Four equal capacitors are connected to a $10 \, V$ battery as shown in the figure. The potentials of $A$ and $B$ are:

$(a)$ $A$ $900 \; pF$ capacitor is charged by a $100 \; V$ battery [Figure $(a)$]. How much electrostatic energy is stored by the capacitor?
$(b)$ The capacitor is disconnected from the battery and connected to another $900 \; pF$ capacitor [Figure $(b)$]. What is the electrostatic energy stored by the system?

$100$ capacitors,each having a capacity of $10\,\mu F$,are connected in parallel and are charged by a potential difference of $100\,kV$. The energy stored in the capacitors and the cost of charging them,if electrical energy costs $108\;paise\;per\;kWh$,will be

The material filled between the plates of a parallel plate capacitor has resistivity $200 \, \Omega \, m$. The capacitance of the capacitor is $2 \, pF$. If a potential difference of $40 \, V$ is applied across the plates,the leakage current flowing through the capacitor is (given the relative permittivity of the material is $50$):

$A$ $1\,\mu F$ capacitor is connected in the circuit shown below. The $EMF$ of the cell is $3\,V$ and its internal resistance is $0.5\,\Omega$. The resistors $R_1$ and $R_2$ have values $4\,\Omega$ and $1\,\Omega$ respectively. The charge on the capacitor in steady state is.......$\mu C$.

$A$ series combination of $N_1$ capacitors (each of capacity $C_1$) is charged to a potential difference $3V$. Another parallel combination of $N_2$ capacitors (each of capacity $C_2$) is charged to a potential difference $V$. The total energy stored in both combinations is the same. The value of $C_1$ in terms of $C_2$ is: