q$ separated by distance $2a$.

Dipole Moment

$$\vec{p} = q \cdot 2a \cdot \hat{p}$$

Direction: From negative to positive charge

Electric Field Due to Dipole

On axial line (at distance $r$ from center):

$$E_{axial} = \frac{2kp}{r^3} \quad (r >> a)$$

On equatorial line:

$$E_{eq} = \frac{kp}{r^3} \quad (r >> a)$$

Torque on Dipole in Uniform Field

$$\boxed{\vec{\tau} = \vec{p} \times \vec{E}}$$ $$\tau = pE\sin\theta$$

Gauss’s Law

The electric flux through a closed surface equals the enclosed charge divided by $\varepsilon_0$:

$$\boxed{\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\varepsilon_0}}$$

Applications

graph LR
    A[Gauss's Law] --> B[Infinite Wire]
    A --> C[Infinite Plane]
    A --> D[Spherical Shell]
    A --> E[Solid Sphere]
  1. Infinite Line Charge ($\lambda$ = linear charge density):

    $$E = \frac{\lambda}{2\pi\varepsilon_0 r}$$
  2. Infinite Plane Sheet ($\sigma$ = surface charge density):

    $$E = \frac{\sigma}{2\varepsilon_0}$$
  3. Uniformly Charged Spherical Shell:

    • Outside: $E = \frac{kQ}{r^2}$
    • Inside: $E = 0$
Common Mistake
Gauss’s law is always true, but it’s useful for calculating fields only when there’s symmetry (spherical, cylindrical, or planar).

Electric Potential

Electric potential at a point is the work done per unit charge to bring a test charge from infinity:

$$V = \frac{W_{\infty \to P}}{q_0} = \frac{kQ}{r}$$

Relation with Electric Field

$$\vec{E} = -\nabla V = -\frac{dV}{dr}\hat{r}$$

For uniform field: $V = -\int \vec{E} \cdot d\vec{r}$

Potential Due to Various Systems

ConfigurationPotential
Point charge$V = \frac{kQ}{r}$
Electric dipole (axial)$V = \frac{kp\cos\theta}{r^2}$
System of charges$V = k\sum\frac{q_i}{r_i}$
Uniformly charged sphere (surface)$V = \frac{kQ}{R}$

Equipotential Surfaces

  • Surfaces with constant potential
  • $\vec{E}$ is perpendicular to equipotential surfaces
  • Work done along equipotential surface = 0

Electric Potential Energy

Two Point Charges

$$U = k\frac{q_1 q_2}{r}$$

System of Charges

$$U = k\sum_{iDipole in External Field$$U = -\vec{p} \cdot \vec{E} = -pE\cos\theta$$

Capacitors

A capacitor stores electric charge and energy.

Capacitance

$$C = \frac{Q}{V}$$

Unit: Farad (F)

Parallel Plate Capacitor

$$\boxed{C = \frac{\varepsilon_0 A}{d}}$$

With dielectric: $C = \frac{K\varepsilon_0 A}{d}$

Combinations

Series:

$$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + ...$$

Parallel:

$$C_{eq} = C_1 + C_2 + ...$$

Energy Stored

$$\boxed{U = \frac{1}{2}CV^2 = \frac{1}{2}QV = \frac{Q^2}{2C}}$$

Energy density: $u = \frac{1}{2}\varepsilon_0 E^2$

Practice Problems

  1. Two charges +4μC and -4μC are placed 20 cm apart. Find the electric field at the midpoint.

  2. A charge of 5μC is placed in a uniform field of 2000 N/C. Find the force on it.

  3. Three capacitors of 2μF, 3μF, and 6μF are connected in series. Find equivalent capacitance.

  4. Find the potential at a point 30 cm from a charge of 4nC.

Quick Check
Why is electric field zero inside a conductor? What happens to the field at the surface?

Further Reading