QJEEM23Pha5cb: JEE Main 2023 Physics - Class 12 JEEM23Pha5cb Solution: The electric field due to a short electric dipole at a large…

At a large distance $r$ from the center of a short electric dipole, the electric field on the equatorial plane can be approximated as:

$$E=\frac{1}{4\pi {ϵ}_0}\frac{2p}{r^3}$$

where $p$ is the dipole moment of the electric dipole, and $\epsilon_0$ is the permittivity of free space.

This formula is derived using the concept of electric dipole moment, which is defined as:

$$\overrightarrow{p}=q\overrightarrow{d}$$

where $q$ is the magnitude of the electric charge, and $\vec{d}$ is the separation vector between the positive and negative charges of the dipole. The electric field at a point on the equatorial plane of the dipole is due to the electric field of the positive and negative charges at that point.

Since the charges are equal in magnitude and opposite in sign, their electric fields at a point on the equatorial plane cancel out along the axis of the dipole, leaving only the component perpendicular to the axis.

This perpendicular component of the electric field is proportional to the dipole moment $p$ and inversely proportional to the cube of the distance $r$ from the center of the dipole.

Therefore, the electric field due to a short electric dipole at a large distance $r$ from the center of the dipole on the equatorial plane varies with distance as:

$$\boxed{{\displaystyle E\propto \frac{1}{r^3}}}$$

where the proportionality constant is $\frac{1}{4\pi\epsilon_0}$ .