When a dipole placed parallel to electric field is rotated through $\pi^c$, the work done is $W$. The work done in rotating the dipole through $(\frac{\pi}{3})^c$ is $(\cos 0^\circ = 1, \cos 60^\circ = \frac{1}{2}, \cos 180^\circ = -1)$

In the first configuration $(1)$ as shown in the figure,four identical charges $(q_0)$ are kept at the corners $A, B, C$ and $D$ of a square of side length $a$. In the second configuration $(2)$,the same charges are shifted to the midpoints $G, E, H$ and $F$ of the sides of the square. If $K = \frac{1}{4 \pi \varepsilon_0}$,the difference between the potential energies of configuration $(2)$ and $(1)$ is given by:

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 point charges of $3 \mu C, 4 \mu C$,and $5 \mu C$ are arranged at the three corners of a right-angled triangle $ABC$ as shown in the figure. The work done in moving the charges at $A$ and $C$,so that the three charges are located at the three corners of an equilateral triangle of side $3 \text{ cm}$ is (in $J$)

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 $2 \text{cm}$? $(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{SI units})$

When one electron is moved towards another electron,the electric potential energy of the system: