Potential & Flux

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Electric Potential (Voltage)

Key Concepts

Potential from point charge Q

V = Q / (4πε₀r)

Scalar — Volts (V). Reference at infinity (V=0)

Work done moving charge q

W = q·VAB = q(VA − VB)

Joules (J). Positive W = energy put in by you

Potential difference

VAB = VA − VB = −∫[B→A] E·dl

Path-independent in electrostatics!

E from potential

E = −∇V

E points from high V to low V (downhill)

E between parallel plates (uniform)

|E| = |VB−VA| / d

V vs E Comparison

Property	V	E
Type	Scalar	Vector
Unit	Volt (V)	V/m
At conductor surface	Constant (equipotential)	⊥ to surface
Relation	V = −∫E·dl	E = −∇V

📌 Equipotential Surfaces

Surfaces where V = constant. E is always perpendicular to them. Zero work done moving charge along an equipotential.

Electric Flux Density (D Field)

D Field Formulas

Flux density

D = ε₀E + P = ε₀εᵣE = εE

C/m² — ε = ε₀εᵣ, εᵣ = relative permittivity (dielectric constant)

Gauss's Law (integral form)

∮ D·dS = Q_enclosed

Gauss's Law (differential form)

∇·D = ρᵥ

Always true regardless of material

💡 Why use D instead of E?

D is independent of the medium — ∇·D = ρᵥ always. E changes with ε. Easier for problems with dielectric interfaces.

εᵣ Values & Polarization

Material	εᵣ
Vacuum / Air	1.0
Paper	3.5
Glass	5–10
Silicon	11.7
Sea Water	72–80

🔬 Polarization P

In a dielectric, applied E field shifts +/− charge centers → dipoles form. This is polarization P. The εᵣ captures this effect: D = ε₀E + P.

Capacitors

Capacitor Formulas

Capacitance definition

C = Q / V [Farads]

Parallel-plate capacitor

C = ε₀εᵣA / d

A = plate area, d = separation, εᵣ = dielectric constant

Energy stored

W = ½CV² = Q²/(2C)

Joules (J)

Capacitors in parallel

C_eq = C₁ + C₂ + ... + Cₙ

Same voltage, total charge adds

Capacitors in series

1/C_eq = 1/C₁ + 1/C₂ + ...

Same charge, voltages add

Why Use a Dielectric?

Increases capacitance — C = ε₀εᵣA/d → higher εᵣ means higher C
Increases breakdown voltage — dielectric can withstand larger E before conducting
Mechanical support — keeps plates from touching, allows smaller d → larger C

⚠️ Dielectric Breakdown

If E exceeds the dielectric strength of the material, it starts conducting. Sparks appear. This permanently damages the capacitor. Air: ~3 MV/m.

Tutorial 4 — Potential Problems

Tutorial 4 · Problem 3 — Potential from two charges

Given: q₁=2μC at origin, q₂=−6μC at (0,3)m. Find V at P(4,0)m, then energy to move q₃=3μC from ∞ to P.

r₁ = distance from q₁ to P = √(4²+0²) = 4 m
r₂ = distance from q₂ to P = √(4²+3²) = 5 m
V = q₁/(4πε₀r₁) + q₂/(4πε₀r₂)
V = [2×10⁻⁶/(4π×8.854×10⁻¹²×4)] + [−6×10⁻⁶/(4π×8.854×10⁻¹²×5)]

V = −6.29×10³ V
W = q₃V = 3×10⁻⁶×(−6.29×10³) = −1.89×10⁻² J

Tutorial 4 · Problem 4 — E between parallel plates

Given: 12V battery connected to two parallel plates, separation d = 0.30 cm = 0.003 m.

For uniform field between plates:
E = V/d = 12 / (0.3×10⁻²)

E = 4×10³ V/m (directed from + plate to − plate)

Tutorial 4 · Problem 5 — E from given V(x)

Given: V(x) = (ρ₀/aε₀)(1 − e^(−ax)). Find (a) E, (b) potential difference x=d to x=0.

(a) E = −∇V = −dV/dx · âx
dV/dx = (ρ₀/aε₀)(ae^(−ax)) = (ρ₀/ε₀)e^(−ax)
E = −(ρ₀/ε₀)e^(−ax) · âx

(b) V₀→d = V(0) − V(d) = 0 − (ρ₀/aε₀)(1−e^(−ad))
= −(ρ₀/aε₀)(1−e^(−ad)) V

Tutorial 4 · Problem 6 — D field from point charge

Given: Point charge Q=12nC at origin. Find D at P(0,−3,2).

r̄ = 0âx − 3ây + 2âz
|r| = √(0+9+4) = √13
D = Q·r̄ / (4π|r|³)

D = −61.12ây + 70.75âz pC/m²

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EE-I Exam Prep · GIU Cairo · Created by Omar Nader