To establish an instantaneous current of $2\,A$ through a $1\,\mu F$ capacitor,the potential difference across the capacitor plates should be changed at the rate of:

Find $V_A - V_B$ for the following given circuit in volts.

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)

Four identical capacitors are connected as shown in the diagram. When a battery of $6 \, V$ is connected between $A$ and $B$,the total charge stored is found to be $1.5 \, \mu C$. The value of $C_1$ is ........ $\mu F$.

$A$ particle of mass $0.2 \ gm$ and charge $1 \ \mu C$ is held at a distance of $1 \ m$ from a fixed charge of $1 \ \mu C$. If the particle is released,it will be repelled. The speed of the particle when it is at a distance of $10 \ m$ from the fixed charge is $.... \ m/s$.

The capacities and connection of five capacitors are shown in the adjoining figure. The potential difference between the points $A$ and $B$ is $60\;V$. Then the equivalent capacity between $A$ and $B$ and the charge on $5\;\mu F$ capacitor will be respectively: