$A$ moving positive charge approaches a negative charge. What will happen to the potential energy of the system?

An electric dipole of dipole moment $\vec{p}$ is placed at the origin of the coordinate system along the $z$-axis. The amount of work required to move a charge $q$ from the point $A(a, 0, 0)$ to the point $B(0, 0, a)$ is,

The distance between $H^+$ and $Cl^-$ ions in an $HCl$ molecule is $1.28 \, \mathring{A}$. What will be the electric potential due to this dipole at a distance of $12 \, \mathring{A}$ on the axis of the dipole (in $, V$)?

Two charges $7 \ \mu C$ and $-4 \ \mu C$ are placed at $(-7 \ cm, 0, 0)$ and $(7 \ cm, 0, 0)$ respectively. Given $\varepsilon_0 = 8.85 \times 10^{-12} \ C^2 \ N^{-1} \ m^{-2}$,the electrostatic potential energy of the charge configuration is: (in $J$)

Two electric dipoles $A$ and $B$ with respective dipole moments $\overrightarrow {{d_A}} = - 4\,qa\,\hat i$ and $\overrightarrow {{d_B}} = 2\,qa\,\hat i$ are placed on the $x-$ axis with a separation $R$,as shown in the figure. The distance from $A$ at which both of them produce the same potential is

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 $\left[\sin 0^{\circ}=\cos 90^{\circ}=0, \cos 0^{\circ}=\sin 90^{\circ}=1\right]$