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

$A$ short electric dipole has a dipole moment of $16 \times 10^{-9} \, Cm$. The electric potential due to the dipole at a point at a distance of $0.6 \, m$ from the centre of the dipole,situated on a line making an angle of $60^{\circ}$ with the dipole axis is $.........V$. (Given: $\frac{1}{4 \pi \epsilon_{0}} = 9 \times 10^{9} \, Nm^{2}/C^{2}$)

Derive the expression for the electric potential due to an electric dipole at a general point $P$ at a distance $r$ from its center.

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

Two particles of charges $4 \text{ nC}$ and $Q$ are kept in air with a separation of $10 \text{ cm}$ between them. If the electrostatic potential energy of the system is $1.8 \mu \text{ J}$, then $Q$ is: (in $\text{ nC}$)

Three infinitely long linear charges of charge density $\lambda$,$\lambda$,and $-2\lambda$ are placed in space. $A$ point in space is specified by its perpendicular distances $r_1$,$r_2$,and $r_3$ respectively from the linear charges. For the points which are equipotential: