$A$ negative charge is placed at the centre of a non-conducting sphere. The direction of the electric field at any point on the surface of the sphere is

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

Consider the following statements about electric field intensity and electric potential.
$A$. The electric field intensity due to a charged spherical shell is inversely proportional to the square of its distance from the center for points outside the shell.
$B$. The electric potential due to a point charge is inversely proportional to the distance between the charge and the point.

The electric field at the center of a semi-circle of radius $a$ and linear charge density $\lambda$ is given by:

$A$ negative charge $-q$ is placed at the midpoint between two fixed equal positive charges $Q$, separated by a distance $2d$. If the negative charge is given a small displacement $x$ $(x \ll d)$ perpendicular to the line joining the positive charges, how will the net force $(F)$ acting on it approximately depend on $x$?

$A$ ring of radius $R$ has charge $-Q$ distributed uniformly over it. The charge that should be placed at the center of the ring such that the electric field becomes zero at a point on the axis of the ring at a distance $R$ from the center of the ring will be: