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 parallel plate condenser with a dielectric of dielectric constant $$K$$ between the plates has a capacity $$C$$ and is charged to a potential $$V$$ volt. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is

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