What is the electric potential at a point in the equatorial plane of an electric dipole?

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 $\pi$)

Obtain the expression for the electric field due to an electric dipole at a point on its axial line.

The dipole moment of a system of charge $+q$ distributed uniformly on an arc of radius $R$ subtending an angle $\pi /2$ at its centre, where another charge $-q$ is placed at the centre, is:

An electrical dipole coincides on the $z$-axis and its midpoint is at the origin of the coordinate system. The electric field at an axial point at a distance $z$ from the origin is $\vec{E}(z)$ and the electric field at an equatorial point at a distance $y$ from the origin is $\vec{E}(y)$. Here $z = y \gg a$,so $\left| \frac{\vec{E}(z)}{\vec{E}(y)} \right| = . . . . . . . .$.

An electric dipole consisting of two opposite charges of $2 \times 10^{-6} \, C$ each,separated by a distance of $3 \, cm$,is placed in an electric field of $2 \times 10^5 \, N/C$. The maximum torque on the dipole will be: