In the network shown below,the charge accumulated in the capacitor in steady state will be ........... $\mu C$.

Calculate the charge on the capacitor in the steady state. (in $\mu C$)

Seven capacitors each of capacitance $2 \mu F$ are to be connected in a configuration to obtain an effective capacitance $\left(\frac{10}{11}\right) \mu F$. The combination is

Three identical capacitors $C_1, C_2$ and $C_3$ have a capacitance of $1.0 \mu F$ each and they are uncharged initially. They are connected in a circuit as shown in the figure and $C_1$ is then filled completely with a dielectric material of relative permittivity $\varepsilon_{r}$. The cell electromotive force (emf) $V_0=8 \,V$. First the switch $S_1$ is closed while the switch $S_2$ is kept open. When the capacitor $C_3$ is fully charged,$S_1$ is opened and $S_2$ is closed simultaneously. When all the capacitors reach equilibrium,the charge on $C_3$ is found to be $5 \mu C$. The value of $\varepsilon_{r}$ is:

Three uncharged capacitors of capacitance $C_1$,$C_2$ and $C_3$ are connected as shown in the figure to one another and to points $A$,$B$ and $D$ at potentials $V_A$,$V_B$ and $V_D$. Then the potential at point $O$ will be

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 $\,V$)