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 $60^{\circ}$? $(\cos 0^{\circ}=1, \cos 60^{\circ}=\frac{1}{2}, \cos 180^{\circ}=-1)$

The charges $+q$,$+q$,and $Q$ are placed at the vertices of an equilateral triangle of side $a$. If the net electrostatic potential energy of the system is zero,then $Q$ is equal to:

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

Two particles of charges $4 \text{ nC}$ and $Q$ are kept in air with a separation of $10 \text{ cm}$ between them. If the electrostatic potential energy of the system is $1.8 \mu \text{ J}$, then $Q$ is: (in $\text{ nC}$)

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 ......... $mV$.

Two opposite and equal charges each of magnitude $4 \times 10^{-8} \text{ C}$ form a dipole. Their separation is $2 \times 10^{-2} \text{ cm}$. When this dipole is placed in an external electric field of $4 \times 10^8 \text{ NC}^{-1}$,the value of maximum torque and the work done in rotating it through $180^{\circ}$ respectively will be: