MHT CET · Physics · Capacitance
The earth is assumed to be a charged conducting sphere having volume ' \(\mathrm{V}^{\prime}\) and surface area 'A'. The capacitance of the earth in free space is \(\left(\mathrm{E}_{0}=\right.\) permittivity of free space \()\)
- A \(12 \pi \in_{0} \frac{\mathrm{V}}{\mathrm{A}}\)
- B \(4 \pi \in_{0} \frac{\mathrm{V}}{\mathrm{A}}\)
- C \(2 \pi \in_{0} \frac{\mathrm{V}}{\mathrm{A}}\)
- D \(8 \pi \in_{0} \frac{\mathrm{V}}{\mathrm{A}}\)
Answer & Solution
Correct Answer
(A) \(12 \pi \in_{0} \frac{\mathrm{V}}{\mathrm{A}}\)
Step-by-step Solution
Detailed explanation
If we assume shape of earth be solid spherical.
Then, the volume of earth \(=4 / 3 \pi \mathrm{R}^{3}\)
where \(R\) is radius of earth.
Given, volume of earth \(=\mathrm{V}=4 / 3 \pi \mathrm{R}^{3}\)
surface area of earth \(=\mathrm{A}=4 \pi \mathrm{R}^{2}\)
So, \(V / A=\left(4 / 3 R^{3}\right) / 4 \pi R^{2}\)
or \(\mathrm{V} / \mathrm{A}=\mathrm{R} / 3\)
or, \(\mathrm{R}=3 \mathrm{~V} / \mathrm{A}\)
Now, capacitance of earth \(=4 \pi \in_{0} \mathrm{R}\)
\(=4 \pi \in_{0} \frac{3 \mathrm{V}}{\mathrm{~A}}\)
\(=\frac{12 \pi \in_{0} \mathrm{V}}{\mathrm{~A}}\).
Then, the volume of earth \(=4 / 3 \pi \mathrm{R}^{3}\)
where \(R\) is radius of earth.
Given, volume of earth \(=\mathrm{V}=4 / 3 \pi \mathrm{R}^{3}\)
surface area of earth \(=\mathrm{A}=4 \pi \mathrm{R}^{2}\)
So, \(V / A=\left(4 / 3 R^{3}\right) / 4 \pi R^{2}\)
or \(\mathrm{V} / \mathrm{A}=\mathrm{R} / 3\)
or, \(\mathrm{R}=3 \mathrm{~V} / \mathrm{A}\)
Now, capacitance of earth \(=4 \pi \in_{0} \mathrm{R}\)
\(=4 \pi \in_{0} \frac{3 \mathrm{V}}{\mathrm{~A}}\)
\(=\frac{12 \pi \in_{0} \mathrm{V}}{\mathrm{~A}}\).
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