The figure shows an experimental plot for discharging of a capacitor in an R-C circuit. The time constant $$\tau $$ of this circuit lies between

Let $$C$$ be the capacitance of a capacitor discharging through a resistor $$R.$$ Suppose $${t_1}$$ is the time taken for the energy stored in the capacitor to reduce to half its initial value and $${t_2}$$ is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio $${t_1}/{t_2}$$ will be

A parallel plate capacitor with air between the plates has capacitance of $$9$$ $$pF.$$ The separation between its plates is $$'d'.$$ The space between the plates has dielectric constant $${k_1}$$ $$=3$$ and thickness $${d \over 3}$$ while the other one has dielectric constant $${k_2} = 6$$ and thickness $${{2d} \over 3}$$. Capacitance of the capacitor is now

A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be