An electric dipole has a charge of magnitude $q$ and a dipole moment $p$. It is placed in a uniform electric field $E$. If its dipole moment is aligned with the direction of the electric field, the force acting on it and its potential energy will be, respectively:

The electric field at a point on the equatorial line of a dipole and the direction of the dipole moment:

An electric dipole of moment $p$ is placed at the origin along the $x$-axis. The electric field at a point $P$,whose position vector makes an angle $\theta$ with the $x$-axis,will make an angle $\phi$ with the $x$-axis. If $\tan \alpha = \frac{1}{2} \tan \theta$,where $\alpha$ is the angle between the electric field vector and the position vector,then the angle $\phi$ that the electric field makes with the $x$-axis is:

An electric dipole consists of two particles, each of mass $1 \ kg$, separated by $1 \ m$, carrying charges $1 \ \mu C$ and $-1 \ \mu C$ respectively. It is in equilibrium in a uniform electric field of $2 \times 10^4 \ Vm^{-1}$. If it is deflected by a small angle $2^{\circ}$, the minimum time taken by it to come back again to the mean position is (in seconds): (in $\pi$)

$A$ positive charge $Q$ is distributed uniformly over a circular ring of radius $R$. $A$ point particle having a mass $m$ and a negative charge $-q$ is placed on its axis at a distance $x$ from the centre. Assuming $x \ll R$,find the time period of oscillation of the particle,if it is released from there [neglect gravity].

If the magnitude of the intensity of the electric field at a distance $x$ on the axial line and at a distance $y$ on the equatorial line for a given electric dipole are equal,then the ratio $x:y$ is: