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

The electric potential at a distance of $3 \, m$ on the axis of a short dipole of dipole moment $4 \times 10^{-12} \, C \cdot m$ is ......... $mV$.

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:

If one of the two electrons of a $H_{2}$ molecule is removed, we get a hydrogen molecular ion $H_{2}^{+}$. In the ground state of an $H_{2}^{+}$, the two protons are separated by roughly $1.5 \; Å,$ and the electron is roughly $1 \; Å$ from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy. (in $; eV.$)

Three point charges $+q, +2q$ and $+Q$ are placed at the vertices of an equilateral triangle. If the potential energy of the system of three charges is zero, the value of $Q$ in terms of $q$ is:

An electric dipole is formed by two charges $+q$ and $-q$ located in the $xy$-plane at $(0, 2) \text{ mm}$ and $(0, -2) \text{ mm}$,respectively,as shown in the figure. The electric potential at point $P(100, 100) \text{ mm}$ due to the dipole is $V_0$. The charges $+q$ and $-q$ are then moved to the points $(-1, 2) \text{ mm}$ and $(1, -2) \text{ mm}$,respectively. What is the value of the electric potential at $P$ due to the new dipole?