An electric dipole of moment $p$ is placed in the position of stable equilibrium in a uniform electric field of intensity $E$. It is rotated through an angle $\theta$ from the initial position. The potential energy of the electric dipole in the final position is

Derive the expression for the potential energy of an electric dipole in a uniform electric field.

The work done required to assemble the four charges at the corners of a square of side $a$, as shown in the figure, is:

Two point charges $q_1 = 6 \mu\text{C}$ and $q_2 = 4 \mu\text{C}$ are kept at points $A$ and $B$ in air where distance $AB = 10 \text{cm}$. What is the increase in potential energy of the system when $q_2$ is moved towards $q_1$ by $2 \text{cm}$? $(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{SI units})$

Three point charges $+Q$,$+2q$,and $+q$ are placed at the vertices of a right-angled isosceles triangle as shown in the figure. The net electrostatic potential energy of the configuration is zero,if $Q$ is equal to

$A$ molecule of a substance has a permanent electric dipole moment of magnitude $10^{-29} \; C \; m$. $A$ mole of this substance is polarized (at low temperature) by applying a strong electrostatic field of magnitude $10^{6} \; V \; m^{-1}$. The direction of the field is suddenly changed by an angle of $60^{\circ}$. Estimate the heat released (in $J$) by the substance in aligning its dipoles along the new direction of the field. For simplicity,assume $100 \%$ polarization of the sample.