Expression for Capacitance of a Parallel-Plate Capacitor

Think of a capacitor like a water tank — the plate area (A) is like the base area of the tank (wider tank stores more water for the same height), and plate separation (d) is like the tank's wall thickness restricting flow. A larger base and thinner walls "store more" for the same effort — similarly, a bigger A and smaller d give higher capacitance.

Setup and Assumptions

• Two large plane parallel conducting plates, each of area A
• Separated by a small distance d (so that d << plate dimensions, edge effects ignored)
• Plates carry equal and opposite charges +q and −q
• Medium between plates: vacuum/air

Step 1: Surface Charge Density

• σ = \[\frac {q}{A}\]

Step 2: Electric Field Between the Plates

Each plate produces a field of magnitude \[\frac {σ}{2ε_0}\]​ in the region between them; the fields from both plates add up (same direction) in the gap:

• E = \[\frac{\sigma}{2\varepsilon_{0}}+\frac{\sigma}{2\varepsilon_{0}}=\frac{\sigma}{\varepsilon_{0}}=\frac{Q}{\varepsilon_{0}A}\]

Step 3: Potential Difference Between Plates

Problem: A parallel-plate capacitor has a plate area of 100 cm² and a separation of 2 mm, with vacuum between the plates. If the field between the plates is 100 N/C, find the charge on each plate.

Solution:
Given: A = 100 cm² = 100 × 10⁻⁴ m² = 10⁻² m², E = 100 N/C

Using E = \[\frac {Q}{ε_0A}\]​, so Q = ε₀AE

Answer: Q = 8.85 × 10⁻¹² C = 8.85 pC