Charges $q$,$2q$,$3q$ and $4q$ are placed at the corners $A$,$B$,$C$ and $D$ of a square as shown in the figure. The direction of the electric field at the centre $O$ of the square is along:

Two equal negative charges $-q$ are fixed at points $(0, -a)$ and $(0, a)$ in the $x-y$ plane. $A$ positive charge $Q$ is released from rest at a point $(2a, 0)$. The charge $Q$ will

An electric field is given by $\vec{E} = e_1 \hat{i} + e_2 \hat{j} + e_3 \hat{k}$. $A$ charge $Q$ is moved by a displacement vector $\vec{r} = a \hat{i} + b \hat{j}$. The work done is:

$A$ point charge $+q$ is placed at the origin. $A$ second point charge $+9q$ is placed at $(d, 0, 0)$ in a Cartesian coordinate system. The point in between them where the electric field vanishes is:

Two charges $9e$ and $3e$ are placed at a distance $r$. At what distance will the electric field be zero?

Two point charges $a$ and $b$,whose magnitudes are the same,are positioned at a certain distance from each other with $a$ at the origin. $A$ graph is drawn between the electric field strength $E$ at points between $a$ and $b$ and the distance $x$ from $a$. $E$ is taken as positive if it is along the line joining $a$ to $b$. From the graph,it can be decided that: