Question

The radius of the earth is \[6381km\]. The capacitance of the earth is:
A) $709 \times {10^9}F$
B) $709 \times {10^{ - 9}}F$
C) $709 \times {10^{ - 12}}F$
D) $709 \times {10^{ - 6}}F$

Answer 1

Hint: By using the formula of capacitance for a solid sphere as we can assume that earth is a solid sphere. In this question, we can take the radius of the solid sphere to be equal to the radius of the earth. In this question, we can assume permittivity as the permittivity of the vacuum while solving it.

Complete step by step answer:
According to the question, we have earth whose radius is \[6381km\]. Let us assume that earth is a sphere. As earth is solid and made of materials. So, we assume earth is a solid sphere.
Let us assume that the radius of the solid sphere i.e. earth is $R$. Let the permittivity of the solid sphere i.e. earth is $\varepsilon $. So, the capacitance of the solid sphere i.e. earth is given as-
$C = 4\pi \varepsilon R$ ----(i)
Now, substituting the value of $R = 6381km(6381 \times {10^3}m)$, we get-
$C = 4\pi \varepsilon \times 6381 \times {10^3}$
Let us assume that the permittivity of the solid sphere i.e. earth is the permittivity of the vacuum. Then we know that $\dfrac{1}{{4\pi \varepsilon }} = 9 \times {10^9}$.
Or we can write $\dfrac{1}{{9 \times {{10}^9}}} = 4\pi \varepsilon $
Substituting the above value in equation (i), we get-
$
   C = \dfrac{1}{{9 \times {{10}^9}}} \times 6381 \times {10^3} \\
   \Rightarrow C = 709 \times {10^{ - 6}}F \\
$
Thus, the capacitance of the earth is $709 \times {10^{ - 6}}F$.
Hence, the correct answer is option (D).

Additional information:
Capacitance is the ratio between the electric charge of a capacitor and the change in electric potential of that conductor is known as the capacitance of the capacitor.

Note: Remember that we have to take earth as a solid sphere having radius $R$. We have to keep in mind that while solving the question, we have to change the radius in meters. The permittivity should be assumed as the permittivity \[\varepsilon \] of the vacuum which is always equal to \[\dfrac{1}{{4\pi \times 9 \times {{10}^9}}}\].