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(A) The capacitance of a spherical conductor of radius $R$ is $C_1 = 4\pi \varepsilon_0 R$. Given diameter $D = 20\,cm$, so $R = 10\,cm = 0.1\,m$. Thu

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$4 	imes 10^{-3}\,m$

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(A) The capacitance of a spherical conductor of radius $R$ is $C_1 = 4\pi \varepsilon_0 R$. Given diameter $D = 20\,cm$, so $R = 10\,cm = 0.1\,m$. Thus, $C_1 = 4\pi \varepsilon_0 (0.1)$.The capacitance of a parallel plate capacitor is $C_2 = \frac{\varepsilon_0 A}{d}$, where $A = \pi r^2$ and $r$ is the radius of the plate. Given diameter of the plate is $4\,cm$, so $r = 2\,cm = 0.02\,m$. Thus, $A = \pi (0.02)^2 = 4\pi 	imes 10^{-4}\,m^2$.Equating $C_1 = C_2$:$4\pi \varepsilon_0 (0.1) = \frac{\varepsilon_0 (4\pi 	imes 10^{-4})}{d}$$0.1 = \frac{4 	imes 10^{-4}}{d}$$d = \frac{4 	imes 10^{-4}}{0.1} = 4 	imes 10^{-3}\,m$.

The diameter of each plate of an air-filled parallel plate capacitor is $4\,cm$. To make the capacitance of this capacitor equal to that of a sphere of diameter $20\,cm$, the distance between the plates will be:

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