Calculate the charge on the second capacitor before and after the switch in the circuit is closed.

$A$ capacitor of $50 \,\mu {F}$ is connected in a circuit as shown in the figure. The charge on the upper plate of the capacitor is $......\,\mu {C} .$

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:

In the given circuit,the charge $Q_2$ on the $2 \ \mu F$ capacitor changes as $C$ is varied from $1 \ \mu F$ to $3 \ \mu F$. $Q_2$ as a function of '$C$' is given properly by:

$A$ parallel plate capacitor of plate area $A$ and plate separation $d$ is charged to a potential $V$ and then the battery is disconnected. $A$ slab of dielectric constant $k$ is then inserted between the plates of the capacitor so as to fill the space between the plates. If $Q$,$E$,and $W$ denote respectively the magnitude of charge on each plate,the electric field between the plates (after the slab is inserted),and the work done on the system in the process of inserting the slab,then state the incorrect relation from the following:

The switch $K$ is closed. Now,the switch $K$ is opened and both capacitors are filled with a dielectric of constant $K = 3$. What is the ratio of the energy of the system when the switch is closed to when it is open,$\frac{U_1}{U_2}$?