The radii of two metallic spheres $P$ and $Q$ are $r_1$ and $r_2$ respectively. They are given the same charge. If $r_1 > r_2$,then on connecting them with a thin wire,the charge will flow

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

In the circuit shown below, $C_1 = 10 \, \mu F$, $C_2 = C_3 = 20 \, \mu F$, and $C_4 = 40 \, \mu F$. If the charge on $C_1$ is $20 \, \mu C$, then the potential difference between $X$ and $Y$ is ......... $V$.

$A$ series combination of $n_1$ capacitors,each of value $C_1$,is charged by a source of potential difference $6 \ V$. Another parallel combination of $n_2$ capacitors,each of value $C_2$,is charged by a source of potential difference $2 \ V$. The total energy of both combinations is the same. The value of $C_2$ in terms of $C_1$ 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 $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \ Nm^2 C^{-2}$)

Three capacitors $A, B$ and $C$ are connected to a battery of $25 \text{ V}$ as shown in the figure. The ratio of charges on capacitors $A, B$ and $C$ will be $-$