Electric field in a certain region is given by $\overrightarrow{E} = (\frac{A}{x^2} \hat{i} + \frac{B}{y^3} \hat{j})$. The $SI$ units of $A$ and $B$ are:

The point charges $+q, -q, -q, +q, +Q$ and $-q$ are placed at the vertices of a regular hexagon $ABCDEF$ as shown in the figure. The electric field at the centre of the hexagon '$O$' due to the five charges at $A, B, C, D$ and $F$ is thrice the electric field at the centre '$O$' due to charge $+Q$ at $E$ alone. The value of $Q$ is:

Two charges $Q$ and $-3Q$ are placed at a certain distance $x$. If the electric field at the position of charge $Q$ due to charge $-3Q$ is $E$,then what is the electric field at the position of charge $-3Q$ due to charge $Q$?

The electric field due to a point charge $2q$ at a distance $r$ is $E$. Now, if charge $q$ is uniformly distributed over a thin spherical shell of radius $R$, the electric field at a distance $\frac{r}{2}$ $(r \gg R)$ from the center of the thin spherical shell is $E'=$ . . . . . . .

$A$ positive charge $q$ is placed in a spherical cavity made in a positively charged sphere. The centres of the sphere and the cavity are displaced by a small distance $\vec{l}$. The force on charge $q$ is:

Assertion: For a non-uniformly charged thin circular ring with net charge $0$,the electric field at any point on the axis of the ring is zero.
Reason: For a non-uniformly charged thin circular ring with net charge $0$,the electric potential at each point on the axis of the ring is zero.