Two electric dipoles,each of dipole moment $P$,are placed at points $A(a, 0)$ and $B(-a, 0)$ as shown in the figure. The work done in rotating both the dipoles through $90^{\circ}$ in the clockwise direction is $(E = \text{Electric field})$

Three charges,each of magnitude $3 \mu C$,are placed on the vertices of an equilateral triangle of side $6 \ cm$. The net potential energy of the system will be nearly $\left[\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \ SI \ unit\right]$. (in $J$)

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 $J$)

Three point charges $Q$, $+2q$ and $+q$ are placed at the vertices of a right-angled isosceles triangle of side length $\sqrt{2}a$. The net electrostatic potential energy of the configuration is zero, if $Q$ is equal to

Two equal charges $q$ of opposite sign separated by a distance $2a$ constitute an electric dipole of dipole moment $p$. If $P$ is a point at a distance $r$ from the centre of the dipole and the line joining the centre of the dipole to this point makes an angle $\theta$ with the axis of the dipole,then the potential at $P$ is given by $(r \gg 2a)$ (where $p = 2qa$).

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