Each plate of a parallel plate capacitor has a charge $q$ on it. The capacitor is now connected to a battery. Now,

Two capacitors of $3\,pF$ and $6\,pF$ are connected in series and a potential difference of $5000\,V$ is applied across the combination. They are then disconnected and reconnected in parallel. The potential between the plates is

The escape speed of an electron launched from the surface of a glass sphere of diameter $1 \ cm$ that has been charged to $10 \ nC$ is $x \times 10^7 \ m/s$. The value of $x$ is:

The potential difference that must be applied across the parallel and series combination of three identical capacitors such that the energy stored in them becomes the same. The ratio of potential difference in parallel to series combination is

$A$ source of potential difference $V$ is connected to the combination of two identical capacitors as shown in the figure. When key $K$ is closed,the total energy stored across the combination is $E_{1}$. Now key $K$ is opened and a dielectric of dielectric constant $K_{d} = 5$ is introduced between the plates of both capacitors. The total energy stored across the combination is now $E_{2}$. The ratio $E_{1} / E_{2}$ will be:

In the given electrical circuit,if the switch $S$ is closed,then the maximum energy stored in the inductor is: (in $J$)