Electric Charges And Fields Class 12 Chapter Notes | Physics Chapter 1

Key Terms and Definitions

Electric Charge (q)

The intrinsic property of elementary particles of matter which gives rise to electric force between them. Its SI unit is the Coulomb (C).

Quantization of Charge

The principle that the charge on any body is an integral multiple of the basic unit of charge (e), which is the charge of an electron/proton. Formula: q = ne, where n is an integer and e = 1.6 x 10⁻¹⁹ C.

Coulomb's Law

States that the electrostatic force between two stationary point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

Electric Field Intensity (E)

The electrostatic force experienced by a unit positive test charge placed at a point in an electric field. E = F/q₀. Its SI unit is Newton per Coulomb (N/C).

Electric Dipole

A system of two equal and opposite point charges separated by a very small distance (2a).

Electric Dipole Moment (p)

The product of the magnitude of either charge and the distance between the two charges. p = q × 2a. It is a vector quantity directed from the negative to the positive charge.

Electric Flux (Φ)

A measure of the total number of electric field lines passing normally through a given surface. Φ = E ⋅ A = EA cosθ. Its SI unit is N m²/C or Volt-meter (V m).

Gauss's Law

States that the total electric flux through any closed surface (Gaussian surface) is equal to 1/ε₀ times the net charge enclosed by the surface. Φ_total = q_in / ε₀.

Formula Sheet: Chapter 1

Understanding Gauss's Law

Gauss's Law is a cornerstone of electrostatics, providing a powerful and elegant method to calculate the electric field of symmetric charge distributions. The law relates the electric flux through a closed surface to the net charge enclosed within it. The mathematical statement is Φ_E = ∮ E ⋅ dS = q_enclosed / ε₀. Here, Φ_E is the total electric flux, E is the electric field, dS is the differential area vector on the closed surface, and q_enclosed is the algebraic sum of all charges inside the surface. The constant ε₀ is the permittivity of free space.

The key to applying Gauss's Law is choosing an appropriate imaginary closed surface, called a Gaussian surface. The surface should be chosen such that the electric field E is either parallel to dS (making E⋅dS = E dS) or perpendicular to dS (making E⋅dS = 0), and the magnitude of E is constant over the surface. For a point charge or a sphere of charge, the Gaussian surface is a concentric sphere. For an infinite line of charge, it's a coaxial cylinder. For an infinite plane sheet of charge, it's a small cylinder or a cuboid piercing the sheet. It's crucial to remember that the electric field E in the equation is the net field due to all charges (both inside and outside the surface), but the flux is determined only by the enclosed charge.

Must-Remember Points

• {"point":"Basic Properties of Charge: Additivity, Quantization (q=ne), and Conservation."}
• {"point":"Charging Methods: Conduction (contact), Induction (no contact), and Friction (rubbing)."}
• {"point":"Electric field lines are imaginary curves that start from a positive charge and end on a negative charge. They never intersect."}
• {"point":"The density of electric field lines in a region represents the magnitude of the electric field."}
• {"point":"For a short dipole, the electric field at an axial point is twice the electric field at an equatorial point for the same distance (E_axial = -2 E_equatorial)."}
• {"point":"A dipole in a uniform electric field experiences a net torque (τ = p x E) but zero net force."}
• {"point":"In a non-uniform electric field, a dipole experiences both a net torque and a net force."}
• {"point":"Gauss's Law simplifies E-field calculation for symmetrical charge distributions only."}
• {"point":"The electric field inside a charged hollow spherical conductor (or shell) is always zero."}

Coulomb's Law vs. Newton's Law of Gravitation

Worked Example

• {"problem":"Calculate the force between two charges of 2 μC and -3 μC placed 30 cm apart in air.","solution":"Given: q₁ = 2 μC = 2 × 10⁻⁶ C, q₂ = -3 μC = -3 × 10⁻⁶ C, r = 30 cm = 0.3 m.\nUsing Coulomb's Law: F = k |q₁q₂| / r²\nF = (9 × 10⁹) × |(2 × 10⁻⁶) × (-3 × 10⁻⁶)| / (0.3)²\nF = (9 × 10⁹) × (6 × 10⁻¹²) / 0.09\nF = (54 × 10⁻³) / (9 × 10⁻²)\nF = 6 × 10⁻¹ = 0.6 N\nThe force is attractive because the charges are opposite."}

Board Exam Traps

Be very careful with vector notation, especially for Coulomb's Law and electric field calculations. When using the vector form F₁₂ = k(q₁q₂/r³) r₁₂, remember the position vector r points from the source charge to the test charge. A common mistake is getting the direction wrong. For Gauss's Law, clearly state your choice of Gaussian surface and justify why it's appropriate (symmetry). Also, remember that the E-field due to an infinite plane sheet (σ / 2ε₀) is independent of the distance from the sheet, which can be a tricky 1-mark question.

Quick Revision Check

• Why do electric field lines never cross each other? Because if they did, there would be two different directions of the electric field at the point of intersection, which is physically impossible.
• What is the SI unit and dimension of electric flux? SI unit: Newton meter squared per Coulomb (N m²/C) or Volt-meter (V m). Dimension: [ML³T⁻³A⁻¹].
• What is the net electric field inside a uniformly charged thin spherical shell? Zero. According to Gauss's law, since there is no charge enclosed within a Gaussian surface drawn inside the shell, the net flux and hence the net field is zero.
• What is the orientation of an electric dipole in a uniform electric field corresponding to stable equilibrium? Stable equilibrium occurs when the dipole moment vector p is aligned parallel to the electric field vector E (θ = 0°), where the torque is zero and potential energy is minimum.

Frequently Asked Questions

Frequently Asked Questions