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Most students can quote C = ε0A/d. Fewer can derive the capacitance of two concentric spheres or a coaxial cable, and those are exactly the questions that test whether you understand capacitance rather than remember it. This post gives one method that works for every shape, applies it to spheres and cylinders, checks the answers against limits you already know, and works two examples with numbers.
Is this on the 2026 syllabus?
The JEE (Main) 2026 syllabus lists "capacitors and capacitance", combinations, and "capacitance of a parallel plate capacitor with and without dielectric medium". The NEET (UG) 2026 syllabus uses the same words. Neither names spherical or cylindrical capacitors.
The current NCERT Class 12 Chapter 2 (reprint 2026-27) derives only the parallel plate case, and its exercises stop at 2.11. The 2022-23 edition (archived copy) had exercise 2.29, deriving the spherical capacitor, and 2.32, a cylindrical capacitor with numbers. Both were removed. Yet every step below uses Gauss's law and potential, which are on the syllabus, so treat these as practice in using listed tools.
What is the four-step method?
Put charge +Q on the inner conductor and −Q on the outer one.
Use Gauss's law to find the field E in the gap.
Integrate E from the inner to the outer conductor to get the potential difference V.
Divide: C = Q/V.
The charge cancels at step 4. If it does not, you have made an error.
How do you derive the spherical capacitor?
Take an inner sphere of radius a and an outer shell of radius b. A Gaussian sphere of radius r between them encloses +Q only, so E = Q/(4πε0r²).
The potential difference is V = (Q/4πε0)(1/a − 1/b) = Q(b − a)/(4πε0ab).
So C = 4πε0ab/(b − a).
Check two limits:
Let b go to infinity. C becomes 4πε0a, the capacitance of an isolated sphere.
Let the gap d = b − a be tiny. C becomes about 4πε0a²/d = ε0A/d, with A = 4πa², which is the parallel plate formula.
Worked example 1: two spheres (sample numbers)
Inner radius a = 10.0 cm, outer radius b = 11.0 cm, air between them, with ε0 = 8.854 × 10⁻¹² F/m from NIST.
C = 4π × 8.854 × 10⁻¹² × (0.10 × 0.11)/0.01 = 1.22 × 10⁻¹⁰ F, about 122 pF.
An isolated 10 cm sphere has 4πε0a = 11.1 pF. The ratio is b/(b − a) = 11: the nearby shell multiplies the capacitance elevenfold.
With 2.0 nC on the inner sphere, V = Q/C = 16.3 V.
For scale, treat Earth as an isolated sphere of radius 6,371 km, the NASA mean radius: C = 4πε0R is only about 709 µF.
A common twist: earth the inner sphere and charge the outer shell. Now the shell also has its outer surface facing infinity, so the two capacitances add: 4πε0ab/(b − a) + 4πε0b = 4πε0b²/(b − a). For the same spheres that is 135 pF.
How do you derive the cylindrical capacitor?
Take an inner cylinder of radius a and outer radius b, length L much greater than b, so end effects can be ignored. With charge per length λ = Q/L, a coaxial Gaussian cylinder gives E = λ/(2πε0r).
Integrating, V = (λ/2πε0) ln(b/a). So C = 2πε0L/ln(b/a). With a dielectric of constant K filling the gap, multiply by K.
Worked example 2: a coaxial cable (sample numbers)
Inner radius 0.50 mm, outer radius 1.75 mm, filled with a dielectric of K = 2.3.
ln(1.75/0.50) = ln 3.5 = 1.253.
C per metre = 2π × 8.854 × 10⁻¹² × 2.3/1.253 = 1.02 × 10⁻¹⁰ F/m, about 102 pF/m.
For 10 m of cable, C = 1.02 nF. With air instead, it would be 44.4 pF/m.
Only the ratio b/a matters, so doubling both radii leaves C per metre unchanged.
Which mistakes cost marks?
Using diameters in the spherical formula. ab/(b − a) doubles if you double every length, so your answer is off by a factor of 2.
Using log base 10. log₁₀ 3.5 = 0.544, not 1.253, and C comes out about 2.3 times too big.
Swapping limits. If C comes out negative, you integrated the wrong way.
Forgetting K when a dielectric fills the gap.
Using 4πε0ab/(b − a) when the inner sphere is earthed.
If your integral keeps going wrong, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can find the step where it slipped.
A practice routine
Derive both formulas on a blank page, saying each of the four steps aloud.
Check both limits for the spherical result.
Try the old NCERT exercises 2.30 and 2.32. You should get about 5.5 nF and 450 V, then about 121 pF and 2.9 × 10⁴ V.
Revise networks and dielectrics with our capacitors and dielectrics guide, and surface choice with our Gauss's law guide.
Tomorrow, derive C = 2πε0L/ln(b/a) in under five minutes without notes.
