Question

Assuming Earth as a spherical conductor of diameter 12,800 km, calculate its capacity.

Hint:

Use \( C = 4\pi\varepsilon_0 R = R/(9\times10^{9}) \) with radius \( R = 6.4\times10^{6}\ \text{m} \) (half the given diameter).

Answer 1

Step 1: Write capacitance in terms of \(\varepsilon_0\).
For an isolated sphere, \(C = 4\pi\varepsilon_0 R\). Take \(\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/\text{N·m}^2\), so \(4\pi\varepsilon_0 = 1.112\times10^{-10}\ \text{F/m}\).

Step 2: Radius of the Earth.
Diameter \(= 12800\ \text{km}\), therefore radius \(R = 6400\ \text{km} = 6.4\times10^{6}\ \text{m}\).

Step 3: Multiply.
\[C = (1.112\times10^{-10})(6.4\times10^{6})\]
\[C = 1.112\times6.4\times10^{-10+6} = 7.12\times10^{-4}\ \text{F}\]

Step 4: Convert to microfarad.
\[C = 7.12\times10^{-4}\ \text{F} = 712\ \mu\text{F} \approx 711\ \mu\text{F}\]
Both routes (using \(9\times10^{9}\) or \(4\pi\varepsilon_0\)) give the same answer, showing the Earth behaves like a capacitor of about \(711\ \mu\text{F}\).

\[\boxed{C \approx 711\ \mu\text{F}}\]