What is the potential energy of two equal positive point charges of $1\,\mu C$ each,held $1\, m$ apart in air?

Three charges $Q, +q$ and $+q$ are placed at the vertices of a right-angled isosceles triangle as shown in the figure. The net electrostatic potential energy of the system is zero. Then $Q$ is equal to:

The work done in rotating a dipole placed parallel to the electric field through $180^{\circ}$ is $w$. What is the work done in rotating it through $60^{\circ}$? $(\cos 0^{\circ}=1, \cos 60^{\circ}=\frac{1}{2}, \cos 180^{\circ}=-1)$

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 $J$)?
$\left(\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ SI units}\right)$

Write the $SI$ unit of electrostatic potential energy and its dimensional formula.

An electric dipole of dipole moment $6 \times 10^{-6} \ Cm$ is placed in a uniform electric field of magnitude $10^6 \ V/m$. Initially,the dipole moment is parallel to the electric field. The work that needs to be done on the dipole to make its dipole moment opposite to the field will be . . . . . . $J$.