Playground
Parallel-plate capacitor: adjust plate area and separation to see capacitance and stored E-field change.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Capacitanceoutput Charge stored per unit voltage | F | I²·T⁴·M⁻¹·L⁻² | 1e-12 – 0.000001 | |
| Plate area Area of one capacitor plate | m² | L² | 0.0001 – 1 | |
| Plate separation Distance between the two parallel plates | m | L | 0.00001 – 0.1 |
Deep dive
Derivation
Apply Gauss's law to one plate: E = σ/ε₀ = Q/(ε₀A). Voltage across the gap: V = Ed = Qd/(ε₀A). Therefore C = Q/V = ε₀A/d.
Experimental verification
Measured using bridge circuits since the 19th century. Modern MEMS capacitors confirm the formula down to micrometer-scale separations. Deviations at nanometer scales reveal quantum capacitance effects.
Common misconceptions
- Capacitance depends only on geometry and the dielectric, not on charge or voltage
- The formula assumes a uniform field — it fails near the plate edges
- Real capacitors have parasitic inductance and resistance not captured by this formula
Real-world applications
- Touchscreen sensors in smartphones and tablets
- DRAM memory cells in computers (capacitor stores 1 bit)
- Power factor correction in AC circuits
- Energy storage in camera flash units
Worked examples
Vacuum capacitor
Given:
- A:
- 0.01
- d:
- 0.001
Find: C
Solution
C = ε₀ × A / d = 8.854×10⁻¹² × 0.01 / 0.001 = 8.854×10⁻¹¹ F = 88.54 pF
Energy stored in a charged capacitor
Given:
- C:
- 8.854e-11
- V_applied:
- 100
Find: Energy U
Solution
U = ½CV² = 0.5 × 88.54×10⁻¹² × 100² = 4.43×10⁻⁷ J = 0.443 μJ
Scenarios
What if…
- scenario:
- What if you insert a glass dielectric (κ = 5)?
- answer:
- Capacitance increases 5×: C = 5 × 88.54 pF = 442.7 pF. The dielectric polarizes and partially cancels the internal field, allowing more charge per volt.
- scenario:
- What if the plate separation doubles?
- answer:
- Capacitance halves: C = 44.27 pF. Wider gap means weaker field for the same charge, so less charge per volt.
- scenario:
- What if you connect two identical capacitors in parallel?
- answer:
- Total capacitance doubles: C_total = 2 × 88.54 = 177.08 pF. Parallel connection effectively doubles the plate area.
Limiting cases
- condition:
- d → 0
- result:
- C → ∞
- explanation:
- Infinitesimally close plates store infinite charge per volt — in practice, dielectric breakdown limits this.
- condition:
- A → 0
- result:
- C → 0
- explanation:
- Zero plate area means no charge storage.
- condition:
- d → ∞
- result:
- C → 0
- explanation:
- Widely separated plates cannot maintain a field to store energy.
Context
Ewald Georg von Kleist / Pieter van Musschenbroek · 1745
The Leyden jar — the first capacitor — was independently discovered by von Kleist and van Musschenbroek. The parallel-plate formula was derived later from Gauss's law and the uniform field between plates.
Hook
How does your phone store charge without any chemical reaction?
A parallel-plate capacitor has plates of area 0.01 m² separated by 0.001 m in vacuum. Find the capacitance.
Dimensions: [C] = [ε₀]·[A]/[d] → (F/m)(m²)(m⁻¹) = F ✓
Validity: Valid for ideal parallel plates where A >> d² (uniform field approximation). Edge fringing effects become significant when plate separation approaches plate dimensions. Add dielectric constant κ for filled capacitors.