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

Three point charges of $2 \ mC$ each are kept at the vertices of an equilateral triangle of side $50 \ cm$. If the system is supplied energy at the rate of $2 \ kW$,the time taken to move one of the charges to the midpoint of the line joining the other two charges is (in $s$)

Two charges $-q$ each are separated by a distance $2d$. $A$ third charge $+q$ is kept at the midpoint $O$. Find the potential energy of $+q$ as a function of a small distance $x$ from $O$ due to the $-q$ charges. Sketch the $P.E.$ versus $x$ graph and convince yourself that the charge at $O$ is in an unstable equilibrium.

An electric dipole of moment $p$ is placed in the position of stable equilibrium in a uniform electric field of intensity $E$. It is rotated through an angle $\theta$ from the initial position. The potential energy of the electric dipole in the final position is

Three point charges shown in the figure lie along a straight line. The energy required to exchange the position of the central charge with one of the negative charges is

Three charges $-q, Q$ and $-q$ are placed at equal distances on a straight line. If the total potential energy of the system of three charges is zero,then the ratio $\frac{Q}{q}$ is