Electrical field intensity due to an electric dipole on its axis at distance $x$ $(x \gg a)$ and on the equatorial line at distance $y$ $(y \gg a)$ are same. What is the ratio of $x$ and $y$?

From which law and principle can the electric field of an electric dipole be obtained?

Two opposite charges each of magnitude $500 \mu C$ are $10 \ cm$ apart. Find the electric field intensity at a distance of $25 \ cm$ from the midpoint on the axial line of the dipole.

Three point charges $q, -2q$ and $q$ are placed along the $x$-axis at $x = -a, 0$ and $a$ respectively. As $a \rightarrow 0$ and $q \rightarrow \infty$ while $qa^2 = Q$ remains finite, the electric field at a point $P$, at a distance $x$ $(x \gg a)$ from $x = 0$ is $E = \frac{\alpha Q}{4 \pi \epsilon_0 x^\beta} \hat{i}$. Then:

The electric intensity due to a dipole of length $10 \, cm$ and having a charge of $500 \, \mu C$,at a point on the axis at a distance $20 \, cm$ from one of the charges in air,is

The ratio of the electric field on the axial line to the equatorial line of a short electric dipole is: