Electric Charges and Fields

What this chapter is about

This chapter introduces electrostatics, the study of electric charges at rest and the forces and fields they create. You will learn how objects acquire charge, the fundamental properties of charge, and how charged bodies interact through Coulomb's law. The chapter then develops the powerful concept of the electric field, which describes how a charge modifies the space around it.

You meet this chapter at the start of Class 12 Physics because electrostatics forms the foundation for understanding electric potential, capacitors, current electricity and electromagnetism in later chapters. The mathematics here—vector addition, integration for continuous charge distributions, and the use of symmetry—prepares you for more advanced topics.

After studying this chapter, you should be able to calculate forces between point charges, sketch electric field lines for simple charge configurations, apply Gauss's law to find the electric field when symmetry permits, and explain everyday phenomena like lightning and electrostatic induction.

Key ideas

• Electric charge is a fundamental property of matter; it is quantised (occurs in multiples of e = 1.6 × 10⁻¹⁹ C) and conserved (total charge in an isolated system remains constant).
• Like charges repel, unlike charges attract; this interaction occurs through the electric field, not by direct contact.
• Coulomb's law states that the force between two point charges q₁ and q₂ separated by distance r is F = k q₁ q₂ / r², directed along the line joining them, where k = 9 × 10⁹ N m² C⁻² in vacuum.
• The electric field at a point is the force per unit positive test charge: E = F / q₀, with SI unit newton per coulomb (N C⁻¹) or equivalently volt per metre (V m⁻¹).
• Electric field lines start on positive charges, end on negative charges, never cross, and their density indicates field strength; they are perpendicular to the surface of a conductor in electrostatic equilibrium.
• For a continuous charge distribution, the net field is found by integrating the contributions dE from each infinitesimal charge element.
• Gauss's law relates the total electric flux through a closed surface to the charge enclosed: Φ = ∮ E · dA = q_enclosed / ε₀, where ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻² is the permittivity of free space.
• Gauss's law simplifies field calculations when the charge distribution has spherical, cylindrical or planar symmetry.

Formulas and facts to remember

• Coulomb's law: F = (1 / 4πε₀) × (q₁ q₂ / r²), with k = 1 / 4πε₀ ≈ 9 × 10⁹ N m² C⁻².
• Electric field of a point charge: E = k q / r², directed radially outward for positive q.
• Principle of superposition: the net force or field equals the vector sum of individual contributions.
• Electric flux through a surface: Φ = E · A = E A cos θ for a uniform field; otherwise integrate.
• Gauss's law: Φ = q_enclosed / ε₀.
• Field due to an infinite plane sheet of surface charge density σ: E = σ / 2ε₀, perpendicular to the sheet.
• Field just outside a charged conductor: E = σ / ε₀, perpendicular to the surface.
• Inside a conductor in electrostatic equilibrium, E = 0; any excess charge resides on the surface.

Worked examples

Example 1: Force between two charges

Two small spheres carry charges q₁ = +3.0 μC and q₂ = −2.0 μC and are held 0.30 m apart in air. Find the magnitude and nature of the force between them.

Solution

Convert charges: q₁ = 3.0 × 10⁻⁶ C, q₂ = 2.0 × 10⁻⁶ C.

Apply Coulomb's law: F = k |q₁ q₂| / r² F = (9 × 10⁹) × (3.0 × 10⁻⁶ × 2.0 × 10⁻⁶) / (0.30)² F = (9 × 10⁹) × (6.0 × 10⁻¹²) / 0.09 F = 54 × 10⁻³ / 0.09 = 0.60 N.

Since the charges are opposite, the force is attractive.

Example 2: Electric field on the axis of a ring

A thin ring of radius R = 0.10 m carries a uniformly distributed charge Q = 5.0 × 10⁻⁸ C. Calculate the electric field at a point P on the axis, 0.10 m from the centre.

Solution

At distance x from the centre along the axis, the field is: E = k Q x / (R² + x²)^(3/2).

Here x = 0.10 m, R = 0.10 m. R² + x² = 0.01 + 0.01 = 0.02 m². (R² + x²)^(3/2) = (0.02)^(1.5) = 0.02 × √0.02 ≈ 0.02 × 0.1414 = 2.83 × 10⁻³ m³.

E = (9 × 10⁹) × (5.0 × 10⁻⁸) × 0.10 / (2.83 × 10⁻³) E = 45 / (2.83 × 10⁻³) ≈ 1.6 × 10⁴ N C⁻¹, directed along the axis away from the ring.

Example 3: Using Gauss's law for a spherical shell

A thin metallic spherical shell of radius 0.20 m carries a charge of +8.0 × 10⁻⁹ C. Find the electric field (a) at a point 0.30 m from the centre, (b) at a point 0.10 m from the centre.

Solution

(a) Outside the shell (r = 0.30 m > R), treat the shell as a point charge at the centre. E = k Q / r² = (9 × 10⁹) × (8.0 × 10⁻⁹) / (0.30)² = 72 / 0.09 = 800 N C⁻¹, radially outward.

(b) Inside the shell (r = 0.10 m < R), the Gaussian surface encloses no charge. E = 0.

Common mistakes

• Using Coulomb's law without converting microcoulombs to coulombs → always write charges in SI base units before calculation.
• Adding electric fields as scalars instead of vectors → resolve each field into components, then add component-wise.
• Forgetting that the test charge in the definition of E must be vanishingly small so it does not disturb the source charge distribution → treat E as a property of the field, not of the test charge.
• Applying Gauss's law to situations without symmetry and assuming E is constant over the surface → Gauss's law always holds, but it only simplifies calculations when E can be factored out of the integral.
• Believing field lines can cross → at any point there is only one direction of the electric field, so lines never intersect.

Quick revision

• Charge is quantised (multiples of 1.6 × 10⁻¹⁹ C) and conserved.
• Coulomb's law: F = k q₁ q₂ / r²; force is along the line joining charges.
• Electric field E = F / q₀; field of a point charge falls as 1 / r².
• Gauss's law: total flux Φ = q_enclosed / ε₀; use it when symmetry makes E constant on a Gaussian surface.
• Inside a conductor in electrostatic equilibrium, E = 0; charges reside on the outer surface.