CBSE Electric Charges And Fields Class 12 One Shot

CBSE Electric Charges And Fields Class 12 One Shot is an important revision topic for students preparing for the Central Board of Secondary Education Class 12 board examination scheduled on 28 February 2026. This chapter lays the foundation of Electrostatics and includes key concepts such as electric charge, Coulomb’s law, electric field, electric field lines, electric dipole, and Gauss’s law. Since numericals and derivations are commonly asked, a focused one-shot revision covering formulas and problem-solving methods can help students perform confidently and score high marks in the exam.

Electric Charges And Fields

The study of Electric Charges and Fields forms a foundational unit in electrostatics, dealing with stationary electric charges. This chapter is crucial for understanding fundamental principles like charge behavior, forces between charges, and the fields they generate. We will explore key concepts, derivations, and problem-solving techniques vital for competitive examinations.

1. Introduction to Electric Charge

Electrostatics is the study of stationary electric charges. An electric charge is a fundamental property of matter, observable through phenomena like rubbing a pipe with silk to attract a soda can, which involves electron transfer.

Types of Electric Charges

Charges are classified by convention:

• Positive Charge: Associated with protons.

• Negative Charge: Associated with electrons.

A body's charge depends on electron transfer:

• Positively Charged Body: Occurs when a body loses electrons.

• Negatively Charged Body: Occurs when a body gains electrons.

(Memory Tip: If you donate something, you are a positive person. Donating electrons makes a body positive. If you take something from someone, you are a negative character. Taking electrons makes a body negative.)

Symbol and Unit of Charge

• Symbol: q or Q.

• SI Unit: Coulomb (C).

• Fundamental Unit (e): 1.6 × 10⁻¹⁹ C.

• Charge of an electron (qₑ): -1.6 × 10⁻¹⁹ C

• Charge of a proton (qₚ): +1.6 × 10⁻¹⁹ C

2. Properties of Electric Charge

I. Additivity of Charge

Electric charge is a scalar quantity, allowing for algebraic addition. The total charge of a system equals the algebraic sum of its individual charges.

• Example: Q = (+2 C) + (+3 C) + (-2 C) = +3 C.

II. Conservation of Charge

The total charge of an isolated system remains constant. Charge can only be transferred, not created or destroyed.

• Example: When a pipe is rubbed with a silk cloth, electrons transfer, but the total charge of the pipe-silk system remains zero.

III. Quantization of Charge

The total charge on any body is always an integral multiple of the elementary charge, e. Charge exists in discrete packets.

• Formula: Q = ±ne

• n: integer (1, 2, 3, …)

• e: elementary charge (1.6 × 10⁻¹⁹ C)

IV. Interaction of Charges

• Like charges repel (e.g., + and +, or - and -).

• Unlike charges attract (e.g., + and -).

2. Worked Examples on Charge Properties

Example 1: Net Charge After Electron Transfer

Problem: An object has an initial charge of +1.0 C. It then gains 5 × 10¹⁸ electrons. What is the net charge?

Solution:

1. Initial Charge (Q_initial): +1.0 C.

2. Charge Gained (ΔQ): (5 × 10¹⁸) × (-1.6 × 10⁻¹⁹ C) = -0.80 C.

3. Net Charge (Q_net): 1.0 C + (-0.80 C) = +0.20 C.

Example 2: Mass Transferred (JEE Mains PYQ)

Problem: Polythene rubbed with wool develops -2 × 10⁻⁷ C. What mass is transferred? (Mass of electron = 9.1 × 10⁻³¹ kg).

Solution:

1. Number of electrons (n): |Q| / e = (2 × 10⁻⁷) / (1.6 × 10⁻¹⁹) = 1.25 × 10¹² electrons.

2. Total Mass: (1.25 × 10¹²) × (9.1 × 10⁻³¹ kg) ≈ 1.1 × 10⁻¹⁸ kg.

3. Methods of Charging

I. Charging by Friction

Rubbing two objects together causes electron transfer due to thermal energy. One object becomes positively charged (loses electrons), the other negatively charged (gains electrons).

II. Charging by Conduction (or Contact)

A neutral conducting object is brought into physical contact with a charged conductor. Charge flows and redistributes. If objects are identical, charge is equally shared. (Analogous to water leveling in connected containers).

III. Charging by Induction

Charging a neutral object without any physical contact.

1. Polarization: A charged object brought near a neutral conductor attracts/repels free electrons, creating charge separation.

2. Grounding: The conductor is connected to Earth, allowing electrons to flow to/from Earth to neutralize one side.

3. Removing Ground: The ground connection is removed while the charging object is still present, trapping the charge.

4. Removing Rod: The charging object is removed, leaving the conductor with a net charge of opposite polarity.

• Example: The Gold-Leaf Electroscope works on induction.

Coulomb's Law

Coulomb's Law quantifies the electrostatic force between two stationary point charges:

1. Directly proportional to the product of charge magnitudes ($F \propto q_1q_2$).

2. Inversely proportional to the square of the distance ($F \propto \frac{1}{r^2}$).

Formula: $F = K \frac{q_1q_2}{r^2}$

The force's direction is along the line joining the two charges.

Principle of Superposition

When multiple charges interact, the force between any two charges is unaffected by other charges. The net force on any charge is the vector sum of individual forces exerted by all other charges.

Force on a Charged Particle in an Electric Field

A charged particle (q) in an electric field (E) experiences a force and accelerates.

• Force: F = qE

• Acceleration: a = qE / m

Electric Dipole and Dipole Moment

• Electric Dipole: Two equal and opposite point charges (+q and -q) separated by a small distance (2l).

• Electric Dipole Moment (p): A vector quantity measuring dipole strength.

• Formula: p = q × (2l)

• 2l is the total separation distance.

• SI Unit: Coulomb-meter (C m).

• Direction: From negative charge to positive charge.

Derivation: Electric Field on the Equatorial Line of a Dipole

To find the electric field at point P on the perpendicular bisector (equatorial line), distance r from the center O:

1. Setup: Charges -q and +q are separated by 2l. Distances from P: √(r² + l²). Magnitudes of fields E₁ and E₂ are equal: E = kq / (r² + l²).

2. Vector Components: Vertical components cancel. Horizontal components (E cosθ) add up.

3. Net Field: E_net = 2Ecosθ. cosθ = l / √(r² + l²).

4. Substitution: E_net = 2 * [kq / (r² + l²)] * [l / (r² + l²)^(1/2)] = k * (q * 2l) / (r² + l²)^(3/2).

5. Ideal Dipole (r >> l): Neglect l². E_equatorial = kp / r³.

• Direction: Opposite to the dipole moment.

Worked Example: Net Dipole Moment

Problem: Dipole d1: -q at (0,0), +q at (a,0). Dipole d2: -q at (0,a), +q at (0,2a). Find net dipole moment.

Solution:

1. P₁ (for d1): Points along x-axis. P₁ = (qa) î.

2. P₂ (for d2): Points along y-axis. P₂ = (qa) ĵ.

3. P_net: Vector sum. P_net = P₁ + P₂ = qa(î + ĵ).

Torque on an Electric Dipole in a Uniform Electric Field

Torque (τ) is the rotational effect of force.

(Memory Tip: "Moment" implies multiplication by distance: Dipole Moment = Charge × Distance; Moment of Force (Torque) = Force × Perpendicular Distance.)

Derivation of Torque

1. Dipole (+q, -q, length 2L) in uniform field E at angle θ.

2. Forces on charges are F = qE (equal and opposite), forming a couple.

3. Torque τ = Force × Perpendicular Distance = (qE) × (2L sinθ).

4. Using p = q × 2L, scalar torque is τ = pE sinθ.

Vector Form and Direction

• Vector Form: τ = p × E

• Direction: Found using the right-hand rule.

Electric Flux

Electric flux (Φ) is the measure of electric field lines passing normally through a surface.

• Area Vector (A): Perpendicular to the surface.

• Formula: Φ = E ⋅ A = EA cosθ

• θ is the angle between the electric field vector (E) and the area vector (A) (normal).

• Scalar Quantity.

• Integral Form: Φ = ∫ E ⋅ dA (for non-uniform fields or curved surfaces).

• SI Unit: N m²/C.

Electric Field of an Infinite Plane Sheet

This derivation is crucial, especially for capacitors.

1. Conceptual Setup: Infinite sheet with surface charge density σ.

2. Gaussian Surface: Cylindrical Gaussian surface piercing the sheet, axis perpendicular to the sheet.

3. Flux Calculation: Flux only passes through the two flat faces (E || dA). Flux through curved surface is zero (E ⊥ dA). Total flux = 2EA (where A is face area).

4. Enclosed Charge: q_enclosed = σA.

5. Gauss's Law: 2EA = σA / ε₀.

6. Result: E = σ / (2ε₀). (Uniform field, independent of distance).

Application Problem: Net Electric Field of Three Parallel Sheets

Problem: Find the net electric field at a point between three parallel sheets: +2σ, -2σ, +σ.

Solution:

1. Magnitudes: E₁ (from +2σ) = σ / ε₀. E₂ (from -2σ) = σ / ε₀. E₃ (from +σ) = σ / 2ε₀.

2. Directions: Assume point is between sheets 2 and 3.

• E₁ (from +2σ, left of point): To the right (repulsive).

• E₂ (from -2σ, right of point): To the left (attractive).

• E₃ (from +σ, right of point): To the right (repulsive).

3. Net Field: E_net = E₁ - E₂ + E₃ = (σ / ε₀) - (σ / ε₀) + (σ / 2ε₀) = σ / (2ε₀).

Electric Field of a Uniformly Charged Spherical Shell

A spherical shell is hollow; charge resides only on its outer surface. (Memory Tip: Like a football, empty inside.)

Total charge Q, radius R.

Case 1: Point Inside the Shell (r < R)

1. Gaussian Surface: Concentric sphere of radius r < R.

2. Enclosed Charge: q_enclosed = 0.

3. Result: E = 0. (No electric field inside a charged spherical shell).

Case 2: Point on the Surface of the Shell (r = R)

1. Gaussian Surface: Sphere of radius r = R.

2. Enclosed Charge: q_enclosed = Q.

3. Result: E = (1 / 4πε₀) * (Q / R²). (Maximum field).

Case 3: Point Outside the Shell (r > R)

1. Gaussian Surface: Concentric sphere of radius r > R.

2. Enclosed Charge: q_enclosed = Q.

3. Result: E = (1 / 4πε₀) * (Q / r²). (Behaves like a point charge at center).

Application Problems & Previous Year Questions (PYQs)

Below is the Application Problems & Previous Year Questions (PYQs) section, which focuses on exam-oriented numericals and concept-based questions frequently asked in the CBSE Class 12 board examinations. Practicing these questions helps students strengthen problem-solving skills, improve accuracy, and understand the latest exam pattern and question trends.

Question 1: Net Electric Field at Vertex C

Problem: Two identical negative charges (-q) are at vertices A and B of an equilateral triangle ABC. What is the direction of the net electric field at vertex C?

Solution: The fields from A and B at C are attractive, along CA and CB respectively. The resultant points along the line bisecting the angle ACB, towards the midpoint of AB. Answer: Along CM (where M is midpoint of AB).

Question 2: Conductor Shielding

Problem: Two point charges (Q) at A and B create field E at P. If a hollow conducting sphere is placed at P, what is the new electric field at P?

Solution: The electric field inside a conductor is always zero. By placing the conducting sphere, P is inside a conductor. Answer: Zero.

Question 3: Acceleration vs. Distance Graph

Problem: Charge 'q' is brought near fixed 'Q' and released. What is the relationship between its acceleration (a) and distance (r) from Q?

Solution: F = kQq/r². Since F = ma, then a = F/m = (kQq/m) * (1/r²). So, a ∝ 1/r². This is a hyperbolic decay.

Question 4: Conservation of Charge

Problem: Ball B3 contacts B1 (-7 pC) and B2 (+4 pC), then separates. Final charge on each of the three balls is -2 pC. Initial charge on B3?

Solution:

1. Initial total (B1+B2) = -7 + 4 = -3 pC.

2. Final total (B1'+B2'+B3') = -2 + (-2) + (-2) = -6 pC.

3. By conservation: Initial (B1+B2) + Initial (B3) = Final Total.
   -3 pC + q_B3 = -6 pC.
   q_B3 = -3 pC.

Question 5: Flux Calculation (5-Mark PYQ)

Problem: Shell S1 has charges (-3μC, -2μC, +9μC). Enclosed by S2. Point charge Q between S1 and S2. Flux through S2 is 4 times flux through S1. Find Q.

Solution:

1. q_enclosed_S1 = (+9 - 3 - 2) μC = 4 μC. So, Φ₁ = 4μC / ε₀.

2. q_enclosed_S2 = q_enclosed_S1 + Q = (4 + Q) μC. So, Φ₂ = (4 + Q)μC / ε₀.

3. Given Φ₂ = 4Φ₁: (4 + Q) / ε₀ = 4 * (4 / ε₀) \implies 4 + Q = 16 \implies \textbf{Q = 12 μC}.

Question 6: Null Point (5-Mark PYQ)

Problem: Two charges +1μC and +4μC are 30 cm apart. At what distance from +1μC is the net electric field zero?

Solution: Let distance be x from +1μC. Then (30-x) from +4μC.

k * (1μC) / x² = k * (4μC) / (30 - x)².

1 / x = 2 / (30 - x) \implies 30 - x = 2x \implies 3x = 30 \implies \textbf{x = 10 cm}.

Question 7: Kinematics & E-Fields (5-Mark PYQ)

Problem: Particle (q=2μC, m=1.6g) with u = 4**i** m/s. Enters E = (80**i** + 60**j**) N/C. Find velocity vector at t=5s.

Solution:

1. F = q**E** = (2x10⁻⁶)(80**i** + 60**j**) = (1.6x10⁻⁴i + 1.2x10⁻⁴j) N.

2. a = F/m (m=1.6x10⁻³ kg) = (0.1**i** + 0.075**j**) m/s².

3. v = u + a**t = 4**i + (0.1**i** + 0.075**j**)5 = 4**i* + 0.5**i** + 0.375**j** = (4.5 i + 0.375 j) m/s.

Question: Properties of Electric Field Lines

Problem: Identify the incorrect statement: Field at A > at B (lines denser at A). Field at C is zero (no lines). Field points away from positive charge. Field is two-dimensional.

Solution: Electric fields are three-dimensional. The incorrect statement is that the field is two-dimensional.

Question: Flux Through Cubes with Shared Edge Charge

Problem: Cubes C1, C2 share vertex. 4q at vertex. 6q inside C2. Ratio of flux through C1 to C2?

Solution:

1. Flux for C1 (due to 4q at vertex): Φ_C1 = (1/4) * (4q / ε₀) = q / ε₀.

2. Flux for C2 (due to 4q at vertex + 6q inside C2): Φ_C2 = (1/4) * (4q / ε₀) + (6q / ε₀) = q / ε₀ + 6q / ε₀ = 7q / ε₀.

3. Ratio: Φ_C1 / Φ_C2 = (q / ε₀) / (7q / ε₀) = \textbf{1/7}.
   (Note: The solution provided in the input calculates Φ_C2 = (1/4) * (10q / ε₀), which implies 6q was also at the edge. However, based on the problem statement, 6q is inside C2 and 4q is at the vertex. The corrected calculation is presented above.)

Question: Flux Through a Sphere with Internal and External Charges

Problem: Charges -2q at (a,0,0) and +q at (4a,0,0). Flux through sphere of radius 3a centered at origin?

Solution: The sphere encloses -2q (at x=a < 3a) but not +q (at x=4a > 3a). By Gauss's Law, flux depends only on enclosed charge.

Φ = q_enclosed / ε₀ = \textbf{-2q / ε₀}.

Question: Calculating Charge from Torque on a Dipole

Problem: Dipole (length 2 cm) in E = 2 x 10⁵ N/C at 30°. Torque = 4 Nm. Find charge on dipole.

Solution:

1. τ = pE sinθ and p = q * (2l). So τ = q * (2l) * E * sinθ.

2. q = τ / [(2l) * E * sinθ].

3. q = 4 / [(2 x 10⁻² m) * (2 x 10⁵ N/C) * sin(30°)] = 4 / [(2 x 10⁻²) * (2 x 10⁵) * (1/2)].

4. q = 4 / (2 x 10³) = 2 x 10⁻³ C = \textbf{2 mC}.

Question: Flux Through a Sheet in a Non-Uniform Field

Problem: Square sheet side 'A' parallel to XY plane at z=A. E = C * z² k̂. Find flux.

Solution:

1. At z=A, E = C * A² k̂.

2. Area vector A = A² k̂ (sheet parallel to XY, normal along z).

3. Φ = E ⋅ A = (C A² k̂) ⋅ (A² k̂) = \textbf{C A⁴}.

Question: Coulomb's Law with Multiple Charges

Problem: W at corner, X at midpoint of adjacent side, Z at midpoint of opposite side. Charges +q each. Force W on X is F. Force W on Z in terms of F?

Solution:

1. Force W on X (F): Distance WX = d/2. F = k q² / (d/2)² = 4kq²/d².

2. Distance WZ: WZ² = d² + (d/2)² = 5d²/4.

3. Force W on Z (F_WZ): F_WZ = k q² / (5d²/4) = (4/5) * (kq²/d²).

4. In terms of F: F_WZ = (1/5) * (4kq²/d²) = \textbf{F/5}.

Question: Equilibrium of a Charged Particle Above a Sheet

Problem: Particle (mass 5 x 10⁻⁶ g) stationary over horizontal sheet with σ = 4 x 10⁻⁶ C/m². Find particle's charge.

Solution:

1. Equilibrium: Upward electric force F_e = qE balances downward gravity F_g = mg.

2. Electric field of sheet: E = σ / (2ε₀).

3. q * (σ / (2ε₀)) = mg \implies q = 2mgε₀ / σ.

4. Substitute values (m=5x10⁻⁹ kg, g=10 m/s², ε₀=8.85x10⁻¹² F/m):
   q = [2 * (5x10⁻⁹) * 10 * (8.85x10⁻¹²)] / (4x10⁻⁶) ≈ \textbf{2.2 x 10⁻¹³ C}.

Question: Tension in a Pendulum in an Electric Field

Problem: Pendulum bob (mass 'm', charge 'q') at rest in horizontal E. Find tension in string.

Solution:

1. Forces: Gravitational (mg, down), Electric (qE, horizontal), Tension (T, along string).

2. Bob is at rest, so T balances the vector sum of mg and qE.

3. Since mg and qE are perpendicular: T = √[ (mg)² + (qE)² ].

Question: Flux Through a Cube in a Non-Uniform Field

Problem: Cube side 10 cm, one face on YZ-plane (x=0). E = (5x² + 2) î N/C. Find net flux.

Solution:

1. Field is only in x-direction. Flux through YZ-parallel faces only.

2. Left Face (x=0): E_left = 5(0)² + 2 = 2 N/C.

3. Right Face (x=0.1m): E_right = 5(0.1)² + 2 = 2.05 N/C.

4. Area A = (0.1)² = 0.01 m².

5. Net Flux Φ_net = (E_right - E_left) * A = (2.05 - 2) * 0.01 = 0.05 * 0.01 = \textbf{5 x 10⁻⁴ Nm²/C}.