What will be the electric field intensity at the point P due to a short dipole if the dipole is placed in air or vaccum?

Question

Question

This question was previously asked in

AAI ATC Junior Executive 25 March 2021 Official Paper (Shift 1)

Answer 1

Given:

A short dipole in air or vacuum

Concept:

An electric dipole is is a system of two-point charges of equal and opposite magnitude placed at a short distance.

Formula:

For a dipole, dipole moment,

p = q(2l)

Calculations:

F2 Savita Engineering 28-6-22 D2

E(+q) = \(\frac{q}{4\pi \epsilon 0(r-l)^2}\) towards BP

E(-q) = \(\frac{q}{4\pi \epsilon 0(r+l)^2}\) towards PA

E = E(+q) - E(-q)

⇒ E = \(\frac{q}{4\pi \epsilon 0(r-l)^2} -\frac{q}{4\pi \epsilon 0(r+l)^2}\)

⇒ E = \(\frac{4qlr}{4\pi \epsilon 0(r^2-l^2)^2}\)

⇒ E = \(\frac{2q(2l)r}{4\pi \epsilon 0(r^2-l^2)^2}\)

But q(2l) = p and for a short dipole r²- l² ≈ r² as l << r

⇒ E = \(\frac{2pr}{4\pi \epsilon 0r^4}\)

⇒ E= \(\frac{2p}{4\pi \epsilon 0r^3}\)