Two charges, each equal to $-q$, are kept at $(-a, 0)$ and $(a, 0)$. $A$ charge $q$ is placed at the origin. If $q$ is given a small displacement $y$ along the $y$-direction, the force acting on $q$ is proportional to:

$A$ charge of $1 \mu C$ each is placed on five corners of a regular hexagon of side $1 \ m$. The electric field at its centre is . . . . . . $N$/$C$.

$A$ ring of radius $R$ is charged with a charge $Q$. The electric field at a point on its axis at a distance $r$ from the circumference of the ring is:

The distance between two charges $25\,\mu C$ and $36\,\mu C$ is $11\,cm$. At what point on the line joining the two charges will the electric field intensity be zero?

Electric field at a distance $r$ from an infinitely long uniformly charged straight conductor,having linear charge density $\lambda$ is $E_1$. Another uniformly charged conductor having same linear charge density $\lambda$ is bent into a semicircle of radius $r$. The electric field at its centre is $E_2$. Then

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: