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(B) The capacitance of a sphere of radius $R$ is given by $C_{sphere} = 4\pi \varepsilon_0 R$.Given $R = 1\,m$,so $C_{sphere} = 4\pi \varepsilon_0 	i

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$0.1$

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(B) The capacitance of a sphere of radius $R$ is given by $C_{sphere} = 4\pi \varepsilon_0 R$.Given $R = 1\,m$,so $C_{sphere} = 4\pi \varepsilon_0 	imes 1 = 4\pi \varepsilon_0$.The capacitance of a parallel plate capacitor is given by $C_{parallel} = \frac{\varepsilon_0 A}{d}$,where $A = \pi r^2$ is the area of the plates and $d$ is the distance between them.The diameter of the plate is $40\,mm$,so the radius $r = 20\,mm = 20 	imes 10^{-3}\,m$.Equating the two capacitances: $4\pi \varepsilon_0 = \frac{\varepsilon_0 \pi r^2}{d}$.Canceling $\pi \varepsilon_0$ from both sides,we get $4 = \frac{r^2}{d}$,which implies $d = \frac{r^2}{4}$.Substituting the value of $r$: $d = \frac{(20 	imes 10^{-3})^2}{4} = \frac{400 	imes 10^{-6}}{4} = 100 	imes 10^{-6}\,m = 0.1 	imes 10^{-3}\,m = 0.1\,mm$.

The distance between the circular plates of a parallel plate capacitor $40\,mm$ in diameter,in order to have the same capacity as a sphere of radius $1\,m$,is....$mm$.

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