In bringing an electron towards another electron,the electrostatic potential energy of the system

If $OP = 1 \, cm$ and $OS = 2 \, cm$,calculate the work done by the electric field in shifting a point charge $q = \frac{4\sqrt{2}}{27} \, \mu C$ from point $P$ to $S$ in the given figure. The dipole moment is $p = 2 \times 10^{-6} \, C \cdot m$.

If one of the two electrons of a $H_{2}$ molecule is removed, we get a hydrogen molecular ion $H_{2}^{+}$. In the ground state of an $H_{2}^{+}$, the two protons are separated by roughly $1.5 \; Å,$ and the electron is roughly $1 \; Å$ from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy. (in $; eV.$)

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 $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \ N \cdot m^{2}/C^{2}$):

An electric dipole of dipole moment $6.0 \times 10^{-6} \, Cm$ is placed in a uniform electric field of $1.5 \times 10^3 \, NC^{-1}$ in such a way that the dipole moment is along the electric field. The work done in rotating the dipole by $180^{\circ}$ in this field will be $......... \, mJ$.

When two electrons $(e^-)$ move towards each other,the electrostatic potential energy of the system will: