See the diagram. The area of each plate is $2.0 \,m^{2}$ and $d=2 \times 10^{-3} \,m$. $A$ charge of $8.85 \times 10^{-8} \,C$ is given to plate $Q$. Then the potential of $Q$ becomes (in $\,V$)

What is the net force on a small electric dipole placed inside a parallel plate capacitor at steady state,if the plates are separated by $1 \ cm$? The circuit diagram is provided.

Two parallel plate capacitors $C_1$ and $C_2$ each having capacitance of $10 \mu F$ are individually charged by a $100 \, V$ $D.C.$ source. Capacitor $C_1$ is kept connected to the source and a dielectric slab is inserted between its plates. Capacitor $C_2$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards,the capacitor $C_1$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be $......... \, V$. (Assuming dielectric constant $K = 10$)

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 $4 V$.

$STATEMENT-1$: For practical purposes,the Earth is used as a reference at zero potential in electrical circuits.
$STATEMENT-2$: The electrical potential of a sphere of radius $R$ with charge $Q$ uniformly distributed on the surface is given by $\frac{Q}{4 \pi \varepsilon_0 R}$.

In the circuit shown in the figure,the total charge is $750\, \mu C$ and the voltage across capacitor $C_{2}$ is $20\, V$. If $C_{3} = 8\, \mu F$,then the charge on capacitor $C_{2}$ is $....\mu C$.