Two capacitors of capacities $1 \mu F$ and $C \mu F$ are connected in series and the combination is charged to a potential difference of $120 \ V$. If the charge on the combination is $80 \mu C$,the energy stored in the capacitor of capacity $C$ in $\mu J$ is

Find the potential difference between points $A$ and $F$,and $F$ and $B$.

Calculate the amount of charge on the capacitor of $4\, \mu \text{F}$ in the given circuit. The internal resistance of the battery is $1\, \Omega$. (in $\mu \text{C}$)

In the circuit shown,initially there is no charge on capacitors and keys $S_1$ and $S_2$ are open. The values of the capacitors are $C_1=10 \mu F$,$C_2=30 \mu F$,and $C_3=C_4=80 \mu F$.
Which of the statement$(s)$ is/are correct?
$(1)$ The key $S_1$ is kept closed for a long time such that capacitors are fully charged. Now key $S_2$ is closed. At this time,the instantaneous current across the $30 \Omega$ resistor (between points $P$ and $Q$) will be $0.2 A$.
$(2)$ If key $S_1$ is kept closed for a long time such that capacitors are fully charged,the voltage difference between points $P$ and $Q$ will be $10 V$.
$(3)$ At time $t=0$,the key $S_1$ is closed,the instantaneous current in the closed circuit will be $25 mA$.
$(4)$ If key $S_1$ is kept closed for a long time such that capacitors are fully charged,the voltage across the capacitor $C_1$ will be $4 V$.

In the circuit shown in the figure, if the point $R$ is earthed and point $P$ is given a potential of $+1800 \, V$, then the charges on $C_2$ and $C_3$ are respectively:

$A$ parallel plate capacitor of capacity $C_0$ is charged to a potential $V_0$.
$(i)$ The energy stored in the capacitor when the battery is disconnected and the separation is doubled is $E_1$.
$(ii)$ The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is $E_2$.
Then the value of $E_1/E_2$ is: