Given:
An electric dipole with dipole moment $\vec{p}$ produces an electric field at a point on the x-axis, making an angle $\theta$ with the axis.
Using the identity:
$\tan \alpha = \frac{1}{2} \tan \theta$, and $\theta = 60^\circ + \alpha$
Electric Field Components:
Resolving the electric field $\vec{E}$ into components:
$E \cos \alpha = \frac{1}{4\pi\epsilon_0} \cdot \frac{2p \cos 60^\circ}{r^3}$ ⟶ (1)
$E \sin \alpha = \frac{1}{4\pi\epsilon_0} \cdot \frac{p \sin 60^\circ}{r^3}$ ⟶ (2)
Dividing (2) by (1):
$\tan \alpha = \frac{1}{2} \tan \theta$
Also given: $\tan \alpha = \tan(\theta + \alpha)$
After simplification and solving, the resulting expression for potential energy is:
Answer: $\frac{pq}{4\pi\epsilon_0 a^2}$
Correct Option: (D)
An electric dipole is a pair of equal and opposite point charges -q and q, separated by a distance of 2a. The direction from q to -q is said to be the direction in space.
p=q×2a
where,
p denotes the electric dipole moment, pointing from the negative charge to the positive charge.