In the given circuit diagram, in the steady state the current through the battery and the charge on the capacitor respectively are

A parallel plate capacitor of capacitance ' $C$ ' is connected to a battery and charged to a potential difference ' $V$ '. Another capacitor of capacitance 3 C is similarly charged to a potential difference 3 V . The charging battery is then disconnected and capacitors are connected in parallel to each other in such a way that positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

A parallel plate air capacitor, with plate separation ' d ' has a capacitance of 9 pF . The space between the plates is now filled with two dielectrics, the first having $\mathrm{K}_1=3$ and thickness $\mathrm{d}_1=\mathrm{d} / 3$, while the $2^{\text {nd }}$ has $\mathrm{K}_2=6$ and thickness $d_2=2 \mathrm{~d} / 3$. The capacitance of the new capacitor is :

A parallel plate capacitor has plate area $40 \mathrm{~cm}^2$ and plate separation 2 mm . The space between the plates is filled with a dielectric medium of thickness 1 mm and dielectric constant 5 . The capacitance of the system is ( $\varepsilon_0=$ permittivity of vacuum)