Two capacitors of $2 \mu F$ and $3 \mu F$ are charged to $150 \ V$ and $120 \ V$ respectively. The plates of the capacitors are connected as shown in the figure. $A$ discharged capacitor of capacity $1.5 \mu F$ is connected to the free ends of the wire. Then:

$A$ positive point charge of $10^{-8} \ C$ is kept at a distance of $20 \ cm$ from the center of a neutral conducting sphere of radius $10 \ cm$. The sphere is then grounded and the charge on the sphere is measured. The grounding is then removed and subsequently the point charge is moved by a distance of $10 \ cm$ further away from the center of the sphere along the radial direction. Taking $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \ N \cdot m^2 / C^2$ (where $\epsilon_0$ is the permittivity of free space),which of the following statements is/are correct:
$(A)$ Before the grounding,the electrostatic potential of the sphere is $450 \ V$.
$(B)$ Charge flowing from the sphere to the ground because of grounding is $5 \times 10^{-9} \ C$.
$(C)$ After the grounding is removed,the charge on the sphere is $-5 \times 10^{-9} \ C$.
$(D)$ The final electrostatic potential of the sphere is $300 \ V$.

In the circuit shown below,the switch $S$ is connected to position $P$ for a long time so that the charge on the capacitor becomes $q_1 \mu C$. Then $S$ is switched to position $Q$. After a long time,the charge on the capacitor is $q_2 \mu C$.
$(1)$ The magnitude of $q_1$ is
$(2)$ The magnitude of $q_2$ is
Give the answer of question $(1)$ and $(2)$

$A$ regular hexagon with side length '$a$' is shown. Find the electric field and electric potential at point '$A$'.

Following operations can be performed on a capacitor: $X$ - connect the capacitor to a battery of $emf$ $E$. $Y$ - disconnect the battery. $Z$ - reconnect the battery with polarity reversed. $W$ - insert a dielectric slab of dielectric constant $K$ in the capacitor. Which of the following statements is correct?

The two plates $X$ and $Y$ of a parallel plate capacitor of capacitance $C$ are given a charge of amount $Q$ each. $X$ is now joined to the positive terminal and $Y$ to the negative terminal of a cell of $emf$ $E = Q/C$.