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UY1: Capacitors And Capacitance

Define capacitance, units, and the parallel-plate formula, with geometric intuition and assumptions.

  • University Physics Year 1
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Learning objectives

  • Analyse capacitance, resistance, energy transfer, and transient circuit behaviour.
Why this matters + quick links

This page gives the UY1 working model/result for Capacitors And Capacitance. You reuse it when you build fields/potentials by symmetry or superposition, and when you connect fields to forces, energy, and circuits.

A capacitor is any pair of conductors separated by an insulator or vacuum that can store equal and opposite charge.

1) At a glance

  • Prerequisites: potential difference (Electric Potential) and the uniform-field plate model (Electric Field Of Two Oppositely Charged Infinite Sheets)
  • Outcomes: define capacitance and derive C = ε₀A/d for ideal parallel plates (and know the assumptions)
  • Definition: C≡Q/(Δ V)
  • Key result (vacuum, ideal large plates): C = ε₀A/d
  • SI unit: farad (F), where 1 F = 1 C/V
  • Common trap: using 2Q in C = Q/Δ V (use the magnitude on one plate) or applying C = ε₀A/d when fringing is not negligible

Motivation / intuition

Capacitance measures how easily a geometry can store separated charge for a given voltage. The key idea is that C depends on geometry/material, not on “how much you charged it” (that dependence cancels in C = Q/Δ V).

2) Setup

Use Q as the magnitude of charge on one plate (plates carry + Q and -Q).

For ideal large parallel plates (area A, separation d):

  • fringe effects neglected (d much smaller than plate dimensions)
  • field between plates approximately uniform

3) Core derivation/explanation

For opposite infinite sheets, field between plates is E = σ/ε₀ = Q/ε₀A Potential difference magnitude: Δ V = Ed = Qd/ε₀A Therefore, C = Q/(Δ V) = ε₀A/d

Why this step matters

Expressing Δ V in terms of Q, geometry, and ε₀ makes the charge cancel in C = Q/Δ V. That cancellation shows capacitance is a geometry/material property, not a “how much you charged it” property.

Key interpretation:

  • larger area A gives larger C
  • larger spacing d gives smaller C
  • for a fixed geometry and material, C is constant (independent of Q and Δ V)

Useful constants/units:

  • ε₀ ≈ 8.854 × 10⁻¹² F/m
  • C in farads is often small, so practical units include μF, nF, pF
Assumptions + pitfalls
  • The parallel-plate formula assumes a nearly uniform field between plates (large plates, small separation), so fringing is neglected.
  • Use Q as the magnitude on one plate in C = Q/Δ V (do not use 2Q).
  • Unit slips are common: convert d from mm to m before using C = ε₀A/d.
Quick checks (units + limits/sign)
  • Units: [C] = C/V = F and [ε₀A/d] = (F/m)m²/m = F.
  • Limits/signs: C must be positive; increasing A increases C, increasing d decreases C; if Δ V → 0 with fixed C, then Q = CΔ V → 0.

4) Worked example(s)

Parallel plates: A = 0.020 m², d = 1.0 mm.

C = ε₀A/d = ((8.854 × 10⁻¹²)(0.020))/(1.0 × 10⁻³) ≈ 1.77 × 10⁻¹⁰ F So C ≈ 177 pF.

If charged to Δ V = 120 V: Q = CΔ V ≈ (1.77 × 10⁻¹⁰)(120) = 2.12 × 10⁻⁸ C

5) Practice set (with hints + answers)

  1. If plate area doubles and d unchanged, how does C change?
  2. If d triples and A unchanged, how does C change?
  3. A capacitor has C = 10 μF at 12 V. Find Q.

Hints

  1. Use C ∝ A.
  2. Use C ∝ 1/d.
  3. Use Q = CΔ V.

Answers

  1. It doubles.
  2. It becomes one-third.
  3. Q = 120 μC.

6) Summary + next steps

  • Capacitance quantifies charge stored per unit potential difference.
  • For ideal parallel plates in vacuum, C = ε₀A/d.
  • Geometry and material set capacitance; charging state does not.

Next: Capacitors In Series And In Parallel Previous: Using Gauss’s Law For Common Charge Distributions Back To Electromagnetism (UY1)