Four capacitors,each of capacity $3\,\mu F$,are connected as shown in the adjoining figure. The ratio of the equivalent capacitance between $A$ and $B$ to that between $A$ and $C$ will be:

$n$ small drops of the same size are charged to $V$ volts each. If they coalesce to form a single large drop,then its potential will be:

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:

Two identical charged spherical drops,each at a potential $V$,coalesce to form a single larger drop. If the capacitance of each small drop is $C$,find the potential of the larger drop.

In the circuit shown in the figure,$K_1$ is open. The charge on capacitor $C$ in steady state is $q_1$. Now,the key $K_1$ is closed and at steady state,the charge on $C$ is $q_2$. The ratio of charges $q_1/q_2$ is

In the circuit below,$C_1 = 20 \, \mu F$,$C_2 = 40 \, \mu F$,and $C_3 = 50 \, \mu F$. If no capacitor can sustain more than $50 \, V$,then the maximum potential difference between $X$ and $Y$ is .......... $V$.