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:

Four charges are placed on the corners of a square as shown in the figure,with a side length of $5 \, cm$. If $Q = 1 \, \mu C$,then the electric field intensity at the centre will be:

$4 \times 10^{10}$ electrons are removed from a neutral metal sphere of diameter $20 \text{ cm}$ placed in air. The magnitude of the electric field (in $\text{N C}^{-1}$) at a distance of $20 \text{ cm}$ from its centre is:

Two charges $q$ and $3q$ are separated by a distance $r$ in air. At a distance $x$ from charge $q$,the resultant electric field is zero. The value of $x$ is:

Two point charges $q_{A} = 3\; \mu C$ and $q_{B} = -3\; \mu C$ are located $20\; cm$ apart in vacuum.
$(a)$ What is the electric field at the midpoint $O$ of the line $AB$ joining the two charges?
$(b)$ If a negative test charge of magnitude $1.5 \times 10^{-9}\; C$ is placed at this point,what is the force experienced by the test charge?

$A$ semicircular ring of radius $0.5 \ m$ is uniformly charged with a total charge of $1.4 \times 10^{-9} \ C$. The electric field intensity at the center of the ring is ........ $V/m$.