Parallel Plate Capacitor Calculator
Enter area, separation and relative permittivity to compute capacitance, charge, energy, field and surface charge density.
Input Data
Results
At a glance:A parallel plate capacitor stores charge Q on two plates separated by distance d and area A. Its capacitance is C = ε₀·ε_r·A/d, where ε₀ = 8.854×10⁻¹² F/m and ε_r the relative permittivity of the dielectric (1 for vacuum/air, ~2–80 for ceramics/water). With voltage V: Q = C·V, stored energy U = ½·C·V² = ½·Q·V, uniform field E = V/d (between plates, edge effects ignored) and surface charge density σ = Q/A = ε₀·ε_r·E. Inserting a dielectric increases C by ε_r and, if isolated, reduces V for fixed Q. This tool returns C, Q, U, E and σ from A, d, ε_r and V.
Formula
C = ε₀·ε_r·A/d
Q = C·V
U = ½·C·V²
E = V/d
σ = Q/A
$$C = \frac{\varepsilon_0 \varepsilon_r A}{d}, \quad Q = CV, \quad E = \frac{1}{2}CV^2, \quad E_{\text{field}} = \frac{V}{d}$$How to Use
- Enter plate area A and separation d.
- Enter relative permittivity ε_r and voltage V.
- The calculator returns C, Q, U, E and σ.
Case Studies
Ceramic capacitor
A = 0.01 m², d = 1e-4 m, ε_r = 1000, V = 12 V.
C = 8.854e-12×1000×0.01/1e-4 ≈ 0.885 µF.
Q = 1.06e-5 C, U = 6.4e-5 J.
FAQ
How does a dielectric affect capacitance?
C scales with ε_r, so a dielectric (ε_r>1) increases capacitance and energy storage at the same voltage (or lets you keep charge with lower voltage if isolated).
Why is the field E = V/d?
For a uniform field between large parallel plates, the potential drops linearly: V = E·d, so E = V/d (ignoring fringing at the edges).
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References
Content review: Calculatorism Science Team. Results are for reference only; please refer to the relevant authorities for the official figures.