Capacitance Calculator - Parallel Plate Capacitor
Calculate the capacitance of a parallel plate capacitor from plate area, plate separation, and dielectric constant.
Capacitance
88.593 pF
0.088593 nF | 8.8593e-11 FHow to Calculate Parallel Plate Capacitance
A parallel plate capacitor's capacitance depends on the plate area, the distance between the plates, and the dielectric material filling the gap. Enter the plate area, plate separation, and pick a dielectric preset (or enter a custom relative permittivity) to get the capacitance.
Example
Two plates with 100 cm² of area, separated by 1mm of air (εr ≈ 1), give a capacitance of C = 8.854×10⁻¹² × 1 × 0.01 ÷ 0.001 ≈ 8.854×10⁻¹¹ farads, or about 88.5 picofarads (pF).
Common Use Cases
- Designing a simple parallel-plate capacitor for a physics lab or hobby electronics project.
- Comparing how different dielectric materials affect capacitance for the same geometry.
- Physics or electrical engineering coursework on capacitors and electric fields.
FAQs
What is a dielectric constant (relative permittivity)?
It's the ratio of a material's permittivity to that of a vacuum, describing how much better that material is at storing electric field energy compared to empty space. A higher dielectric constant means more capacitance for the same plate area and spacing.
Does this account for real-world fringing effects?
No — this uses the idealized formula that assumes a uniform electric field between infinite parallel plates. Real capacitors have fringing fields at the plate edges, which make measured capacitance somewhat higher than this calculation, especially when the plate separation isn't small compared to the plate size.
How does the distance between plates affect capacitance?
Capacitance is inversely proportional to plate separation, so halving the gap between the plates doubles the capacitance, while doubling the gap halves it.