$A$ parallel plate capacitor is charged fully by using a battery. Then,without disconnecting the battery,the plates are moved further apart. Then,

For the shown situation, in the steady state condition, what is the ratio of the charge stored in the first capacitor to the charge stored in the last $(n^{th})$ capacitor?

As shown in the figure,what will be the charge on each capacitor and the potential difference across each,respectively?

Statement $(A)$: Inside a charged hollow metal sphere,$E = 0$ and $V \neq 0$. ($E$ = electric field,$V$ = electric potential).
Statement $(B)$: The work done in moving a positive charge on an equipotential surface is zero.
Statement $(C)$: When two like charges are brought closer,their mutual electrostatic potential energy will increase.

For the given circuit,calculate the charge on each capacitor in steady state in $\mu C$.

$A$ 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 $300 \ V$.