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(B) The capacitance $C$ of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$,where $A = \pi r^2$.Given: $r = 0.08\,m$,$d = 1.0

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$1.8 	imes 10^{-8}\,C$

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(B) The capacitance $C$ of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$,where $A = \pi r^2$.Given: $r = 0.08\,m$,$d = 1.0 	imes 10^{-3}\,m$,$V = 100\,V$,and $\varepsilon_0 = 8.854 	imes 10^{-12}\,F/m$.First,calculate the area $A = \pi (0.08)^2 = \pi 	imes 0.0064\,m^2$.Now,calculate the capacitance $C = \frac{8.854 	imes 10^{-12} 	imes \pi 	imes 0.0064}{1.0 	imes 10^{-3}} \approx 1.78 	imes 10^{-10}\,F$.The charge $Q$ is given by $Q = CV = (1.78 	imes 10^{-10}) 	imes 100 = 1.78 	imes 10^{-8}\,C \approx 1.8 	imes 10^{-8}\,C$.

$A$ parallel plate capacitor has circular plates of $0.08\,m$ radius and $1.0 \times 10^{-3}\,m$ separation. If a potential difference of $100\,V$ is applied,the charge on the capacitor will be:

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