Parallel-Plate Capacitance
Capacitance from plate area and separation.
Formula
Variables
How it works
For two large, closely spaced, parallel conducting plates with air or vacuum between them, the capacitance is set by geometry: larger plate area lets the plates hold more opposite charge, while a smaller gap makes it easier for charge to build up a strong electric field. The formula comes from combining the electric field between ideal parallel plates with the definition of capacitance, C = Q/V, using the fact that the potential difference grows in proportion to separation d and falls with increasing area A. In this idealized case, the result depends only on the plate dimensions and the permittivity of free space, not on the amount of charge already stored.
When to use it
Use this for ideal or nearly ideal parallel-plate capacitors, especially in physics and engineering problems involving air/vacuum gaps and plate geometry. It is also the starting point for estimating how changing plate area or spacing affects capacitance before introducing dielectric materials or edge effects.
Worked example, step by step
A parallel-plate capacitor has plate area A = 0.025 m^2 and plate separation d = 1.5 mm. Convert the separation to meters: d = 1.5 × 10^-3 m Use ε0 = 8.85 × 10^-12 F/m Substitute into C = ε0A/d: C = (8.85 × 10^-12)(0.025) / (1.5 × 10^-3) C = 2.2125 × 10^-13 / 1.5 × 10^-3 C = 1.475 × 10^-10 F So the capacitance is C ≈ 1.48 × 10^-10 F, or about 148 pF.
Common mistakes to avoid
- Forgetting to convert the separation into meters before substituting
- Using the formula for a capacitor with a dielectric without multiplying by the material's relative permittivity
- Assuming the result is exact even when the plates are not much wider than the gap, where edge effects matter
Did You Know?
Bringing capacitor plates closer boosts capacitance, the trick that shrinks the huge capacitors of a century ago.
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