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,$\frac{U_1}{U_2}$?

Four equal capacitors are connected to a $10 \, V$ battery as shown in the figure. The potentials of $A$ and $B$ are:

$64$ drops,each having the capacity $C$ and potential $V$,are combined to form a big drop. If the charge on each small drop is $q$,then the charge on the big drop will be: (in $,q$)

$A$ capacitor,$C_1 = 6 \ \mu F$ is charged to a potential difference of $V_0 = 5 \ V$ using a $5 \ V$ battery. The battery is removed and another capacitor,$C_2 = 12 \ \mu F$ is inserted in place of the battery. When the switch $S$ is closed,the charge flows between the capacitors for some time until an equilibrium condition is reached. What are the charges ($q_1$ and $q_2$) on the capacitors $C_1$ and $C_2$ when the equilibrium condition is reached?

Three point charges $1 \ C$,$-2 \ C$,and $-2 \ C$ are placed at the vertices of an equilateral triangle of side $1 \ m$. The work done by an external force to increase the separation of the charges to $2 \ m$ in joules is ($\varepsilon_0 =$ permittivity of air).

Consider two identical metallic spheres of radius $R$ each having charge $Q$ and mass $m$. Their centers have an initial separation of $4 R$. Both the spheres are given an initial speed of $u$ towards each other. The minimum value of $u$, so that they can just touch each other is: (Take $k=\frac{1}{4 \pi \epsilon_0}$ and assume $k Q^2 > G m^2$ where $G$ is the Gravitational constant)