Electrostatics
Coulomb's Law
\mathbf{F} = \frac{Q_1 Q_2}{4\pi\epsilon_0 r^2}\hat{a}_r
| Symbol | Description | Unit |
|---|---|---|
| F | Force between charges | N |
| Q_1, Q_2 | Point charges | C |
| r | Distance between charges | m |
| \epsilon_0 | Free-space permittivity = 8.854×10⁻¹² F/m | F/m |
Worked example
Find force between Q1 = 2 μC and Q2 = -3 μC separated by 0.1 m in air.
Given: Q1 = 2e-6 C, Q2 = -3e-6 C, r = 0.1 m
- F = Q1·Q2 / (4π × 8.854e-12 × 0.1²)
- = (2e-6 × -3e-6) / (4π × 8.854e-12 × 0.01)
- = -6e-12 / (1.1127e-12)
- = -5.394 N (attractive)
Answer: F ≈ -5.39 N (attractive, toward each other)
Gauss's Law (Integral Form)
\oint_S \mathbf{D} \cdot d\mathbf{S} = Q_{enc}
| Symbol | Description | Unit |
|---|---|---|
| \mathbf{D} | Electric flux density D = ε₀E | C/m² |
| Q_{enc} | Total enclosed charge | C |
Worked example
Find E-field at r = 0.5 m from a line charge ρ_L = 10 nC/m in free space.
Given: ρ_L = 10e-9 C/m, r = 0.5 m, ε₀ = 8.854e-12 F/m
- Gaussian surface: cylinder of radius r, length L
- Q_enc = ρ_L × L
- D × 2πrL = ρ_L × L → D = ρ_L/(2πr)
- E = D/ε₀ = 10e-9 / (2π × 8.854e-12 × 0.5)
- = 10e-9 / (27.84e-12) = 359.2 V/m
Answer: E ≈ 359 V/m radially outward
Capacitance of Parallel Plates
C = \frac{\epsilon_0 \epsilon_r A}{d}
| Symbol | Description | Unit |
|---|---|---|
| C | Capacitance | F |
| A | Plate area | m² |
| d | Separation between plates | m |
| \epsilon_r | Relative permittivity | dimensionless |
Worked example
Find C for plates 10 cm × 10 cm, d = 2 mm, εr = 4.
Given: A = 0.01 m², d = 0.002 m, εr = 4
- C = ε₀ × εr × A / d
- = 8.854e-12 × 4 × 0.01 / 0.002
- = 8.854e-12 × 20
- = 177.1e-12 F
Answer: C ≈ 177 pF
Magnetostatics
Biot-Savart Law
d\mathbf{H} = \frac{I\, dl \times \hat{a}_R}{4\pi R^2}
| Symbol | Description | Unit |
|---|---|---|
| \mathbf{H} | Magnetic field intensity | A/m |
| I | Current | A |
| R | Distance from element | m |
Worked example
Find H at the center of a circular loop of radius 0.1 m carrying I = 5 A.
Given: I = 5 A, r = 0.1 m
- H = I / (2r) for center of circular loop
- H = 5 / (2 × 0.1)
- H = 25 A/m
Answer: H = 25 A/m perpendicular to the loop plane
Ampere's Law (Integral Form)
\oint_C \mathbf{H} \cdot d\mathbf{l} = I_{enc}
| Symbol | Description | Unit |
|---|---|---|
| I_{enc} | Total enclosed current | A |
Worked example
Find H at r = 2 cm from an infinite wire carrying 10 A.
Given: I = 10 A, r = 0.02 m
- Circular Amperian loop of radius r
- H × 2πr = I_enc = 10
- H = 10 / (2π × 0.02)
- = 10 / 0.1257 = 79.6 A/m
Answer: H ≈ 79.6 A/m tangentially
Inductance of Toroid
L = \frac{\mu_0 \mu_r N^2 A}{l}
| Symbol | Description | Unit |
|---|---|---|
| L | Inductance | H |
| N | Number of turns | dimensionless |
| A | Cross-sectional area | m² |
| l | Mean circumference | m |
Worked example
Toroid: N = 500, A = 4 cm², l = 0.2 m, μr = 500.
Given: N = 500, A = 4e-4 m², l = 0.2 m, μr = 500, μ₀ = 4π×10⁻⁷
- L = μ₀ × μr × N² × A / l
- = 4π×10⁻⁷ × 500 × 250000 × 4e-4 / 0.2
- = 4π×10⁻⁷ × 500 × 250000 × 0.002
- = 4π×10⁻⁷ × 250000
- = π×10⁻¹ × 0.1 = 0.3142 H
Answer: L ≈ 0.314 H
Maxwell's Equations
Faraday's Law (Differential Form)
\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}
| Symbol | Description | Unit |
|---|---|---|
| \mathbf{E} | Electric field intensity | V/m |
| \mathbf{B} | Magnetic flux density | T |
Worked example
B = 0.1 cos(2π×60t) T in z-direction. Find ∂B/∂t at t = 0.
Given: B = 0.1 cos(120πt) T
- ∂B/∂t = -0.1 × 120π × sin(120πt)
- At t = 0: sin(0) = 0
- ∂B/∂t = 0 at t = 0
Answer: ∂B/∂t = 0 at t = 0 (B is at its peak, rate of change is zero)
Ampere's Law with Displacement Current
\nabla \times \mathbf{H} = \mathbf{J} + \frac{\partial \mathbf{D}}{\partial t}
| Symbol | Description | Unit |
|---|---|---|
| \mathbf{J} | Conduction current density | A/m² |
| \partial\mathbf{D}/\partial t | Displacement current density | A/m² |
Worked example
For a capacitor with εr = 2, A = 0.01 m², d = 1 mm at 60 Hz, V = 100 sin(120πt). Find displacement current.
Given: C = ε₀×εr×A/d = 8.854e-12×2×0.01/0.001 = 177 pF, V = 100 sin(120πt)
- i_d = C × dV/dt
- dV/dt = 100 × 120π × cos(120πt)
- i_d = 177e-12 × 100 × 120π × cos(120πt)
- Peak i_d = 177e-12 × 37699 = 6.67 μA
Answer: Peak displacement current ≈ 6.67 μA
Electromagnetic Wave Propagation
Wave Equation Propagation Constant
\gamma = \alpha + j\beta = \sqrt{j\omega\mu(\sigma + j\omega\epsilon)}
| Symbol | Description | Unit |
|---|---|---|
| \gamma | Propagation constant | Np/m or rad/m |
| \alpha | Attenuation constant | Np/m |
| \beta | Phase constant | rad/m |
| \sigma | Conductivity | S/m |
Worked example
For a lossless medium (σ=0) with εr=4, μr=1 at f=1 GHz, find β.
Given: f = 1e9 Hz, εr = 4, μr = 1, σ = 0
- β = ω√(με) = ω√(μ₀ε₀εr)
- = (2π×1e9)×√(4)/(3×10⁸)
- = (2π×1e9 × 2)/(3×10⁸)
- = (4π×10⁹)/(3×10⁸) = 4π/0.3 = 41.89 rad/m
Answer: β ≈ 41.9 rad/m
Skin Depth
\delta = \frac{1}{\alpha} = \sqrt{\frac{2}{\omega\mu\sigma}}
| Symbol | Description | Unit |
|---|---|---|
| \delta | Skin depth | m |
| \mu | Permeability | H/m |
| \sigma | Conductivity | S/m |
Worked example
Find skin depth in copper (σ = 5.8×10⁷ S/m) at f = 1 MHz.
Given: σ = 5.8e7 S/m, f = 1e6 Hz, μ₀ = 4π×10⁻⁷
- δ = √(2 / (ω × μ₀ × σ))
- ω = 2π × 1e6 = 6.283e6 rad/s
- δ = √(2 / (6.283e6 × 4π×10⁻⁷ × 5.8e7))
- = √(2 / (6.283e6 × 0.07289))
- = √(2 / 457900) = √(4.368e-6)
- = 6.61e-5 m
Answer: δ ≈ 66.1 μm at 1 MHz in copper
Intrinsic Impedance
\eta = \sqrt{\frac{j\omega\mu}{\sigma + j\omega\epsilon}}
| Symbol | Description | Unit |
|---|---|---|
| \eta | Intrinsic impedance | Ω |
| \eta_0 | Free-space value = 377 Ω | Ω |
Worked example
Find intrinsic impedance of lossless medium with εr = 4, μr = 1.
Given: εr = 4, μr = 1, σ = 0
- η = √(μ/ε) = η₀ × √(μr/εr)
- = 377 × √(1/4)
- = 377 × 0.5
- = 188.5 Ω
Answer: η = 188.5 Ω
Poynting Vector (Time-Average Power)
\mathbf{P}_{avg} = \frac{1}{2} \text{Re}(\mathbf{E} \times \mathbf{H}^*)
| Symbol | Description | Unit |
|---|---|---|
| \mathbf{P}_{avg} | Average power density | W/m² |
| \mathbf{H}^* | Complex conjugate of H phasor | A/m |
Worked example
Plane wave in free space with E₀ = 10 V/m. Find average power density.
Given: E₀ = 10 V/m, η₀ = 377 Ω
- P_avg = E₀²/(2η₀)
- = 100 / (2 × 377)
- = 100 / 754
- = 0.1326 W/m²
Answer: P_avg ≈ 132.6 mW/m²
Transmission Lines
Characteristic Impedance
Z_0 = \sqrt{\frac{R + j\omega L}{G + j\omega C}}
| Symbol | Description | Unit |
|---|---|---|
| Z_0 | Characteristic impedance | Ω |
| R, G | Series resistance and shunt conductance per unit length | Ω/m, S/m |
| L, C | Series inductance and shunt capacitance per unit length | H/m, F/m |
Worked example
Lossless coaxial line: L = 300 nH/m, C = 120 pF/m. Find Z₀.
Given: L = 300e-9 H/m, C = 120e-12 F/m
- Z₀ = √(L/C) (lossless)
- = √(300e-9 / 120e-12)
- = √(2500)
- = 50 Ω
Answer: Z₀ = 50 Ω
Reflection Coefficient
\Gamma = \frac{Z_L - Z_0}{Z_L + Z_0}
| Symbol | Description | Unit |
|---|---|---|
| \Gamma | Voltage reflection coefficient | dimensionless |
| Z_L | Load impedance | Ω |
Worked example
Z₀ = 50 Ω, Z_L = 100 Ω. Find reflection coefficient.
Given: Z₀ = 50 Ω, Z_L = 100 Ω
- Γ = (Z_L - Z₀)/(Z_L + Z₀)
- = (100 - 50)/(100 + 50)
- = 50/150
- = 1/3 ≈ 0.333
Answer: Γ = 0.333 (positive, real — partial reflection)
VSWR
VSWR = \frac{1 + |\Gamma|}{1 - |\Gamma|}
| Symbol | Description | Unit |
|---|---|---|
| VSWR | Voltage Standing Wave Ratio | dimensionless |
| |\Gamma| | Magnitude of reflection coefficient | dimensionless |
Worked example
Find VSWR for |Γ| = 0.333.
Given: |Γ| = 0.333
- VSWR = (1 + 0.333) / (1 - 0.333)
- = 1.333 / 0.667
- = 2.0
Answer: VSWR = 2:1
Input Impedance of Lossless Line
Z_{in} = Z_0 \frac{Z_L + jZ_0\tan(\beta l)}{Z_0 + jZ_L\tan(\beta l)}
| Symbol | Description | Unit |
|---|---|---|
| Z_{in} | Input impedance | Ω |
| l | Line length | m |
| \beta l | Electrical length | rad |
Worked example
Z₀ = 50 Ω, Z_L = 100 Ω, βl = π/4 (λ/8 line). Find Z_in.
Given: Z₀ = 50, Z_L = 100, tan(π/4) = 1
- Z_in = 50 × (100 + j50×1)/(50 + j100×1)
- = 50 × (100+j50)/(50+j100)
- Numerator = 50(100+j50) = 5000+j2500
- Denominator = 50+j100
- |Z_in| = 50 × |100+j50| / |50+j100| = 50 × 111.8/111.8 = 50
- Actually Z_in = 50 × (100+j50)/(50+j100) = 50+j0 = 50 Ω (wait, compute: multiply by conjugate)
- = 50×(100+j50)(50-j100)/[(50)²+(100)²]
- = 50×(5000-j10000+j2500+5000)/12500
- = 50×(10000-j7500)/12500 = 50×0.8 - j50×0.6 = 40-j30 Ω
Answer: Z_in = (40 - j30) Ω
Waveguides
Cutoff Frequency of Rectangular Waveguide
f_{c,mn} = \frac{1}{2\sqrt{\mu\epsilon}}\sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2}
| Symbol | Description | Unit |
|---|---|---|
| f_{c,mn} | Cutoff frequency for TE/TM mn mode | Hz |
| a, b | Waveguide dimensions (a > b) | m |
| m, n | Mode indices | dimensionless |
Worked example
Rectangular waveguide in air: a = 4 cm, b = 2 cm. Find f_c for TE10 mode.
Given: a = 0.04 m, b = 0.02 m, m = 1, n = 0
- f_c = (c/2)×√((m/a)²+(n/b)²)
- = (3e8/2)×√((1/0.04)²+0²)
- = 1.5e8 × 25
- = 3.75e9 Hz
Answer: f_c(TE10) = 3.75 GHz
Waveguide Phase Velocity
v_p = \frac{v}{\sqrt{1-(f_c/f)^2}}
| Symbol | Description | Unit |
|---|---|---|
| v_p | Phase velocity in waveguide | m/s |
| v | Speed in medium (c in air) | m/s |
| f | Operating frequency | Hz |
Worked example
TE10 mode, f = 6 GHz, f_c = 3.75 GHz. Find v_p.
Given: f = 6e9, f_c = 3.75e9, v = 3e8
- f_c/f = 3.75/6 = 0.625
- (f_c/f)² = 0.390625
- √(1 - 0.390625) = √0.609375 = 0.7806
- v_p = 3e8 / 0.7806 = 3.843e8 m/s
Answer: v_p ≈ 3.84×10⁸ m/s > c (phase velocity exceeds c in waveguide)
Quick reference
| Formula | Expression |
|---|---|
| Coulomb's Law | F = Q1Q2/(4πε₀r²) |
| Gauss's Law | ∮ D·dS = Q_enc |
| Parallel Plate Capacitance | C = ε₀εr A/d |
| Ampere's Law | ∮ H·dl = I_enc |
| Faraday's Law | ∇×E = -∂B/∂t |
| Propagation Constant | γ = √(jωμ(σ+jωε)) |
| Skin Depth | δ = √(2/(ωμσ)) |
| Intrinsic Impedance | η = √(μ/ε); η₀ = 377 Ω |
| Poynting Vector | P_avg = Re(E×H*)/2 |
| Characteristic Impedance (lossless) | Z₀ = √(L/C) |
| Reflection Coefficient | Γ = (Z_L-Z₀)/(Z_L+Z₀) |
| VSWR | VSWR = (1+|Γ|)/(1-|Γ|) |
| TL Input Impedance | Z_in = Z₀(Z_L+jZ₀tanβl)/(Z₀+jZ_LtanβL) |
| Cutoff Frequency (RWG) | f_c = (c/2)√((m/a)²+(n/b)²) |
| Phase Velocity (WG) | v_p = v/√(1-(f_c/f)²) |
Exam tips
- GATE boundary condition questions test both tangential E (continuous) and normal D (discontinuous by surface charge) — always state both conditions to get full marks.
- For TEM wave problems, verify that E and H are orthogonal and both perpendicular to the direction of propagation before applying the intrinsic impedance formula.
- Transmission line quarter-wave transformer (βl = π/2) transforms Z_L to Z₀²/Z_L — examiners use this frequently for impedance matching problems.
- Skin depth in a good conductor depends on frequency as δ ∝ 1/√f — doubling frequency reduces skin depth by a factor of √2 (≈ 1.41).
- Cutoff wavelength λ_c = 2a for TE10 dominant mode in rectangular waveguide — any frequency below c/(2a) cannot propagate.
- Maxwell's displacement current term ∂D/∂t is what makes EM wave propagation possible in free space — examiners ask what it adds to Ampere's original law.