Spherical & Cylindrical Capacitors
Curved plates, and the limits that connect them.
Lesson at a glanceShow the key points
- Spherical: C = 4πε₀R₁R₂/(R₂ − R₁) (31.3).
- Isolated sphere: C = 4πε₀R.
- Large radii, fixed gap: C → ε₀A/d.
- Cylindrical: C = 2πε₀l/ln(R₂/R₁) (31.4).
The same recipe (charge the plates, find E, integrate to V) works for curved conductors. Spheres and cylinders are the two shapes symmetric enough to do by hand.
Spherical capacitor
A conducting sphere of radius R₁ sits inside a concentric hollow conducting sphere of radius R₂. Give the inner one +Q and the outer one −Q. Between them, a concentric Gaussian sphere of radius r encloses Q, so E = Q/4πε₀r², radially outward. Integrating from R₂ in to R₁:
Concentric spheres of radii 9.0 cm and 10 cm form a capacitor. Find C.
Two limits worth knowing
Isolated sphere. Send the outer sphere off to infinity (R₂ → ∞). Then R₁R₂/(R₂ − R₁) → R₁ and
Parallel-plate limit. Make both radii large but keep the gap d = R₂ − R₁ fixed. Then 4πR₁R₂ ≈ 4πR² = A, the area of either sphere, and C → ε₀A/d: the curved plates look flat from up close.
An isolated metal sphere has a capacitance of 20 pF. (a) Find its radius. (b) How much charge raises it to 50 kV?
Cylindrical capacitor
A conducting cylinder of radius R₁ inside a coaxial hollow cylinder of radius R₂, both of length l, much longer than the gap. With +Q inside and −Q outside, the charge per length is Q/l. A coaxial Gaussian cylinder of radius r and length x gives E · 2πrx = (Q/l)x/ε₀, so E = Q/2πε₀rl. Integrating across the gap:
A coaxial cable has an inner wire of radius 0.50 mm and an outer sheath of radius 2.0 mm, with vacuum (or air) between. Find its capacitance per metre.
- 1Spherical: C = 4πε₀R₁R₂/(R₂ − R₁) (31.3).
- 2Isolated sphere: C = 4πε₀R. Thin gap: reduces to ε₀A/d.
- 3Cylindrical: C = 2πε₀l/ln(R₂/R₁) (31.4).
Check yourself
Keep exploring
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