In the given circuit, if point $C$ is connected to the earth and a potential of $+2000 \, V$ is applied to point $A$, the potential at point $B$ is.....$V$.

$A$ capacitor of capacitance $C = 900 \, pF$ is charged fully by a $100 \, V$ battery as shown in figure $(a)$. Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance $C = 900 \, pF$ as shown in figure $(b)$. The electrostatic energy stored by the system in figure $(b)$ is $\dots \times 10^{-6} \, J$.

An electrician requires a capacitance of $6 \mu F$ in a circuit across a potential difference of $1.5 kV$. $A$ large number of $2 \mu F$ capacitors which can withstand a potential difference of not more than $500 V$ are available. The minimum number of capacitors required for the purpose is

The escape speed of an electron launched from the surface of a glass sphere of diameter $1 \ cm$ that has been charged to $10 \ nC$ is $x \times 10^7 \ m/s$. The value of $x$ is:

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

If the potential of $A$ is $10\,V$,then the potential of $B$ is