The potential energy of a system of two equal negative point charges of $2 \mu C$ each,held $1 \ m$ apart in air,is . . . . . . . $\left( k = 9 \times 10^9 \text{ SI units} \right)$

$A$ 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.

$A$ system of two charges separated by a certain distance apart stores electrical potential energy. If the distance between them is increased,the potential energy of the system,

An electric dipole of moment $\overrightarrow{p}$ is lying along a uniform electric field $\overrightarrow{E}$. The work done in rotating the dipole through $90^{\circ}$ is $\left[\sin 0^{\circ}=\cos 90^{\circ}=0, \cos 0^{\circ}=\sin 90^{\circ}=1\right]$

Three particles,each having a charge of $10 \mu C$,are placed at the corners of an equilateral triangle of side $10 \ cm$. The electrostatic potential energy of the system is (Given $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \ N \cdot m^{2}/C^{2}$):

Three charges are placed along the $x$-axis at $x = -a$,$x = 0$,and $x = a$ as shown in the figure. The potential energy of the system is