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 $P$

As shown in the figure,a particle $A$ of mass $2m$ and carrying charge $q$ is connected by a light rigid rod of length $L$ to another particle $B$ of mass $m$ and carrying charge $-q$. The system is placed in an electric field $\vec E$. The electric force on a charge $q$ in an electric field $\vec E$ is $\vec F = q \vec E$. After the system settles into equilibrium,one particle is given a small push in the transverse direction so that the rod makes a small angle $\theta_0$ with the electric field. Find the maximum tension in the rod.

Two charges $+q$ and $-q$ are kept at a certain distance apart. Then, at any point on the perpendicular bisector of the line joining the two charges:

For the shown situation of two dipoles,the nature of forces between them is:

$A$ molecule of a substance has a permanent electric dipole moment of magnitude $10^{-29} \text{ C m}$. $2 \text{ mole}$ of this substance is polarized (at low temperature) by applying a strong electrostatic field of magnitude $10^6 \text{ V m}^{-1}$. What is the potential energy of the system? ($1 \text{ mole}$ of substance contains $6 \times 10^{23}$ molecules.)

$A$ point electric dipole placed at the origin has a potential given by $V(r, \theta) = \frac{p \cos \theta}{4 \pi \varepsilon_0 r^2}$,where $\theta$ is the angle made by the position vector with the direction of the dipole. Then,