PHYS7021 · Electricity, Magnetism & Wave Optics

Concept: field inside a polarized medium.

1. When a field E is applied, the dielectric polarizes; bound surface charges create an opposing internal field, so the net field is E = E₀ − P/ε₀.
2. The electric displacement is defined to contain only free charges: D = ε₀E + P
3. For a linear, isotropic medium P = ε₀χ_eE, hence D = ε₀(1+χ_e)E = ε₀KE.
4. Gauss's law then reads ∮D·dA = q_free (bound charges not needed).

∴ D = ε₀E + P (and D = ε₀KE in a linear medium)

1. Polarization is the process in which an external electric field aligns the permanent dipoles (polar molecules) or induces dipoles (non-polar molecules), so that bound (+ and −) charges appear on the surfaces of the dielectric.
2. Polarization vector: P = total dipole moment / volume = Σp/ΔV
3. Unit: C m⁻²; in a linear medium P = ε₀χ_eE.

∴ P = dipole moment per unit volume (C/m²)

1. Take a slab of dielectric of area A and thickness d, uniformly polarized; induced bound charges ±q_b appear on the two faces.
2. The slab is then equivalent to a dipole of moment p = q_b × d.
3. Polarization P = dipole moment/volume = (q_bd)/(A·d) = q_b/A.
4. But q_b/A is exactly the surface density of induced charge σ_p.

∴ P = σₚ (proved)

1. Polar dielectrics: molecules have permanent dipole moment even without a field, because centres of + and − charge do not coincide. Examples: H₂O, HCl, NH₃, CO.
2. Non-polar dielectrics: centres of + and − charge coincide; no permanent dipole; a field merely induces a dipole. Examples: O₂, N₂, CO₂, CH₄.
3. In both, an applied field produces net polarization, but polar molecules additionally need orientation against thermal agitation.

∴ Polar: permanent dipole (H₂O, HCl) · Non-polar: induced only (O₂, CH₄)

1. Let E₀ = field between plates with air (or vacuum).
2. Introducing the slab polarizes it; the bound surface charges set up an opposing field E_p = P/ε₀.
3. Net field: E = E₀ − P/ε₀. Using P = ε₀(K−1)E ⇒ E = E₀ − (K−1)E ⇒ KE = E₀.
4. E = E₀/K and since K > 1, E < E₀.

∴ Field reduces K times: E = E₀/K

Given: C = 2 pF = 2×10⁻¹² F, V = 1 kV = 10³ V. Formula: U = ½CV².

1. U = ½ × (2×10⁻¹²) × (10³)²
2. = ½ × 2×10⁻¹² × 10⁶ = 1×10⁻⁶ J.

∴ U = 1×10⁻⁶ J = 1 μJ

1. While charging, when charge on the plates is q, potential v = q/C.
2. Work to bring further charge dq: dW = v·dq = (q/C) dq.
3. Total work: W = ∫₀^Q (q/C) dq = Q²/2C.
4. This work is stored as energy: U = Q²/2C; substituting Q = CV gives U = ½CV² (also = ½QV).

∴ U = ½CV² (proved)

1. Field between plates carrying ±Q: E = σ/ε₀ = Q/(ε₀A).
2. Potential difference V = E·d = Qd/(ε₀A).
3. Therefore C = Q/V = ε₀A/d.
4. With a dielectric slab filling the gap, E becomes E/K, so V becomes V/K and C becomes C′ = Kε₀A/d — capacitance increases K times.

∴ C = ε₀A/d ; with slab C′ = Kε₀A/d

Given: K = 1.6; half area in oil ⇒ two capacitors in parallel, each of area A/2 and gap d.

1. C₁ (air half) = ε₀(A/2)/d = C/2.
2. C₂ (oil half) = Kε₀(A/2)/d = KC/2.
3. Parallel: C′ = C/2 + KC/2 = (C/2)(1+K) = (C/2)(2.6).

∴ C′ = 1.3 C

Given: a = 10 cm = 0.10 m, b = 15 cm = 0.15 m. Formula: C = 4πε₀ab/(b−a).

1. ab = 0.10 × 0.15 = 0.015 m²; b − a = 0.05 m.
2. C = 0.015 / (9×10⁹ × 0.05) = 0.015 / 4.5×10⁸.
3. = 3.33×10⁻¹¹ F.

∴ C ≈ 3.33×10⁻¹¹ F ≈ 33.3 pF

1. Let inner radius a, outer b, charge per unit length λ. By Gauss's law, between cylinders E = λ/(2πε₀r).
2. Potential difference V = ∫_a^b E dr = (λ/2πε₀) ln(b/a).
3. Capacitance per unit length = λ/V = 2πε₀/ln(b/a).

∴ C/L = 2πε₀ / ln(b/a)

1. Total charge conserved: Q = C₁V₁ + C₂V₂; common potential V = Q/(C₁+C₂).
2. Initial energy U_i = ½C₁V₁² + ½C₂V₂²; final U_f = ½(C₁+C₂)V².
3. Loss ΔU = U_i − U_f. Substituting V and simplifying: ΔU = (C₁C₂/2(C₁+C₂)) (V₁−V₂)²
4. (Lost as heat and spark/radiation; zero only when V₁ = V₂.)

∴ ΔU = [C₁C₂/2(C₁+C₂)](V₁−V₂)²

1. Electric flux Φ = ∮E·dA — the total number of field lines crossing a surface; unit N m² C⁻¹ (or V·m).
2. A dipole contains +q and −q ⇒ net enclosed charge q_enc = 0.
3. By Gauss's law Φ = q_enc/ε₀ = 0.

∴ Φ = 0 for a closed surface enclosing a dipole

1. Statement: ∮E·dA = q_enc/ε₀ for any closed surface.
2. Proof (charge q at centre of sphere of radius r): E = q/4πε₀r², radial and constant over the sphere.
3. ∮E dA = (q/4πε₀r²) × 4πr² = q/ε₀. ✔
4. By superposition the result holds for any distribution and (since flux depends only on enclosed charge) any closed surface.
5. Differential form: apply the divergence theorem ∮E·dA = ∫(∇·E)dV with q_enc = ∫(ρ/ε₀)dV ⇒ ∇·E = ρ/ε₀.

∴ ∮E·dA = q/ε₀ ; differential form ∇·E = ρ/ε₀

1. Outside (r ≥ R): Gaussian sphere encloses whole charge q ⇒ E(4πr²) = q/ε₀ ⇒ E = q/4πε₀r² (like a point charge).
2. Inside, volume-charged solid sphere: enclosed charge q_enc = q(r³/R³) ⇒ E = q r/4πε₀R³ — field rises linearly from centre.
3. If the charge lies only on the surface (conducting/hollow sphere): for r < R, q_enc = 0 ⇒ E = 0.
4. Graph: E ∝ r inside (solid sphere) or zero (shell) up to r = R; beyond R, E ∝ 1/r²; maximum at the surface.

∴ Outside q/4πε₀r² ; inside (solid) qr/4πε₀R³ ; inside shell = 0

Given: Z = 92, r = 7×10⁻¹⁵ m. Formula: E = (1/4πε₀)(Ze/r²).

1. Charge q = Ze = 92 × 1.6×10⁻¹⁹ = 1.472×10⁻¹⁷ C.
2. E = (9×10⁹ × 1.472×10⁻¹⁷) / (7×10⁻¹⁵)² = 1.325×10⁻⁷ / 4.9×10⁻²⁹.
3. = 2.7×10²¹ V/m.

∴ E ≈ 2.7×10²¹ V/m

1. Let λ = charge per unit length. Choose a coaxial Gaussian cylinder of radius r and length l.
2. Flux passes only through the curved surface: Φ = E(2πrl).
3. Gauss: E(2πrl) = λl/ε₀ ⇒ E = λ/2πε₀r, directed radially.

∴ E = λ/2πε₀r

1. An electric dipole = a system of two equal and opposite point charges ±q separated by a small distance 2a.
2. Dipole moment: p = q × 2a, directed from −q to +q.
3. Unit: coulomb-metre (C·m); it measures the strength of the dipole.

∴ p = q·(2a), from −q to +q, unit C·m

1. Distances of P(r,θ) from +q and −q are ≈ (r − a cosθ) and (r + a cosθ).
2. V = (q/4πε₀)[1/(r−a cosθ) − 1/(r+a cosθ)] = (q/4πε₀)·(2a cosθ)/(r²−a²cos²θ).
3. For r ≫ a: V = p cosθ/4πε₀r² (p = 2aq).
4. Field components: E_r = −∂V/∂r = 2p cosθ/4πε₀r³; E_θ = −(1/r)∂V/∂θ = p sinθ/4πε₀r³.
5. Magnitude E = (p/4πε₀r³)√(1+3cos²θ). Axial (θ=0): 2p/4πε₀r³; equatorial (θ=90°): p/4πε₀r³.

∴ V = p cosθ/4πε₀r² ; E = (p/4πε₀r³)√(1+3cos²θ)

Formula: E = −∇v.

1. E_x = −∂v/∂x = +2kx; similarly E_y = 2ky, E_z = 2kz.
2. At (1,1,1): E = 2k(î + ĵ + k).
3. Magnitude |E| = 2k√3.

∴ E = 2k(î + ĵ + k), |E| = 2√3·k

1. r = √(x²+y²+z²); ∂r/∂x = x/r (etc.).
2. ∂(1/r)/∂x = −(1/r²)(x/r) = −x/r³; similarly for y, z.
3. ∇(1/r) = −(xî + yĵ + zk)/r³ = −r/r³ = −r̂/r².

∴ ∇(1/r) = −r̂/r²

Method: field is conservative ⇒ find φ with ∇φ = F.

1. ∫F_x dx = ∫2xy dx = x²y + f(y,z) (using z²x term from F_x: ∫z²dx = xz²).
2. So try φ = x²y + xz² + C.
3. Check: ∂φ/∂x = 2xy + z² ✔; ∂φ/∂y = x² ✔; ∂φ/∂z = 2xz ✔.

∴ φ = x²y + xz² + constant

1. In electrostatics E = −∇V (V = potential).
2. Work around any closed path: W = ∮E·dl = −∮∇V·dl = 0, because V returns to its initial value.
3. Equivalently ∇×E = 0; hence work depends only on end points.

∴ ∮E·dl = 0 ⇒ electrostatic field is conservative

Given: q₁ = 8 μC, q₂ = 2 μC, separation 0.5 m. Let the point be at x from 8 μC (between the charges, where the two fields oppose).

1. Set magnitudes equal: k·8/x² = k·2/(0.5−x)².
2. ⇒ 8(0.5−x)² = 2x² ⇒ 2(0.5−x) = x (taking positive roots).
3. ⇒ 1 − 2x = x ⇒ x = 1/3 m.

∴ x = 1/3 m ≈ 33.3 cm from 8 μC (16.7 cm from 2 μC)