A capacitor is connected to a 12 V battery through a resistance of 10Ω . It is found that the potential difference across the capacitor rises to 4.0 V in 1µS . Find the capacitance of the capacitor. 

(Given  $\ln \left(\frac{3}{2}\right)=0.405$)

The charge on the capacitor during charging is given by $\mathrm{Q}={\mathrm{Q}}_0\left(1-{\mathrm{e}}^{-\mathrm{t}/\mathrm{RC}}\right)$ .
Hence, the potential difference across the capacitor is
$\mathrm{V}=\mathrm{Q}/\mathrm{C}={\mathrm{Q}}_0/\mathrm{C}\left(1-{\mathrm{e}}^{-\mathrm{t}/\mathrm{RC}}\right)$
Here at t = 1µs , the potential difference is 4 V whereas the steady state potential difference is ${\mathrm{Q}}_0/\mathrm{C}=12\mathrm{V}.\mathrm{S}0,4\mathrm{V}=12\mathrm{V}\left(1-{\mathrm{e}}^{-\mathrm{t}/\mathrm{RC}}\right)$ .Or,   $1-{\mathrm{e}}^{-\mathrm{t}/\mathrm{RC}}=\frac{1}{3}$    Or,  ${\mathrm{e}}^{-\mathrm{t}/\mathrm{RC}}=\frac{2}{3}$
Or,    $\frac{\mathrm{t}}{\mathrm{RC}}=\ln \left(\frac{3}{2}\right)=0.405$ Or,  $\mathrm{RC}=\frac{\mathrm{t}}{0.405}=\frac{1\mathrm{\mu s}}{0.405}=2.469\mathrm{\mu s}$
Or,   $\mathrm{C}=\frac{2.469}{10\mathrm{\Omega }}\approx 0.25\mathrm{\mu F}$