Class 12 · Derivations

An infinitely long wire has uniform linear charge density λ. Find E at a distance r.

1. Choose a coaxial cylindrical Gaussian surface of radius r and length ℓ. By symmetry, E is radial.
2. The flux through the two end caps is zero (E is parallel to their surface). Flux through the curved surface:
   Φ = E × 2πrℓ
3. Charge enclosed:
   q = λℓ
4. Apply Gauss's law, Φ = q/ε₀:
   E · 2πrℓ = λ/ε₀  ⇒  E = λ / (2πε₀r)

Two plates of area A are separated by distance d, carrying charges +Q and −Q.

1. Surface charge density and field between the plates:
   σ = Q/A,   E = σ/ε₀ = Q/(ε₀A)
2. Potential difference across the gap:
   V = E d = Qd/(ε₀A)
3. Capacitance is the ratio of charge to potential difference:
   C = Q/V = ε₀A / d
4. With a dielectric of constant K filling the gap: C = Kε₀A/d.

Free electrons (number density n, relaxation time τ) drift in a conductor of length ℓ and area A.

1. Acceleration of an electron in field E, and the drift velocity gained in time τ:
   a = eE/m  ⇒  v_d = aτ = eEτ/m
2. Current through the conductor:
   I = n e A v_d = (n e²τA/m) E
3. Put E = V/ℓ and compare with V = IR:
   R = mℓ / (n e²τA)
4. Since R = ρℓ/A, the resistivity is:
   ρ = m / (n e²τ)

A loop of radius R carries current I. Find B at a point on the axis at distance x from the centre.

1. By the Biot–Savart law, for a small element dl (always perpendicular to the line to the point, distance √(R² + x²)):
   dB = (μ₀/4π) · I dl / (R² + x²)
2. Components perpendicular to the axis cancel by symmetry. The axial component is dB cos φ, where cos φ = R/√(R² + x²).
3. Integrate around the loop, using ∮dl = 2πR:
   B = (μ₀I / 4π) · R/(R² + x²)^3/2 · 2πR
4. Simplify:
   B = μ₀ I R² / [2 (R² + x²)^3/2]
5. At the centre (x = 0): B = μ₀I / 2R.

A rod of length ℓ slides with speed v on rails, perpendicular to a uniform field B.

1. Flux through the loop when the rod is at distance x:
   Φ = B ℓ x
2. Faraday's law gives the induced emf:
   ε = −dΦ/dt = −Bℓ (dx/dt)
3. Since dx/dt = v, the magnitude is:
   ε = B ℓ v
4. Check: the Lorentz force on charges in the rod is qvB; work per unit charge along length ℓ is vBℓ — the same result.

Slits S₁, S₂ are separated by d; the screen is at distance D (d ≪ D). P is at height y from the central point.

1. Path difference at P:
   Δ = S₂P − S₁P ≈ d y / D
2. Bright fringes (constructive) need Δ = nλ:
   y_n = n λ D / d
3. Dark fringes (destructive) need Δ = (n + ½)λ.
4. Fringe width is the spacing between consecutive bright fringes:
   β = y_n+1 − y_n = λD / d

A photon of frequency ν strikes a metal with work function φ.

1. The photon gives all its energy hν to one electron (one-to-one interaction).
2. Part of it (φ) frees the electron; the rest becomes kinetic energy:
   hν = φ + K_max
3. Rearrange, with threshold frequency ν₀ = φ/h:
   K_max = hν − φ = h(ν − ν₀)
4. With stopping potential V₀: eV₀ = K_max, so V₀ = (h/e)ν − φ/e — a straight line of slope h/e.

An electron (mass m, charge e) orbits a proton in a circle of radius r.

1. Coulomb force supplies the centripetal force:
   mv²/r = e² / (4πε₀r²)
2. Bohr's quantisation of angular momentum:
   mvr = n h / 2π
3. Eliminate v and solve for r:
   r_n = n² a₀,   a₀ = ε₀h²/(πme²) ≈ 0.529 Å
4. Total energy = K + U = −e²/(8πε₀r), which gives:
   E_n = −13.6 / n² eV