Two point charges of same magnitude and opposite sign are fixed at points $A$ and $B$. $A$ third small point charge is to be balanced at point $P$ by the electrostatic force due to these two charges. The point $P$

Two charges $+q$ and $-q$ are placed at a distance $b$ apart as shown in the figure. The electric field at a point $P$ on the perpendicular bisector as shown is

Charges $-q$ and $+q$ located at $A$ and $B$,respectively,constitute an electric dipole. Distance $AB = 2a$,$O$ is the midpoint of the dipole and $OP$ is perpendicular to $AB$. $A$ charge $Q$ is placed at $P$ where $OP = y$ and $y >> 2a$. The charge $Q$ experiences an electrostatic force $F$. If $Q$ is now moved along the equatorial line to $P'$ such that $OP' = \frac{y}{3}$,the force on $Q$ will be close to: $\left( \frac{y}{3} >> 2a \right)$

$A$ neutral ammonia $(NH_3)$ molecule in its vapour state has an electric dipole moment of magnitude $5 \times 10^{-30} \ C \cdot m$. How far apart are the molecule's centres of positive and negative charge?

An electric dipole is placed at an angle of $30^\circ$ with an electric field intensity of $2 \times 10^5 \, \text{NC}^{-1}$. It experiences a torque equal to $4 \, \text{Nm}$. The charge on the dipole,if the dipole length is $2 \, \text{cm}$,is:

$A$ point $Q$ lies on the perpendicular bisector of an electric dipole with dipole moment $p$. If the distance of $Q$ from the dipole is $r$ (where $r$ is much larger than the size of the dipole),then the electric field at $Q$ is proportional to: