In the following diagram,the charge and potential difference across $8\, \mu F$ capacitance will be respectively

Consider two charged metallic spheres $S_{1}$ and $S_{2}$ of radii $R_{1}$ and $R_{2},$ respectively. The electric fields $E_{1}$ (on $S_{1}$) and $E_{2}$ (on $S_{2}$) on their surfaces are such that $E_{1} / E_{2} = R_{1} / R_{2}.$ Then the ratio $V_{1} / V_{2}$ of the electrostatic potentials on each sphere 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 $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \ Nm^2 C^{-2}$)

Four capacitors with capacitances $C_1 = 1 \ \mu F, C_2 = 1.5 \ \mu F, C_3 = 2.5 \ \mu F$ and $C_4 = 0.5 \ \mu F$ are connected as shown and are connected to a $30 \ V$ source. The potential difference between points $a$ and $b$ is: (in $V$)

Two capacitors $C_1 = 4\ \mu F$ and $C_2 = 2\ \mu F$ are charged to the same potential $V = 500\ V$,but with opposite polarity as shown in the figure. The switches $S_1$ and $S_2$ are closed. Determine the correct statement.

In the following diagram,the charge and potential difference across the $8\,\mu F$ capacitor will be nearly: