$A$ parallel plate capacitor with a dielectric of dielectric constant $K$ between the plates has a capacitance $C$ 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

$A$ parallel plate capacitor of capacity $C_0$ is charged to a potential $V_0$. $(i)$ The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is $E_1$. (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is $E_2$. Then,the value of $E_1 / E_2$ is:

Two particles of equal mass $m$ have charges $+q$ and $+4q$. If they are accelerated through the same potential difference $V$,the ratio of their velocities $\frac{v_A}{v_B}$ is:

$A$ thin metallic partition of negligible thickness is inserted between two shaded metallic plates as shown. The remaining ends are then packed with insulating plates to form a container-like structure. Two taps shown are opened at $t = 0$ and finally closed at $t = 5 \ s$. Find the capacitance of the system between $A$ and $B$ after closing the taps. (Assume the liquid to be non-conducting). Volumetric flow rates and dielectric constants of the liquid are given.

Four capacitors of capacitance $10\,\mu F$ and a battery of $200\, V$ are arranged as shown. How much charge will flow through $AB$ after the switch $S$ is closed? (in $\mu C$)

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)