Question

Iaking earth to be a metallic spheres, its capacity will approximately be (A) \(6.4 \times 10^{6} \mathrm{~F}\) (B) \(700 \mathrm{pF}\) (C) \(711 \mu \mathrm{F}\) (D) \(700 \mathrm{pF}\)

Short answer

The capacitance of Earth, considering it to be a metallic sphere, can be calculated using the formula \(C = 4 \pi \varepsilon_0 R\), where \(\varepsilon_0\) is the permittivity of free space and \(R\) is the radius of Earth. Plugging in the values \(\varepsilon_0 \approx 8.85 \times 10^{-12} \mathrm{F/m}\) and \(R \approx 6.37 \times 10^6 \mathrm{m}\), we get \(C \approx 712.49 \times 10^{-6} \mathrm{F}\). Comparing this to the given options, the closest value is (C) \(711 \mu \mathrm{F}\), which is the correct answer.

Step-by-step solution

Insert the values into the formula

Now we will insert the values \(\varepsilon_0\) and \(R\) into the capacitance formula: \[C = 4 \pi (8.85 \times 10^{-12} \mathrm{F/m}) (6.37 \times 10^6 \mathrm{m})\]

Solve for capacitance

To solve for capacitance, we just need to do the arithmetic: \[C = 4 \pi(8.85 \times 10^{-12} \mathrm{F/m}) (6.37 \times 10^6 \mathrm{m})\] \[C = 4 \pi(56.4855 \times 10^{-6} \mathrm{F})\] Next, we multiply the values: \[C = 4 \pi(56.4855 \times 10^{-6} \mathrm{F}) \approx 712.49 \times 10^{-6} \mathrm{F}\]

Choose the correct option

The calculated value of Earth's capacitance is approximately \(712.49 \mu \mathrm{F}\). Comparing the calculated value to the given options, we find that option (C) \(711 \mu \mathrm{F}\) is the closest to our result. Hence, the answer is: (C) \(711 \mu \mathrm{F}\)