The capacity of a parallel plate air capacitor is $10 \mathrm{\mu F}$. As shown in the figure, this capacitor is divided into two equal parts. These parts are filled by media of dielectric constants $K_1=2$ and $K_2=4.$ The capacity of this arrangement will be

Three capacitors, each of value, $1 \mu \mathrm{F}$ are so combined that the resultant capacity is $1.5 \mu \mathrm{F}$. Then:

A capacitor is charged with a battery and energy stored is $\mathrm{U}$. After disconnecting the battery another capacitor of same capacity is connected in parallel with it. The energy stored in each capacitor is:

Three capacitors of capacitances, $3 \mu \mathrm{F}, 10 \mu \mathrm{F}$ and $15 \mu \mathrm{F}$are connected in series to a voltage source of, $100 \mathrm{V}$. The charge on $15 \mu \mathrm{F}$ is:

Two capacitors $A$ and $B$ are connected in series with a battery as shown in the figure. When the switch $S$ is closed and the two capacitors get charged fully, then 

The charge on each capacitors shown in figure and the potential difference across them will be respectively :-

Three capacitance $2 \mu \mathrm{F}, 3 \mu \mathrm{F}$ and $6 \mu \mathrm{F}$ are connected in series with a $10 \mathrm{volt}$ battery, then charge on $3 \mu \mathrm{F}$ capacitor is:

A solid conducting sphere of radius R₁ is surrounded by another concentric hollow conducting sphere of radius R₂. The capacitance of this assembly is proportional to :