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 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 $, V$)?

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 $......... \, mJ$.

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

Three point charges of $3 \mu C, 4 \mu C$,and $5 \mu C$ are arranged at the three corners of a right-angled triangle $ABC$ as shown in the figure. The work done in moving the charges at $A$ and $C$,so that the three charges are located at the three corners of an equilateral triangle of side $3 \text{ cm}$ is (in $J$)

An electric dipole with dipole moment $\vec p$ is placed in a uniform electric field $\vec E$. The work done in rotating the dipole by $90^\circ$ from its initial equilibrium position is: