Answer
5.00 nF
Work Step by Step
(a) According to the relation $C=\frac{\varepsilon_{0}A}{d},$ $A\propto C$. Therefore, a larger plate area results in a larger capacitance value.
(b) $C=2.50\times10^{-9}\,m,$ $A=0.514\,m^{2}$
We know that $C=\frac{\varepsilon_{0}A}{d}$
$\implies d=\frac{\varepsilon_{0}A}{C}=\frac{8.854\times10^{-12}\,F/m\times0.514\,m^{2}}{2.50\times10^{-9}\,F}=1.82\times10^{-3}\,m$
Given $A'=2A=0.514\,m^{2}\times2$
In that case, $C'=\frac{\varepsilon_{0}A'}{d}=\frac{8.854\times10^{-12}\,F/m\times0.514\,m^{2}\times2}{1.82\times10^{-3}\,m}=5.00\times10^{-9}\,F$
$=5.00\,nF$
(C'=2C, capacitance doubled when area is doubled keeping other things constant)
