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 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