Capacitance of parallel plate, coaxial and spherical capacitors

Why it matters

Capacitive transducers measure displacement, pressure, level, humidity and proximity by changing the plate area, gap or dielectric of a capacitor. Coaxial cables connecting sensors to instruments have a capacitance per metre that loads high-impedance sources and limits bandwidth. Being able to derive C for the three standard geometries — and to see how C changes with each dimension — is core instrumentation knowledge.

Key ideas

Definition. A capacitor is two conductors carrying equal and opposite charges +Q and −Q. Capacitance is C = Q/V, where V is the potential difference between them. It depends only on geometry and the permittivity of the material between the conductors — not on Q or V (for a linear dielectric). Unit: farad (F = C/V); practical values are pF to μF.

General recipe (works for every geometry):

1. Assume charge +Q on one conductor and −Q on the other.
2. Find E between them, usually with Gauss's law.
3. Find V = −∫E·dl from the negative to the positive conductor.
4. C = Q/V — the Q cancels.

Parallel plates. Field E = ρS/ε = Q/(εA) is uniform if fringing is neglected (valid when the gap d is much smaller than the plate dimensions). V = Ed, so C = εA/d. For a displacement sensor, C ∝ 1/d is non-linear in d but linear in 1/d; a change of area or of εr gives a linear response.

Coaxial (cylindrical) capacitor. Inner radius a, outer radius b, length L. E = Q/(2πεLρ) between conductors; V = (Q/2πεL) ln(b/a). The field is highest at the surface of the inner conductor, which is where insulation stress is greatest.

Spherical capacitor. Inner radius a, outer radius b. E = Q/(4πεr²); V = (Q/4πε)(1/a − 1/b). As b → ∞ this becomes an isolated sphere, C = 4πεa.

Combinations. Dielectric layers stacked across the gap (field passing through each in turn) behave as capacitors in series: D is common, and the voltage divides in inverse proportion to each layer's capacitance. Dielectrics placed side by side (each spanning the full gap) behave as capacitors in parallel: E is common.

Fixed charge versus fixed voltage. Inserting a dielectric of εr multiplies C by εr. With the battery connected (V fixed), Q rises εr times and E is unchanged. With the capacitor isolated (Q fixed), V and E fall by εr.

Formulas

• C = Q / V — capacitance (F); Q in C, V in V.
• C = ε₀εr A / d — parallel plates; A = plate area (m²), d = gap (m); fringing neglected (d ≪ plate size).
• C = 2πε₀εr L / ln(b/a) — coaxial; a, b = inner and outer radii (m), L = length (m). Per metre: C/L = 2πε / ln(b/a).
• E_max = V / (a ln(b/a)) — coaxial, at ρ = a.
• C = 4πε₀εr ab / (b − a) — concentric spheres, radii a < b (m).
• C = 4πε₀εr a — isolated sphere of radius a.
• C = A / (d1/ε1 + d2/ε2) — two dielectric layers in series between parallel plates.
• C = (ε1A1 + ε2A2) / d — two dielectrics side by side.
• ε₀ = 8.854 × 10⁻¹² F/m.

Worked examples

Example 1 (standard). A parallel-plate capacitor has A = 0.02 m², d = 1 mm and a dielectric of εr = 2.5. Find C, and the charge and field at 100 V.

1. C = ε₀εr A / d = 8.854 × 10⁻¹² × 2.5 × 0.02 / 0.001 = 4.427 × 10⁻¹⁰ F = 442.7 pF.
2. Q = CV = 442.7 × 10⁻¹² × 100 = 44.3 nC.
3. E = V/d = 100 / 0.001 = 100 kV/m (0.1 kV/mm).

Example 2 (coaxial cable). A coaxial cable has a = 0.5 mm, b = 1.75 mm and polyethylene insulation, εr = 2.25. Find the capacitance per metre and the maximum field when 1 kV is applied.

1. ln(b/a) = ln 3.5 = 1.2528.
2. C/L = 2πε₀εr / ln(b/a) = 2π × 8.854 × 10⁻¹² × 2.25 / 1.2528 = 99.9 pF/m.
3. E_max = V / (a ln(b/a)) = 1000 / (0.5 × 10⁻³ × 1.2528) = 1.60 MV/m (1.6 kV/mm), at the inner conductor surface. A 10 m run therefore adds about 1 nF across the sensor — important for piezoelectric and other high-impedance sources.

Example 3 (GATE level, layered dielectric). Plates of area 0.01 m² are 3 mm apart: 2 mm of glass (εr = 4) and a 1 mm air gap in series. 6 kV is applied. Find C and the field in each layer; does the air break down (take its dielectric strength as 3 MV/m)?

1. C = ε₀A / (d1/εr1 + d2/εr2) = 8.854 × 10⁻¹² × 0.01 / (0.002/4 + 0.001/1) = 8.854 × 10⁻¹⁴ / 0.0015 = 59.0 pF.
2. D is the same in both layers: εr1E1 = εr2E2, so E_air = 4E_glass.
3. Voltage: E_glass × 0.002 + 4E_glass × 0.001 = 6000 → E_glass × 0.006 = 6000 → E_glass = 1 MV/m, E_air = 4 MV/m.
4. Check: 1 × 10⁶ × 0.002 + 4 × 10⁶ × 0.001 = 2000 + 4000 = 6000 V ✓.
5. E_air = 4 MV/m exceeds 3 MV/m, so the air gap breaks down even though the glass is lightly stressed — the reason voids in insulation are dangerous.

Common mistakes

• Using log base 10 instead of the natural log in ln(b/a).
• Using diameters where radii are needed (the ratio b/a is the same, but E_max = V/(a ln(b/a)) needs the radius).
• Forgetting unit conversions: cm² to m² is ×10⁻⁴, mm to m is ×10⁻³.
• Treating layered dielectrics as parallel instead of series (or vice versa).
• Assuming a bigger outer sphere increases capacitance. For fixed a, increasing b reduces C towards the isolated-sphere value 4πεa.
• Assuming the dielectric always lowers E: at fixed voltage E = V/d is unchanged.

For GATE IN

Expect NAT items computing C for the three geometries, capacitance per metre of a coaxial line, the effect of partially filling a gap with a dielectric (series or parallel), field in each layer, and sensitivity of capacitive displacement sensors (dC/dd = −εA/d², so sensitivity rises as the gap falls). Practise the four-step recipe so you can derive any formula you forget.

Quick check

1. What happens to C of a parallel-plate capacitor if the gap is doubled?
2. A coaxial cable is made twice as long; how does its capacitance change?
3. What is C of an isolated sphere of radius 1 m in air?
4. A capacitor is charged and disconnected, then a dielectric of εr = 3 is inserted. What happens to V?
5. In a series air-and-glass stack, which layer has the higher field?

Answers: 1. It halves. 2. It doubles. 3. About 111 pF. 4. It falls to one third. 5. The air layer.