Class 12 · CBSE / NCERT · Physics

Class 12 Physics — All Derivations

Every board-favourite derivation, one step at a time. Try to recall the next line yourself, then reveal step by step — the fastest way to actually remember a derivation instead of cramming it.

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Electric Charges and Fields

Electric field on the axis of a dipole

Aim: Show that the field at a point on the axis of a short dipole is E = 2kp / r³.

A dipole has charges +q and −q separated by 2a; take a point P on the axis at distance r from the centre. Set up the geometry.
E₊ = kq / (r − a)² (away from +q), E₋ = kq / (r + a)² (towards −q). Field of each charge at P.
E = E₊ − E₋ = kq [ 1/(r−a)² − 1/(r+a)² ] They point in opposite directions along the axis.
E = kq · [ (r+a)² − (r−a)² ] / (r² − a²)² = kq · (4ar) / (r² − a²)² Simplify the brackets.
For a short dipole (r ≫ a): E ≈ kq·4ar / r⁴ = k·(2qa)·2 / r³ Neglect a compared to r.

Field due to an infinite charged sheet

Aim: Use Gauss's law to show E = σ / 2ε₀.

Take a cylindrical Gaussian "pillbox" piercing the sheet, with flat faces of area A on each side. By symmetry E is perpendicular to the sheet.
Flux through the curved surface = 0; through the two flat faces = E·A + E·A = 2EA. E is parallel to the curved side, so no flux there.
Charge enclosed = σ·A (σ = surface charge density). Only the patch of sheet inside the pillbox.
Gauss's law: 2EA = σA / ε₀ Total flux = q_enclosed / ε₀.

Electrostatic Potential and Capacitance

Capacitance of a parallel-plate capacitor

Aim: Show that C = ε₀A / d.

Two plates of area A separated by distance d carry charge +Q and −Q. Surface charge density σ = Q/A.
Field between the plates: E = σ / ε₀ = Q / (ε₀A). Uniform field of a parallel-plate capacitor.
Potential difference: V = E·d = Q·d / (ε₀A). V = E times the gap.
Capacitance C = Q / V = Q / [ Q·d / (ε₀A) ]. Definition of capacitance.

Energy stored in a capacitor

Aim: Show that U = ½CV² = Q²/2C = ½QV.

To add a small charge dq when the charge is already q, work done dW = V·dq = (q/C)·dq. Potential at that instant is V = q/C.
Total work U = ∫₀^Q (q/C) dq. Integrate as charge builds from 0 to Q.
U = (1/C) · [q²/2]₀^Q = Q² / 2C. Evaluate the integral.
Using Q = CV: U = (CV)² / 2C = ½CV² = ½QV. Express in the other standard forms.

Current Electricity

Drift velocity and Ohm's law

Aim: Derive I = nAev_d and hence V = IR.

Under field E each electron gains drift velocity v_d = (eE/m)·τ. τ = relaxation time between collisions.
In time t, charge crossing area A = (n·A·v_d·t)·e, so current I = nAev_d. n = number density of electrons.
Substitute v_d: I = nAe·(eEτ/m) = (nAe²τ/m)·E. Combine the two results.
With E = V/L: I = (nAe²τ/m)·(V/L) ⇒ V/I = mL / (nAe²τ) = R (a constant). R is independent of V ⇒ Ohm's law.

Moving Charges and Magnetism

Force between two parallel currents

Aim: Show that F/L = μ₀I₁I₂ / 2πd.

Wire 1 produces a field at wire 2 (distance d): B₁ = μ₀I₁ / (2πd). Field of a long straight wire.
Force on length L of wire 2 carrying I₂: F = B₁·I₂·L. F = BIL (wires are perpendicular to the field).
F = [ μ₀I₁ / (2πd) ] · I₂ · L. Substitute B₁.

Electromagnetic Induction

Motional EMF

Aim: Show that a rod of length L moving at speed v across a field B develops EMF = BLv.

A free charge q in the rod feels a magnetic force F = qvB along the rod. Force on a moving charge.
Work done per unit charge in moving it along length L = (qvB·L)/q. EMF = work per unit charge.
Alternatively, flux Φ = B·L·x, so EMF = −dΦ/dt = −BL·(dx/dt). Faraday's law, with dx/dt = v.

Ray Optics and Optical Instruments

Mirror formula

Aim: Show that 1/v + 1/u = 1/f for a concave mirror.

Draw an object AB, its image A′B′, the pole P, focus F and centre C on the principal axis. Use two standard rays to locate the image.
Triangles A′B′F and (mirror) are similar ⇒ A′B′/AB = B′F/PF. Ray through the focus.
Triangles A′B′P and ABP are similar ⇒ A′B′/AB = B′P/BP. Ray through the pole.
Equate and apply the sign convention (all distances from P): (v−f)/f = v/u ... simplify. Substitute PF = f, B′P = v, BP = u.

Refraction at a spherical surface

Aim: Show that n₂/v − n₁/u = (n₂ − n₁)/R.

Light goes from medium n₁ to n₂ through a spherical surface of radius R; use small-angle (paraxial) rays. Angles ≈ their tangents.
Snell's law for small angles: n₁·i = n₂·r. i, r are the angles of incidence and refraction.
Express i and r using the geometry (exterior-angle relations) in terms of u, v and R. i = α + θ, θ = r + β.
Substitute and apply the sign convention. Object distance u, image distance v measured from the pole.

Wave Optics

Fringe width in Young's double-slit experiment

Aim: Show that the fringe width β = λD / d.

Two slits distance d apart, screen at distance D; consider a point at distance y from the centre. D ≫ d, so rays are nearly parallel.
Path difference Δ = d·sinθ ≈ d·(y/D). For small θ, sinθ ≈ tanθ = y/D.
Bright fringe: Δ = nλ ⇒ y_n = nλD/d. Constructive interference.
Fringe width β = y_{n+1} − y_n = (n+1)λD/d − nλD/d. Spacing between consecutive bright fringes.

Dual Nature of Radiation and Matter

de Broglie wavelength

Aim: Show that a particle of momentum p has wavelength λ = h/p = h/mv.

For a photon: energy E = hf = hc/λ. Planck's relation.
Also E = pc for a photon ⇒ pc = hc/λ ⇒ p = h/λ. Momentum–energy relation for light.
de Broglie proposed the same relation for matter: λ = h/p. Wave nature of particles.

Atoms

Bohr's radius and energy of the nth orbit

Aim: Show that r_n ∝ n² and E_n = −13.6/n² eV (hydrogen).

Coulomb force provides the centripetal force: ke²/r² = mv²/r. Electron in a circular orbit.
Bohr quantisation: mvr = nh/2π. Angular momentum is quantised.
Eliminate v between the two ⇒ r_n = n²h²ε₀ / (πme²) ∝ n². Radius grows as n².
Total energy E = KE + PE = ½mv² − ke²/r = −ke²/2r. PE is twice the (negative) magnitude of KE.
Substitute r_n ⇒ E_n = −(13.6 / n²) eV for hydrogen. Put in the constants.

Nuclei

Law of radioactive decay

Aim: Show that N = N₀ e^(−λt) and half-life T = 0.693/λ.

The rate of decay is proportional to the number present: dN/dt = −λN. λ = decay constant.
Separate variables: dN/N = −λ dt. Prepare to integrate.
Integrate: ln(N/N₀) = −λt. From N₀ at t = 0 to N at time t.
For half-life, put N = N₀/2 ⇒ ln(½) = −λT ⇒ T = 0.693/λ. ln 2 = 0.693.