If a capacitor of capacitance $100 \mu F$ is charged at a steady rate of $100 \mu C s^{-1}$,then the time taken to produce a potential difference of $100 \ V$ between the capacitor plates is (in $s$)

$A$ circuit is connected as shown in the figure with the switch $S$ open. When the switch is closed,the total amount of charge that flows from $Y$ to $X$ is

$A$ parallel-plate capacitor is connected to a cell. Its positive plate $A$ and its negative plate $B$ have charges $+Q$ and $-Q$ respectively. $A$ third plate $C$, identical to $A$ and $B$, with charge $+Q$, is now introduced midway between $A$ and $B$, parallel to them. Which of the following are correct?

Two thin dielectric slabs of dielectric constants $K_1$ and $K_2$ $(K_1 < K_2)$ are inserted between the plates of a parallel plate capacitor,as shown in the figure. The variation of the electric field $E$ between the plates with distance $d$ as measured from plate $P$ is correctly shown by:

In the adjoining figure,three capacitors $C_1$,$C_2$,and $C_3$ are connected to a battery of potential $V$. The correct condition is (Symbols have their usual meanings):

Six point charges are kept at the vertices of a regular hexagon of side $L$ and centre $O$,as shown in the figure. Given that $K = \frac{1}{4 \pi \varepsilon_0} \frac{q}{L^2}$,which of the following statement$(s)$ is (are) correct?
$(A)$ The electric field at $O$ is $6K$ along $OD$
$(B)$ The potential at $O$ is zero
$(C)$ The potential at all points on the line $PR$ is same
$(D)$ The potential at all points on the line $ST$ is same.