Top 20 Physics Derivations Class 12: Must Revise Before Exam

Top 20 Physics Derivations Class 12 is a must-revise list for students appearing in the CBSE Board Class 12 Physics exam tomorrow. If your exam is just a day away, focusing on these most important derivations will help you score direct theory marks and solve competency-based and numerical questions confidently. These derivations cover high-weightage chapters like Electrostatics, Current Electricity, Magnetism, Optics, and Modern Physics, which are frequently asked in board exams.

This quick revision of the Top 20 Physics Derivations Class 12 strengthens conceptual clarity and improves answer presentation in long-answer questions. Important topics include Gauss’s Law, dipole electric field and potential, capacitance with dielectric, drift velocity and current relation (I = neAvd), Wheatstone bridge condition (P/Q = R/S), magnetic field due to circular loop, RMS value of AC, mirror formula, lens maker’s formula, prism formula, and Bohr’s model radius and energy derivations. 

Revising these board-sure derivations today will boost your confidence, reduce exam stress, and help you attempt derivation-based and numericals accurately in tomorrow’s CBSE Physics paper.

Top 20 Physics Derivations Class 12

Physics derivations are fundamental for Class 12 students, not merely for marks but for building a strong conceptual foundation. They unveil the core concepts of each chapter, enhancing problem-solving abilities and preparing students for competency-based questions. A solid grasp of derivations is crucial for excelling in numerical problems.

1. Application of Gauss's Law: Electric Field of an Infinite Charged Wire

This derivation uses Gauss's Law to determine the electric field generated by an infinitely long, uniformly charged straight wire.

Recap of Gauss's Law: The net electric flux through any closed surface (Gaussian surface) equals the net charge enclosed divided by ε₀.

• Formula: ∮ E ⋅ dA = Q_inside / ε₀

Derivation Setup:

1. System: An infinitely long, uniformly charged wire. Due to infinite length, linear charge density (λ) is used, defined as charge per unit length (λ = Charge / Length).

2. Objective: To find the electric field (E) at a point P at a perpendicular distance r from the wire.

3. Gaussian Surface: A cylindrical Gaussian surface of radius r and length L, with the wire as its axis, is constructed. Point P lies on this cylinder's surface.

4. Vector Analysis:

• The electric field (E) points radially outward from a positively charged wire.

• The area vector (dA) for the curved surface is also radially outward. Thus, E and dA are in the same direction (angle = 0°).

• The flux through the flat circular ends is zero because the E field is parallel to these surfaces (angle = 90°, cos(90°)=0).

Mathematical Derivation:

1. Gauss's Law: ∮ E ⋅ dA = Q_inside / ε₀

2. For the curved surface: ∫ E dA cos(0°) = Q_inside / ε₀ => E ∫ dA = Q_inside / ε₀

3. Q_inside = λL (charge enclosed in length L).

4. The integral ∫ dA is the curved surface area: 2πrL.

5. Substitute: E (2πrL) = λL / ε₀

6. Cancel L and solve for E: E = λ / (2πε₀r)

2. Electric Field of a Dipole on its Axial Line

Dipole Fundamentals: A dipole comprises two equal and opposite charges (+q and -q) separated by a distance 2L. Its dipole moment (p) is p = q ⋅ (2L), directed from negative to positive charge.

Derivation Setup:

1. Objective: Find the net electric field at a point P on the axis of the dipole, at a distance r from its center.

2. A positive test charge (+1) is placed at P.

• +q creates repulsive field E₂ (away from dipole).

• -q creates attractive field E₁ (towards dipole).

3. Since P is closer to +q, E₂ is stronger than E₁. The net electric field (E_net) is in E₂'s direction: E_net = E₂ - E₁.

Mathematical Derivation:

1. Distances: P from +q is (r - L); P from -q is (r + L).

2. Expressions for E₁ and E₂:

• E₁ = k|q| / (r + L)²

• E₂ = kq / (r - L)²

3. Net field: E_net = kq [ 1/(r - L)² - 1/(r + L)² ]

4. Simplify brackets: E_net = kq [ 4rL / (r² - L²)² ]

5. Rearrange using p = q ⋅ 2L: E_net = k (q ⋅ 2L) (2r) / (r² - L²)²

6. E_net = k p (2r) / (r² - L²)²

The Ideal Dipole Approximation: For r >> L, L² is neglected.

• E_net = 2kpr / r⁴

• Final formula: E_net = 2kp / r³

3. Electric Field of a Dipole at an Equatorial Point

Derivation Setup:

1. Objective: Find the net electric field at point P on the equatorial line (perpendicular bisector) of the dipole, at distance r from its center.

2. P is equidistant from +q and -q: √(r² + L²).

3. Vector Analysis:

• +q creates repulsive field E₂.

• -q creates attractive field E₁.

• Magnitudes are equal: |E₁| = |E₂| = E.

• Resolving Components: Vertical components (E₁sinθ, E₂sinθ) cancel out. Horizontal components (E₁cosθ, E₂cosθ) add up in the direction opposite to the dipole moment.

• E_net = E₁cos(θ) + E₂cos(θ) = 2Ecos(θ).

Mathematical Derivation:

1. Magnitude of E: E = kq / (r² + L²)

2. cos(θ) = L / √(r² + L²)

3. Substitute: E_net = 2 * [kq / (r² + L²)] * [L / √(r² + L²)]

4. E_net = k (2qL) / (r² + L²)^(3/2)

5. Recognize p = 2qL: E_net = kp / (r² + L²)^(3/2)

The Ideal Dipole Approximation: For r >> L, L² is negligible.

• Denominator becomes (r²)^(3/2) = r³.

• Final formula: E_net = kp / r³
  Note: The magnitude of the electric field on the axial line (2kp/r³) is twice the magnitude of the field at an equatorial point (kp/r³) for the same distance r.

4. Torque on a Dipole in a Uniform External Field

Concept: An electric dipole in a uniform external electric field experiences equal and opposite forces, forming a couple that exerts a torque, causing it to align with the field.

Derivation Setup:

1. A dipole (+q, -q, separation 2L) is in a uniform electric field E at an angle θ.

2. Force on +q: F = qE (direction of E).

3. Force on -q: F = -qE (opposite to E).

4. Torque (τ) is force times the perpendicular distance between force lines.

5. From geometry, perpendicular distance is 2Lsin(θ).

Mathematical Derivation:

1. τ = Force × Perpendicular Distance

2. τ = (qE) × (2Lsin(θ))

3. Rearrange using p = q(2L): τ = (q ⋅ 2L) E sin(θ)

4. Magnitude: τ = pE sin(θ)

5. Vector form: τ = p × E

5. Electric Potential due to a Dipole on its Axial Line

Concept: Electric potential is a scalar quantity, so the net potential is the simple algebraic sum of potentials from individual charges.

Derivation Setup:

1. Same as the axial electric field: dipole, point P on its axis, distance r from center.

2. Distance of P from +q: (r - L).

3. Distance of P from -q: (r + L).

Mathematical Derivation:

1. Net potential V_net = V₁ + V₂.

2. V₁ = -kq / (r + L) (for -q).

3. V₂ = +kq / (r - L) (for +q).

4. Sum: V_net = [kq / (r - L)] - [kq / (r + L)]

5. Simplify: V_net = kq [ 2L / (r² - L²) ]

6. Recognize q ⋅ 2L = p: V_net = kp / (r² - L²)

The Ideal Dipole Approximation: For r >> L, L² is negligible.

• Denominator becomes r².

• Final formula: V = kp / r²
  Note: The potential at any point on the equatorial line of a dipole is zero, as points are equidistant from equal and opposite charges.

6. Potential Energy of a Dipole in an External Field

Concept: The potential energy (U) of a dipole in an electric field equals the external work done (W_ext) to rotate it from a reference position (where U=0) to its current position.

Derivation of Work Done:

1. Work done dW for infinitesimal rotation dθ: dW = τ dθ.

2. Substitute τ = pE sin(θ): dW = pE sin(θ) dθ.

3. Integrate from θ₁ to θ₂: W_ext = ∫(θ₁ to θ₂) pE sin(θ) dθ

4. W_ext = pE [-cos(θ)] from θ₁ to θ₂

5. W_ext = pE [cos(θ₁) - cos(θ₂)]

Deriving Potential Energy (U): The formula depends on the reference point.

• Case 1: Standard Reference (θ₁ = 90°)

• U=0 when dipole is perpendicular to E field.

• Let θ₂ = θ.

• U = pE [cos(90°) - cos(θ)]

• Since cos(90°) = 0: U = -pE cos(θ)

• Vector form: U = -p ⋅ E

• Case 2: Alternative Reference (θ₁ = 0°)

• U=0 when dipole is aligned with E field.

• U = pE [cos(0°) - cos(θ)]

• U = pE (1 - cos(θ))

Pedagogical Emphasis: Understanding the reference orientation is critical for applying the potential energy formula correctly.

7. Capacitance of a Parallel Plate Capacitor

Concept: A parallel plate capacitor stores charge per unit potential difference.

Derivation:

1. Electric Field (E): Between two oppositely charged parallel plates: E = σ / ε₀.

2. Potential Difference (V): V = E ⋅ d (where d is plate separation).

3. Substitute E: V = (σ / ε₀) ⋅ d.

4. Charge Density (σ): σ = Q / A (charge Q, area A).

5. Substitute σ: V = (Qd) / (Aε₀).

6. Capacitance (C = Q / V): Rearrange to Q / V = Aε₀ / d.

7. Therefore: C = Aε₀ / d

8. Capacitance of a Capacitor with a Dielectric Slab

Concept of Dielectrics: An insulating material that, when placed in an external electric field, polarizes and reduces the net electric field inside by a factor K (dielectric constant). E_new = E_old / K.

Special Case: Fully Filled Dielectric:

1. Original (Air/Vacuum): V_old = E_old ⋅ d, C_old = Aε₀ / d.

2. With Dielectric:

• Electric field reduces: E_new = E_old / K.

• New potential difference: V_new = E_new ⋅ d = (E_old / K) ⋅ d = V_old / K.

3. New Capacitance (C_new): Charge Q remains constant.

• C_new = Q / V_new = Q / (V_old / K)

• C_new = K ⋅ (Q / V_old)

• Final formula: C_new = K ⋅ C_old

Comparative Structure: Effect of a Dielectric (K > 1)

9. Electron Motion and Drift Velocity

This is a highly important concept for understanding electric current.

Electron Motion Without an Electric Field:

• Free electrons exhibit continuous, random thermal velocity due to thermal energy.

• The net average thermal velocity of all electrons is zero (Σuᵢ / N = 0), as motion is chaotic.

• Thus, no electric current flows without an external electric field.

Electron Motion With an Electric Field:

• When an electric field is applied, electrons experience a force and accelerate.

• Frequent collisions with atoms cause them to lose momentum, resulting in a "stop-and-start" motion.

• The average velocity with which electrons slowly drift in one direction is called drift velocity (v_d).

• The average time between successive collisions is the relaxation time (τ).

Derivation of Drift Velocity (v_d):

1. Using v = u + at, average final velocity v_d for n electrons:
   v_d = (Σvᵢ) / n = (Σuᵢ) / n + a * (Σtᵢ) / n

2. Average initial thermal velocity (Σuᵢ) / n is zero.

3. Average time (Σtᵢ) / n is relaxation time (τ).

4. So, v_d = a * τ.

5. Acceleration a is from electric force: a = F/m.

6. Force on electron (charge -e): F = qE = -eE.

7. Thus, a = -eE / m.

8. Final expression for drift velocity (vector form): v⃗_d = (-eE⃗ / m) * τ

10. Relation between Electric Current and Drift Velocity

This derivation establishes the fundamental relationship between macroscopic current I and microscopic drift velocity v_d.

1. Electron Density (n): Number of free electrons N per unit volume V. n = N / V.

• For conductor of length L and area A: V = A * L.

• Total electrons N = n * A * L.

2. Total Charge (Q): Q = N * e = (n * A * L) * e.

3. Current (I = dq / dt): Consider charge dq in length dx: dq = (n * A * dx) * e.

• I = d/dt [ (n * A * dx) * e ] = n * A * e * (dx / dt).

4. dx / dt is the drift velocity (v_d).

5. Therefore, the relationship is: I = n * e * A * v_d
   [Memory Tip]: This formula is often remembered by the mnemonic "V-E-N-A" or "VINA": I = v_d * e * n * A. This is a very frequently tested topic.

11. Wheatstone Bridge and Kirchhoff's Law

A Wheatstone bridge measures unknown resistance by balancing two legs. Its balance condition is a key application of Kirchhoff's Laws.

Circuit Setup: Four resistors (P, Q, R, S), galvanometer (G) between B and D. Current I splits into I₁ (through P) and I₂ (through R). I_G flows through galvanometer.

Applying Kirchhoff's Loop Law (Sign Convention):

• Traversing in same direction as current through resistor: potential decreases (negative potential difference).

• Traversing in opposite direction of current: potential increases (positive potential difference).

1. Loop ABDA (clockwise):
   -I₁P - I_G*G + I₂R = 0 (Equation 1)

2. Loop BCDB (clockwise):
   -(I₁ - I_G)Q + (I₂ + I_G)S + I_G*G = 0 (Equation 2)

Condition for a Balanced Wheatstone Bridge: The bridge is balanced when I_G = 0 (no current through galvanometer).

1. From Equation 1 (with I_G = 0): -I₁P + I₂R = 0 => I₁P = I₂R (Simplified Eq 1')

2. From Equation 2 (with I_G = 0): -(I₁ - 0)Q + (I₂ + 0)S = 0 => -I₁Q + I₂S = 0 => I₁Q = I₂S (Simplified Eq 2')

3. Deriving Final Condition: Divide Eq 1' by Eq 2': (I₁P) / (I₁Q) = (I₂R) / (I₂S).
   Canceling I₁ and I₂ yields: P / Q = R / S
   This means if the resistance ratio in one arm (P/Q) equals that in the other (R/S), the bridge is balanced.

12. Magnetic Field on the Axis of a Circular Current Loop

This derivation uses the Biot-Savart Law to find the magnetic field at a point P on the axis of a circular loop of radius R carrying current I, at a distance x from its center.

Derivation Steps:

1. Biot-Savart Law: dB = (μ₀ / 4π) * (I dl sinθ / r²).

• dl is tangential to loop, r is distance from dl to P.

• Angle θ between dl and r is 90°, so sin(90°) = 1.

• r = √(R² + x²).

2. Direction of dB: Perpendicular to both dl and r.

3. Symmetry and Component Cancellation:

• Consider diametrically opposite dl elements. Their dB vectors have components perpendicular to the axis that cancel each other out.

• Components along the axis add up.

4. Net Magnetic Field: B_net = ∫ dB_axial = ∫ dB sinα (where sinα = R/r).

• B_net = ∫ [ (μ₀ I dl) / (4π (R² + x²)) ] * [ R / (R² + x²)¹/² ]

• B_net = [ μ₀ I R / (4π (R² + x²)^(3/2)) ] ∫ dl.

• ∫dl over the loop is 2πR.

5. Final Formula: B = μ₀ I R² / (2 * (R² + x²)^(3/2))

13. Force Between Two Parallel Current-Carrying Wires

This principle explains the interaction between two long, straight, parallel wires carrying currents.

Scenario: Wire 1 carries I₁, Wire 2 carries I₂, separated by distance d.

Derivation:

1. Field Produced by Wire 1 (at Wire 2's location): For a long straight wire, B₁ = (μ₀ I₁) / (2πd).

2. Force on Wire 2: Wire 2 (current I₂, length L) in field B₁ experiences force F = I L B.

• F₂ = I₂ * L * B₁

3. Substitute B₁: F₂ = I₂ * L * [ (μ₀ I₁) / (2πd) ]

4. Force per Unit Length (F/L): F/L = (μ₀ I₁ I₂) / (2πd)
   By Newton's third law, Wire 2 exerts an equal and opposite force on Wire 1.

Comparative Structure: Direction of Force

14. RMS Value of an Alternating EMF

For AC, the average value is zero. The Root Mean Square (RMS) value quantifies its effective value.

• RMS Definition: Square Root of the Mean (average) of the Square of the function. E_rms = √[ (average of E²) ].

Derivation for a Sinusoidal EMF (E = E₀ sin(ωt)):

1. Square the EMF: E² = E₀² sin²(ωt).

2. Mean (Average) of E² over a Full Cycle (T = 2π/ω):
   Average(E²) = (1/T) ∫₀ᵀ E₀² sin²(ωt) dt

3. Use identity: sin²(θ) = (1 - cos(2θ)) / 2.
   Average(E²) = (E₀²/2T) [ ∫₀ᵀ 1 dt - ∫₀ᵀ cos(2ωt) dt ]

4. Evaluate Integrals: ∫₀ᵀ 1 dt = T. ∫₀ᵀ cos(2ωt) dt = 0 (over a complete cycle).

5. Simplifies to: Average(E²) = (E₀²/2T) * [ T - 0 ] = E₀² / 2.

6. Take the Square Root: E_rms = √[ E₀² / 2 ].

7. Final result: E_rms = E₀ / √2.
   Similarly, for alternating current I = I₀ sin(ωt), I_rms = I₀ / √2.

15. Mirror Formula Derivation

The mirror formula relates object distance (u), image distance (v), and focal length (f) of a spherical mirror using similar triangles.

Setup: Object AB, concave mirror, image A'B'. C is center of curvature, F is focal point. Small aperture approximation for mirror.

Derivation using Similar Triangles:

1. ΔABC ~ ΔA'B'C: (Angles at 90°, vertically opposite angles)
   AB / A'B' = AC / A'C (Equation 1)

2. ΔPQF ~ ΔA'B'F: (Ray parallel to axis from B strikes Q, passes through F. PQ is approximated as AB)
   PQ / A'B' = PF / A'F => AB / A'B' = PF / A'F (Equation 2)

3. Equate Ratios: From (1) and (2): AC / A'C = PF / A'F (Equation 3)

4. Substitute Distances (using magnitudes):

• AC = u - R (or u - 2f)

• A'C = R - v (or 2f - v)

• PF = f

• A'F = v - f

5. Solve (cross-multiply): (u - 2f)(v - f) = f(2f - v) => uv - uf - 2fv + 2f² = 2f² - fv
   uv - uf - 2fv = -fv => uv - uf = fv

6. Final Step: Divide by uvf: 1/f - 1/v = 1/u.
   Rearranging gives the standard Mirror Formula: 1/f = 1/v + 1/u

16. Refraction at a Single Spherical Surface

This formula describes light refraction from medium n₁ to n₂ through a curved surface with radius R, using paraxial ray approximation.

Setup: Object O in n₁, image I in n₂. C is center of curvature.

Derivation:

1. Geometric Relations (Exterior Angle Theorem):

• i = α + β

• β = r + γ => r = β - γ

2. Snell's Law (for small angles): n₁ sin(i) = n₂ sin(r) => n₁ * i = n₂ * r.

3. Substitute Geometric Relations: n₁ (α + β) = n₂ (β - γ).

4. Angle-to-Distance Relation (Small Angle Approx., let point of incidence be h from axis):

• α ≈ h / u (u is object distance)

• β ≈ h / R (R is radius of curvature)

• γ ≈ h / v (v is image distance)

5. Substitute Distances and simplify (h cancels): n₁/u + n₁/R = n₂/R - n₂/v.

6. Rearrange: Group v and u terms, and R terms.
   * (n₂ / v) - (n₁ / u) = (n₂ - n₁) / R *
   (Note: This standard formula uses Cartesian sign convention.)

17. Lens Maker's Formula

The Lens Maker's formula relates a lens's focal length (f) to its refractive index (n₂), surrounding medium (n₁), and radii of curvature (R₁, R₂) by applying the single spherical surface refraction formula twice.

Setup: Thin lens (n₂) in medium (n₁). R₁ for first surface, R₂ for second.

Derivation:

1. Refraction at the First Surface (n₁ to n₂): Object at u, forms intermediate image at v'.
   (n₂ / v') - (n₁ / u) = (n₂ - n₁) / R₁ (Equation A)

2. Refraction at the Second Surface (n₂ to n₁): v' acts as a virtual object. Forms final image at v.
   (n₁ / v) - (n₂ / v') = (n₁ - n₂) / R₂ (Equation B)

3. Combine Equations (Add A and B):
   (n₁ / v) - (n₁ / u) = (n₂ - n₁) / R₁ + (n₁ - n₂) / R₂
   n₁ (1/v - 1/u) = (n₂ - n₁) (1/R₁ - 1/R₂)

4. Introduce Focal Length (1/f = 1/v - 1/u):
   n₁ (1/f) = (n₂ - n₁) (1/R₁ - 1/R₂)

5. Final Formula: Divide by n₁.
   * 1/f = (n₂/n₁ - 1)  (1/R₁ - 1/R₂) **

18. Prism Formula and Minimum Deviation

This derivation finds the relationship between a prism's refractive index (n), apex angle (A), and angle of minimum deviation (δ_min).

Setup: A (apex angle), i (incidence), e (emergence), r₁, r₂ (refraction angles), δ (deviation).

Derivation:

1. Geometric Relations:

• From quadrilateral formed by normals and faces: A = r₁ + r₂

• From external deviation: δ = i + e - A

2. Condition for Minimum Deviation (δ_min): Occurs when ray passes symmetrically.

• i = e and r₁ = r₂ = r.

3. Applying Conditions:

• From A = r₁ + r₂: A = 2r => r = A / 2

• From δ = i + e - A: δ_min = 2i - A => i = (A + δ_min) / 2

4. Using Snell's Law (n = sin(i) / sin(r)): Substitute i and r.
   * n = sin[ (A + δ_min) / 2 ] / sin[ A / 2 ] *

19. Bohr's Model: Radius of the nth Orbit

This derivation determines the radius of allowed electron orbits in the Bohr model for hydrogen-like atoms (atomic number Z).

Core Principles:

1. Force Balance: Electrostatic attraction = centripetal force.

2. Quantization of Angular Momentum: L = mvr = n * (h / 2π).

Derivation:

1. Equating Forces:

• mv² / r = kZe² / r² => mv² = kZe² / r (Equation 1)

2. Bohr's Condition: From mvr = n * (h / 2π), solve for v: v = nh / (2πmr) (Equation 2)

3. Substitute Velocity: Square (2) to get v² = n²h² / (4π²m²r²). Substitute into (1):
   m * [n²h² / (4π²m²r²)] = kZe² / r

4. Solve for Radius (r_n): Simplify and rearrange.
   * r_n = [h² / (4π²mk e²)]  (n² / Z) **

5. Final Expression: The bracketed term is a constant (Bohr radius a₀ ≈ 0.529 Å).
   * r_n = a₀  (n² / Z) **

20. Bohr's Model: Energy in the nth Orbit

This derivation finds the total energy (sum of kinetic and potential) of an electron in the nth orbit of a hydrogen-like atom.

Derivation:

1. Kinetic Energy (KE): From force balance (mv² = kZe² / r), KE = ½ mv².
   * KE = kZe² / 2r *

2. Potential Energy (PE): Between nucleus (+Ze) and electron (-e).

• PE = k q₁q₂ / r = k (Ze) (-e) / r.

• PE = -kZe² / r (negative due to attraction).

3. Total Energy (E = KE + PE):

• E = (kZe² / 2r) + (-kZe² / r)

• E = kZe² * (1/2r - 1/r) = kZe² * (-1 / 2r)

• E = -kZe² / 2r

Comparative Structure: Relationships Between Energy Types

For a hydrogen atom (where Z=1), the total energy in the nth orbit is E = -13.6 eV / n².