NCERT Solutions Class 12 Physics Chapter 1 Electric Charges and Fields

Master the fundamental concepts of electrostatics with our comprehensive NCERT Solutions Class 12 Physics Chapter 1 Electric Charges and Fields. Designed strictly in alignment with the latest CBSE syllabus and NCERT textbook guidelines, these step-by-step solutions provide conceptual clarity, complete mathematical derivations, and board-level problem-solving methodologies for Class 12 board aspirants and competitive test takers (JEE and NEET).

Chapter Overview

Chapter 1 of NCERT Class 12 Physics, Electric Charges and Fields, marks the foundation of Unit 1 (Electrostatics). Electrostatics deals with the study of forces, fields, and potentials arising from static electric charges. This chapter transitions students from microscopic charge properties to macroscopic field configurations and flux calculations.

The core syllabus covers the following major thematic areas:

• Electric Charge & Basic Properties: Quantization of charge (q = ±ne), conservation of electric charge, and additivity of charges.
• Coulomb’s Law: Electrostatic force between two point charges in scalar and vector forms, principle of superposition for multi-charge systems, and permittivity of free space (ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻²).
• Electric Field & Field Lines: Electric field intensity E due to a point charge, system of charges, physical significance and geometric properties of electric field lines.
• Electric Dipole: Dipole moment (p = q × 2a), electric field on axial and equatorial positions, torque acting on a dipole in a uniform electric field (τ = p × E), and potential energy of an electric dipole.
• Electric Flux & Gauss’s Law: Definition of electric flux (Φ = ∮ E · dA), statement and proof of Gauss’s Theorem (Φ = q_enclosed / ε₀).
• Applications of Gauss’s Law: Field due to an infinitely long straight uniformly charged wire, field due to a uniformly charged infinite plane sheet, and field due to a uniformly charged thin spherical shell (inside and outside).

Important Formulas and Theorems

Having instant recall of essential electrostatics formulas is vital for solving numerical problems and writing derivations accurately in CBSE board examinations:

• Quantization of Electric Charge:
  q = ± n e (where n = 1, 2, 3... and e = 1.6 × 10⁻¹⁹ C)
• Coulomb’s Law (in vacuum):
  F = (1 / 4πε₀) × (|q₁ q₂| / r²)
  where 1 / (4πε₀) ≈ 9 × 10⁹ N·m²/C²
• Electric Field Intensity of a Point Charge:
  E = (1 / 4πε₀) × (q / r²) r̂
• Electric Dipole Moment:
  p = q × 2a (directed from negative charge −q to positive charge +q)
• Electric Field due to a Short Dipole (r >> a):
  Axial line: E_axial = (1 / 4πε₀) × (2p / r³) (along p)
  Equatorial line: E_equatorial = (1 / 4πε₀) × (p / r³) (opposite to p)
  Relation: E_axial = 2 × E_equatorial
• Torque on a Dipole in Uniform Electric Field:
  τ = p × E = p E sin θ
• Gauss’s Law:
  Φ_E = ∮ E · dA = q_enclosed / ε₀
• Field due to Infinitely Long Straight Charged Wire (Linear Charge Density λ):
  E = λ / (2πε₀ r)
• Field due to Uniform Infinite Plane Sheet of Charge (Surface Charge Density σ):
  E = σ / (2ε₀) (independent of distance r)
• Field due to Uniformly Charged Thin Spherical Shell (Radius R, Total Charge q):
  Outside (r > R): E = (1 / 4πε₀) × (q / r²)
  On surface (r = R): E = σ / ε₀
  Inside (r < R): E = 0

Key Derivations Explained

1. Electric Field on the Axial Line of an Electric Dipole

Consider an electric dipole consisting of charges −q and +q separated by a distance 2a. Let P be a point on the axial line at a distance r from the centre O of the dipole.

1. Distance of P from +q = (r − a). Electric field at P due to +q: E₊ = [1 / (4πε₀)] × [q / (r − a)²] (directed away from +q).
2. Distance of P from −q = (r + a). Electric field at P due to −q: E₋ = [1 / (4πε₀)] × [q / (r + a)²] (directed towards −q).
3. Net Electric Field E_net = E₊ − E₋ = [q / (4πε₀)] × [1/(r − a)² − 1/(r + a)²]
4. Simplifying the algebraic term: [(r + a)² − (r − a)²] / (r² − a²)² = 4ar / (r² − a²)²
5. Therefore, E_net = [1 / (4πε₀)] × [2(q × 2a)r / (r² − a²)²] = [1 / (4πε₀)] × [2pr / (r² − a²)²]
6. For a short dipole where r >> a, neglecting a² gives: E_axial = [1 / (4πε₀)] × (2p / r³).

2. Electric Field due to an Infinitely Long Straight Charged Wire using Gauss’s Law

Consider an infinitely long thin wire having uniform linear charge density λ. To calculate the electric field at a perpendicular distance r:

1. Construct a coaxial cylindrical Gaussian surface of radius r and length l around the wire.
2. The total flux Φ through the cylinder consists of three surfaces: two circular flat caps and one curved cylindrical surface.
3. For the two circular end caps, the area vector dA is perpendicular to E (θ = 90°), so Φ_ends = ∫ E dA cos 90° = 0.
4. For the curved surface, E is parallel to dA (θ = 0°) everywhere, and |E| is constant: Φ_curved = E × (2πrl).
5. Total electric flux: Φ = E(2πrl).
6. By Gauss’s Law: Φ = q_enclosed / ε₀ = (λl) / ε₀.
7. Equating the expressions: E(2πrl) = (λl) / ε₀ ⇒ E = λ / (2πε₀r).

NCERT Step-by-Step Exercise Solutions

NCERT Exercise 1.1

Question: What is the force between two small charged spheres having charges of 2 × 10⁻⁷ C and 3 × 10⁻⁷ C placed 30 cm apart in air?

1. Given: Charge q₁ = 2 × 10⁻⁷ C, Charge q₂ = 3 × 10⁻⁷ C, Distance r = 30 cm = 0.3 m, Electrostatic constant k = 1 / (4πε₀) = 9 × 10⁹ N·m²/C².
2. To Find: Electrostatic force F between the two spheres.
3. Formula: F = k × (|q₁ q₂| / r²)
4. Substitution & Calculation:
   F = (9 × 10⁹ × 2 × 10⁻⁷ × 3 × 10⁻⁷) / (0.3)²
   F = (54 × 10⁻⁵) / 0.09 = 6 × 10⁻³ N
5. Answer: The electrostatic force between the spheres is 6 × 10⁻³ N (Repulsive in nature since both charges are positive).

NCERT Exercise 1.2

Question: The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge −0.8 μC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

1. Given: q₁ = 0.4 μC = 0.4 × 10⁻⁶ C, q₂ = −0.8 μC = −0.8 × 10⁻⁶ C, Force F = 0.2 N.
2. To Find: (a) Distance r, (b) Force on q₂ due to q₁.
3. Formula: F = (1 / 4πε₀) × (|q₁ q₂| / r²) ⇒ r² = (1 / 4πε₀) × (|q₁ q₂| / F)
4. Substitution & Calculation:
   (a) r² = (9 × 10⁹ × 0.4 × 10⁻⁶ × 0.8 × 10⁻⁶) / 0.2
   r² = (2.88 × 10⁻³) / 0.2 = 14.4 × 10⁻³ = 1.44 × 10⁻² m²
   r = √(1.44 × 10⁻²) = 0.12 m = 12 cm.
   (b) According to Newton’s third law of motion and Coulomb’s law, electrostatic force is mutually interactive and forms an action-reaction pair. Hence, the force on the second sphere due to the first is equal in magnitude and opposite in direction.
5. Answer: (a) The distance between the spheres is 12 cm (0.12 m). (b) The force on the second sphere is 0.2 N (Attractive, directed towards the first sphere).

NCERT Exercise 1.3

Question: Check that the ratio e² / (G m_e m_p 4πε₀) is dimensionless. Look up a table of physical constants and determine the value of this ratio. What does the ratio signify?

1. Dimensional Analysis:
   Unit of e² = C².
   Unit of 1 / (4πε₀) = N·m²/C².
   Unit of G = N·m²/kg².
   Unit of m_e m_p = kg².
   Dimension of numerator = (N·m²/C²) × C² = N·m².
   Dimension of denominator = (N·m²/kg²) × kg² = N·m².
   Ratio = [M⁰ L⁰ T⁰ A⁰] (Dimensionless quantity).
2. Numerical Value Calculation:
   e = 1.6 × 10⁻¹⁹ C, k = 9 × 10⁹ N·m²/C², G = 6.67 × 10⁻¹¹ N·m²/kg², m_e = 9.11 × 10⁻³¹ kg, m_p = 1.67 × 10⁻²⁷ kg.
   Ratio = [9 × 10⁹ × (1.6 × 10⁻¹⁹)²] / [6.67 × 10⁻¹¹ × 9.11 × 10⁻³¹ × 1.67 × 10⁻²⁷] ≈ 2.27 × 10³⁹.
3. Significance: This ratio represents the ratio of electrostatic force to gravitational force between an electron and a proton. Its immense value (≈ 10³⁹) signifies that gravitational forces are negligible in comparison to electrostatic forces at atomic dimensions.

NCERT Exercise 1.6

Question: Four point charges q_A = 2 μC, q_B = −5 μC, q_C = 2 μC, and q_D = −5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?

1. Given: Square of side a = 10 cm. Diagonal AC = BD = 10√2 cm. Distance of centre O from each corner r = d / 2 = 5√2 cm = 5√2 × 10⁻² m. Charge at centre q_O = 1 μC.
2. Force Analysis:
   Force F_A (due to q_A) is directed along OC (repulsive).
   Force F_C (due to q_C) is directed along OA (repulsive). Since q_A = q_C = 2 μC and distances are equal, |F_A| = |F_C|. These forces act along the same diagonal in opposite directions: F_A + F_C = 0.
   Similarly, force F_B (attractive towards B) and force F_D (attractive towards D) are equal in magnitude and opposite in direction: F_B + F_D = 0.
3. Net Force: F_net = F_A + F_B + F_C + F_D = 0 N.
4. Answer: The net electrostatic force on the 1 μC charge placed at the centre of the square is 0 N.

Official CBSE Board Exam Questions & Solutions

To support your holistic board exam revision, here are standard official CBSE question bank problems with exhaustive answers and step-by-step marking keys:

CBSE Question 1: Total Internal Reflection

Q: What is total internal reflection? State the conditions.
A: Total Internal Reflection (TIR) is the phenomenon in which a ray of light travelling at an angle of incidence greater than the critical angle from an optically denser medium to an optically rarer medium is completely reflected back into the denser medium.
Conditions for TIR:

1. The light ray must travel from an optically denser medium to an optically rarer medium.
2. The angle of incidence in the denser medium must be strictly greater than the critical angle (i > i_c) for the given pair of media.

CBSE Question 2: Lens Formula Numerical

Q: A concave lens of focal length 15 cm forms an image at 10 cm from the lens. Find the object distance.
A:

1. Given: For a concave lens, focal length f = −15 cm, image distance v = −10 cm (virtual image formed on the same side).
2. Formula: 1/v − 1/u = 1/f ⇒ 1/u = 1/v − 1/f
3. Substitution: 1/u = −1/10 − (−1/15) = −1/10 + 1/15 = (−3 + 2)/30 = −1/30
4. Answer: u = −30 cm. The object is placed 30 cm in front of the concave lens.

CBSE Question 3: Astronomical Telescope Magnifying Power

Q: An astronomical telescope has an objective of focal length 100 cm and eyepiece of 5 cm. Calculate its magnifying power when the final image is at infinity.
A:

1. Given: Focal length of objective f₀ = 100 cm, Focal length of eyepiece f_e = 5 cm.
2. Formula: Magnifying power in normal adjustment (image at infinity): M = f₀ / f_e
3. Calculation: M = 100 / 5 = 20×
4. Answer: The magnifying power of the telescope in normal adjustment is 20×.

CBSE Question 4: Ray Diagram for Convex Lens

Q: Draw a labeled ray diagram showing image formation by a convex lens when the object is placed between F₁ and 2F₁.
A:

1. Ray 1: Starting from the top of the object parallel to the principal axis passes through the principal focus F₂ on the other side after refraction.
2. Ray 2: Passing through the optical centre O goes straight without suffering any deviation.
3. The two refracted rays intersect at a point beyond 2F₂.
4. Nature of Image: Real, inverted, and magnified (enlarged), formed beyond 2F₂.

CBSE Question 5: Lens Maker’s Formula

Q: State the lens maker’s formula. Derive it for a convex lens.
A:

1. Statement: The lens maker’s formula relates the focal length of a lens to the refractive index of its material and the radii of curvature of its two surfaces: 1/f = (n₂/n₁ − 1)(1/R₁ − 1/R₂), where n₂ is the refractive index of lens material and n₁ is that of the surrounding medium. For a lens in air (n₁ = 1, n₂ = μ): 1/f = (μ − 1)(1/R₁ − 1/R₂).
2. Derivation: Consider refraction at two spherical refracting surfaces of radii R₁ and R₂.
   • Refraction at first surface: n₂/v₁ − n₁/u = (n₂ − n₁)/R₁
   • Refraction at second surface (image I₁ acts as virtual object): n₁/v − n₂/v₁ = (n₁ − n₂)/R₂ = −(n₂ − n₁)/R₂
3. Adding both surface equations: n₁(1/v − 1/u) = (n₂ − n₁)(1/R₁ − 1/R₂)
4. Since 1/v − 1/u = 1/f, we obtain: 1/f = (n₂/n₁ − 1)(1/R₁ − 1/R₂). Using Cartesian sign convention for a biconvex lens, R₁ is positive and R₂ is negative, yielding 1/f = (μ − 1)(1/R₁ + 1/R₂).

CBSE Question 6: Mirror Formula and Magnification

Q: Derive the mirror formula for a concave mirror. Define magnification.
A:

1. Derivation: For a concave mirror forming a real image A′B′ of an object AB placed beyond centre of curvature C:
   • From similar triangles ΔA′B′F and ΔMPF (where MP ≈ AB for small aperture): A′B′ / AB = B′F / FP = (v − f) / f.
   • From similar triangles ΔA′B′P and ΔABP: A′B′ / AB = B′P / BP = v / u.
   • Equating ratios: (v − f) / f = v / u ⇒ v/f − 1 = v/u.
   • Dividing throughout by v gives: 1/f = 1/v + 1/u.
2. Magnification: Linear magnification is defined as the ratio of the height of the image (h′) to the height of the object (h): m = h′/h = −v/u. A negative sign denotes a real and inverted image, while a positive sign signifies a virtual and erect image.

CBSE Question 7: Prism and Minimum Deviation

Q: What is a prism? Derive the expression for the angle of minimum deviation.
A:

1. Definition: A prism is a transparent optical medium bounded by two plane refracting surfaces inclined at a non-zero angle called the angle of prism (A).
2. Derivation: In quadrilateral AQNR of a triangular prism, A + ∠QNR = 180°. In ΔQNR, r₁ + r₂ + ∠QNR = 180°. Hence, A = r₁ + r₂.
3. Total deviation: δ = (i − r₁) + (e − r₂) = (i + e) − (r₁ + r₂) = i + e − A.
4. At the condition of minimum deviation (δ = δ_m), the refracted ray inside the prism travels parallel to the base, resulting in symmetry: i = e and r₁ = r₂ = r.
5. Thus: 2r = A ⇒ r = A / 2 and δ_m = 2i − A ⇒ i = (A + δ_m) / 2.
6. Applying Snell’s law at the first refracting interface: n = sin[(A + δ_m) / 2] / sin(A / 2).

Common Mistakes and Tips

• Vector Addition Overlooked: Electric field and electrostatic force are vector quantities. Never add magnitudes algebraically when multiple charges are present; resolve them into orthogonal components (x and y axes) or use the parallelogram law of vector addition.
• Dipole Moment Direction Confusion: Remember that in Physics, the electric dipole moment vector p points from −q to +q (whereas in Chemistry, dipole moments are conventionally depicted from positive to negative).
• Gauss’s Law Gaussian Surfaces: Gauss’s Law is always true for any closed surface, but it is useful for calculating electric fields only when the charge distribution exhibits high spatial symmetry (spherical, cylindrical, or planar). Ensure the electric field magnitude is constant over the Gaussian surface or perpendicular to the surface vector.
• Sign Convention in Coulomb’s Law: When substituting charges in F = k |q₁ q₂| / r², use absolute values for magnitude calculations and indicate the physical direction (attractive vs. repulsive) explicitly by checking the signs of the charges.
• Charge on a Conductor: In static equilibrium, excess charge resides strictly on the outer surface of a conductor. The electrostatic field inside an empty cavity or interior of a charged conductor is always zero (E = 0).

Board Exam Relevance

In the CBSE Class 12 Physics examination, Unit 1: Electrostatics (comprising Chapter 1: Electric Charges and Fields and Chapter 2: Electrostatic Potential and Capacitance) carries a combined weightage of 8 to 9 marks in the theory paper of 70 marks.

Question distribution typically includes:

• 1-Mark MCQs / Assertion-Reason: Properties of electric field lines, flux through closed geometries, quantization of charge, and ratio of forces.
• 2-Mark Short Answer Questions: Conceptual questions on dipole equilibrium, electrostatic shielding, and vector form of Coulomb’s law.
• 3-Mark Numerical & Derivations: Derivation of electric field on axial/equatorial positions of a dipole, application of Gauss’s law for infinite wire or infinite sheet.
• 5-Mark Long Answer / Case-Based Questions: Comprehensive derivation of Gauss’s law with numerical sub-parts, or analysis of multi-charge systems.

More NCERT Solutions and Practice

Consistent numerical practice and continuous self-assessment are key to scoring 95+ in Class 12 Physics. To test your conceptual understanding, generate customized chapter-wise mock tests, and download authentic CBSE previous year questions (PYQs) with step-by-step marking keys, visit qptool.theorify.in today.