Question

A point dipole \(\vec{p} = -p_0 \hat{x}\) is kept at the origin. The potential and electric field due to this dipole on the y-axis at a distance d are, respectively: (Take V = 0 at infinity)

• A. \(\dfrac{|\vec{p}|}{4\pi \in_0 d^2},\ \dfrac{-\vec{p}}{4\pi \in_0 d^3}\)
• B. \(0,\ \dfrac{\vec{p}}{4\pi \in_0 d^3}\)
• C. \(\dfrac{|\vec{p}|}{4\pi \in_0 d^2},\ \dfrac{\vec{p}}{4\pi \in_0 d^3}\)
• D. \(0,\ \dfrac{-\vec{p}}{4\pi \in_0 d^3}\)

Hint:

The y-axis is the equatorial line of a dipole oriented along x: V = 0 and the field is antiparallel to p with magnitude p/(4 pi eps0 d^3).

Answer 1

The Correct Option is D

The dipole lies along the x-axis, so the y-axis is its equatorial line (perpendicular bisector).
Potential: \(V = \dfrac{\vec{p}\cdot\hat{r}}{4\pi \in_0 r^2}\). On the equatorial line \(\vec{p}\perp\hat{r}\), so \(V = 0\).
Field: \(\vec{E} = \dfrac{3(\vec{p}\cdot\hat{r})\hat{r} - \vec{p}}{4\pi \in_0 r^3}\). With \(\vec{p}\cdot\hat{r} = 0\), \(\vec{E} = \dfrac{-\vec{p}}{4\pi \in_0 d^3}\) (directed opposite to \(\vec{p}\), i.e. along \(+\hat{x}\)).
So potential = 0 and field = \(\dfrac{-\vec{p}}{4\pi \in_0 d^3}\).
Answer: (4) \(0,\ \dfrac{-\vec{p}}{4\pi \in_0 d^3}\)