Derive an expression for the capacitance of a parallel plate capaci...

Question

Derive an expression for the capacitance of a parallel plate capacitor with air present between the two plates.

Accepted Answer

📌 Derivation:

Setup:
• Two parallel plates, each of area A
• Separation between plates = d
• Medium = air (vacuum), permittivity = ε₀

Step 1 — Surface Charge Density:
σ = Q / A

Step 2 — Electric Field Between Plates:
Using Gauss's Law:
E = σ / ε₀ = Q / (ε₀A)

Step 3 — Potential Difference:
(field is uniform between plates)
V = E × d = Qd / (ε₀A)

Step 4 — Capacitance:
C = Q / V = Q / [Qd/(ε₀A)]

┌─────────────────────┐
│ C = ε₀A / d │
└─────────────────────┘

Conclusion:
Capacitance increases with:
• Larger plate area A ↑
• Smaller separation d ↓

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