(c) capacitance C = 4πε₀R for an isolated sphere

The idea

An isolated conducting sphere of radius R stores charge on its surface, and its potential is V = Q/(4πε₀R). Since capacitance is defined by C = Q/V, substituting gives C = 4πε₀R. The capacitance depends only on the radius of the sphere and the permittivity of free space, not on the charge stored. For R = 0.10 m, C = 4π × 8.85 × 10⁻¹² × 0.10 ≈ 1.1 × 10⁻¹¹ F, a very small value. This shows why isolated spheres are poor capacitors compared with parallel plates. The equation applies to a sphere in free space, with R measured from the centre to the outer surface.

What earns the marks

• States that capacitance is charge stored per unit potential difference, C = Q/V.
• Combines V = Q/(4πε₀R) with C = Q/V to derive C = 4πε₀R.
• Uses R as the radius of the isolated sphere and ε₀ = 8.85 × 10⁻¹² F m⁻¹.
• Recognises that capacitance depends only on R and ε₀, not on the charge or potential.
• Evaluates C for a given radius and comments on the very small magnitude compared with typical capacitors.

Common mistakes and how to put them right

• Using the diameter instead of the radius in C = 4πε₀R.

  Put it right: R is the radius from the centre to the surface.

• Thinking capacitance increases as more charge is added.

  Put it right: For an isolated sphere, C is fixed by its geometry.

• Confusing the sphere equation with the parallel-plate equation C = ε₀A/d, which applies to a different geometry.

• Omitting the factor 4π when substituting, which changes the answer by more than an order of magnitude.

Exam tips

• Write C = 4πε₀R, substitute R in metres and ε₀ in F m⁻¹, and check the unit of the result is the farad.
• If asked to compare, calculate both values and state the ratio or order-of-magnitude difference explicitly.
• Keep the full expression until the final step to avoid rounding errors in the 4π factor.