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

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

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

Charges $Q$,$+q$,and $+q$ are placed at the vertices of an isosceles right-angled triangle as shown in the figure. If the total electrostatic potential energy of the system is zero,then $Q = $ . . . . . . .

Consider the configuration of a system of four charges each of value $+q$. The work done by an external agent in changing the configuration of the system from Figure $(1)$ to Figure $(2)$ is:

Two point charges $q_1 = 6 \mu C$ and $q_2 = 4 \mu C$ are kept at points $A$ and $B$ in air where $AB = 10 \ cm$. What is the increase in potential energy of the system when $q_2$ is moved towards $q_1$ by $2 \ cm$ (in $J$)?
$\left(\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ SI units}\right)$