How much work is required to increase the voltage of a $6\,\mu F$ capacitor from $10\ V$ to $20\ V$?

An air-filled parallel plate capacitor has a uniform electric field $E$ in the space between the plates. If the distance between the plates is $d$ and the area of each plate is $A$,the energy stored in the capacitor is ($\epsilon_{0} =$ permittivity of free space).

$A$ parallel plate capacitor has a uniform electric field $E$ in the space between the plates. If the distance between the plates is $d$ and the area of each plate is $A$,the energy stored in the capacitor is (where $\epsilon_{0}$ is the permittivity of free space).

$A$ capacitor has two circular plates,each of radius $8\,cm$,separated by a distance of $1\,mm$. $A$ dielectric slab (dielectric constant $K = 6$) is inserted between the plates. Calculate the energy stored in the capacitor when it is connected to a $150\,V$ potential difference.

Two capacitors,one of capacity $C$ and other of capacity $C/2$ are connected to a $V$ volt battery,as shown in the figure. The work done in charging fully both the capacitors is

Four capacitors each of capacity $2 \mu F$ are connected as shown in the figure. If $V_{A}-V_{B}=10 \,V$,the energy stored in the system is: