From a supply of identical capacitors rated $8 \mu F, 250 V$,the minimum number of capacitors required to form a composite $16 \mu F, 1000 V$ capacitor is:

Two neutral conducting spheres of diameters $8 \ cm$ and $2 \ cm$ separated with a distance of $15 \ cm$ between their centres are joined by a thin conducting wire. $A$ charge of $100 \ nC$ is given to one of the spheres and the system is allowed to reach electrostatic equilibrium. The electric potential at a point on the line joining the centres of the two spheres where the net electric field becomes zero is . . . . . . $V$. (Neglect the charge acquired by the wire and $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \ Nm^2 C^{-2}$)

$A$ network of four capacitors with capacitances $C_1 = C$,$C_2 = 2C$,$C_3 = 3C$,and $C_4 = 4C$ is connected to a battery of potential $V$ as shown in the figure. The ratio of the charges on $C_2$ and $C_4$ 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:

Two thin dielectric slabs of dielectric constants $K_1$ and $K_2$ $(K_1 < K_2)$ are inserted between the plates of a parallel plate capacitor,as shown in the figure. The variation of the electric field $E$ between the plates with distance $d$ as measured from plate $P$ is correctly shown by:

Find the maximum charge on the capacitor $C$ in the following circuit in $\mu C$.