Calculate capacitance of a spherical capacitor from inner/outer radii and dielectric. Includes charge, energy, E-field, surface charge density, cross-section visual, and dielectric comparison table.
A spherical capacitor consists of two concentric conducting spheres separated by a dielectric material. Its capacitance depends on the inner radius a, outer radius b, and dielectric constant κ: C = 4πε₀κ(ab)/(b−a). This geometry appears in Van de Graaff generators, high-voltage components, and idealized electrostatics models.
The electric field between the spheres can be derived exactly from Gauss's law: E = Q/(4πε₀κr²). Because the field is strongest at the inner surface and falls as 1/r², the inner conductor is usually the first place to check for dielectric breakdown.
This calculator computes capacitance, stored charge, energy, maximum electric field, and surface charge density, then shows the geometry in an SVG cross-section so you can see how radius and dielectric choice change the result.
Spherical capacitors are a compact way to study how geometry changes capacitance and field strength. The formulas are simple enough to write down, but the unit conversions and derived values are easy to slip on if you do them by hand.
This calculator keeps the electrical result, the field check, and the dielectric comparison together so you can inspect the full setup before using it in a design or classroom example.
Capacitance: C = 4πε₀κ × (a × b) / (b − a) Isolated Sphere (b → ∞): C = 4πε₀κa Charge: Q = C × V Stored Energy: E = ½CV² Max Electric Field (at r = a): E_max = V × b / [a × (b − a)] Surface Charge Density: σ = Q / (4πa²) Where: ε₀ = 8.854 × 10⁻¹² F/m κ = relative permittivity
Result: C = 11.13 pF
With a = 50 mm = 0.05 m and b = 100 mm = 0.1 m in air (κ ≈ 1): C = 4π × 8.854×10⁻¹² × 1 × (0.05 × 0.1) / (0.1 − 0.05) = 11.13 pF. At 1000 V, the stored charge is Q = 11.13 nC and the stored energy is E = 5.56 μJ.
Applying Gauss's law to a spherical surface of radius r between the conductors: ∮E·dA = Q/ε₀κ, giving E = Q/(4πε₀κr²). Integrating E from a to b gives the potential difference: V = Q/(4πε₀κ) × (1/a − 1/b). Since C = Q/V, we get C = 4πε₀κ(ab)/(b−a). This elegant derivation shows why the spherical capacitor is a classic problem in electrostatics courses.
While perfect concentric spheres are rare in commercial electronics, the spherical capacitor model appears in several practical contexts. Van de Graaff generators use a large conducting sphere to accumulate charge, with the dome acting as one electrode and the ground (or a grounded shell) as the other. High-voltage bushings and feedthroughs sometimes use spherical geometries to minimize electric field concentration.
The three analytically solvable capacitor geometries — parallel plate, cylindrical, and spherical — form a hierarchy of increasing mathematical complexity. The parallel plate (C = ε₀A/d) has a uniform field, the cylindrical (C = 2πε₀L/ln(b/a)) has a 1/r field, and the spherical has a 1/r² field. Understanding all three provides deep insight into electrostatic energy storage.
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The electric field is E = Q/(4πε₀κr²), so it is inversely proportional to r². The smallest radius (inner surface) has the highest field concentration, making it the first location to exceed the dielectric breakdown strength.
The formula simplifies to C = 4πε₀κa — the capacitance of an isolated sphere. This is a useful concept in electrostatics: even a single conductor has capacitance with respect to infinity.
For the same gap and surface area, a spherical capacitor has a non-uniform electric field (stronger near the inner sphere), while a parallel plate has a uniform field. When the gap is much smaller than the radius, the two converge.
This calculator assumes full concentric spheres. For a hemisphere, the capacitance is approximately half. For partial spheres or non-concentric geometries, numerical methods (FEM) are typically needed.
The energy density is u = ½ε₀κE² and varies with position (r). It is highest at the inner surface and decreases as 1/r⁴. The total energy is the integral over the volume: U = ½CV².
Water molecules are strongly polar (permanent dipole moment). In an electric field, they orient to oppose the field, effectively reducing it and increasing capacitance. This high κ = 80 is why water is such a good solvent for ionic compounds.