Two point charges $Q$ and $-3 Q$ are placed at some distance apart. If the electric field due to $-3 Q$ at the location of $Q$ is $E \hat{i}$,then the electric field due to $Q$ at the location of $-3 Q$ is:

$A$ solid sphere of radius $R$ carries a positive charge $Q$ distributed uniformly throughout its volume. $A$ very thin hole is drilled through its center. $A$ particle of mass $m$ and charge $-q$ performs simple harmonic motion about the center of the sphere in this hole. The frequency of oscillation is

$A$ thin,positively charged metal ring of radius $R$ is fixed in the $X-Y$ plane with its center at the origin $O$. $A$ negatively charged particle $P$ is released from rest at the point $(0, 0, Z_0)$,where $Z_0 > 0$. Describe the motion of the particle.

The three charges $q/2, q$ and $q/2$ are placed at the corners $A, B$ and $C$ of a square of side '$a$' as shown in the figure. The magnitude of the electric field $(E)$ at the corner $D$ of the square is:

$A$ point charge of $50 \mu C$ is placed in the $XY$ plane at a location with radius vector $\vec{r}_0 = 2 \hat{i} + 3 \hat{j} \ m$. The electric field strength magnitude at a point with radius vector $\vec{r} = 8 \hat{i} - 5 \hat{j} \ m$ is (Given: $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \ N \ m^2 \ C^{-2}$). (in $kV \ m^{-1}$)

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: