Question

A point dipole with dipole moment, \( \vec{p} = p_0 \hat{k} \), is kept at the origin. An external electric field given by, \( \vec{E} = E_0(2\hat{i} - 3\hat{j} + 4\hat{k}) \), is applied on it. Which one of the following statements is true?

• A. The force on the dipole is zero while torque rotates the dipole on the $xy$-plane
• B. The force on the dipole moves it along the direction of electric field
• C. The interaction energy between the dipole and electric field is zero
• D. The potential due to the dipole alone on the $xy$-plane with $z = 0$ depends on the value of $p_0$
• E. The application of the electric field orients the dipole along the $-\hat{k}$ direction

Hint:

In a uniform field, a dipole only experiences torque, not linear force. Also, for a dipole oriented along the $z$-axis, the $xy$-plane ($z=0$) is an equipotential surface with $V=0$ because the position vector is always perpendicular to the dipole moment.

Answer 1

The Correct Option is A

Concept: For a dipole in a uniform electric field:
• Net Force: $\vec{F} = 0$ (since the field is uniform)
• Torque: $\vec{\tau} = \vec{p} \times \vec{E}$
• Potential Energy: $U = -\vec{p} \cdot \vec{E}$
• Dipole Potential: $V = \frac{k \vec{p} \cdot \hat{r}}{r^2}$

Step 1: {Analyze the net force on the dipole.}
The given electric field $\vec{E} = E_0(2\hat{i} - 3\hat{j} + 4\hat{k})$ is uniform (constant in space). In any uniform electric field, the net force on an electric dipole is always zero. $$\vec{F}_{net} = 0$$

Step 2: {Calculate the torque acting on the dipole.}
The torque is given by $\vec{\tau} = \vec{p} \times \vec{E}$: $$\vec{\tau} = (p_0 \hat{k}) \times [E_0(2\hat{i} - 3\hat{j} + 4\hat{k})]$$ $$\vec{\tau} = p_0 E_0 [2(\hat{k} \times \hat{i}) - 3(\hat{k} \times \hat{j}) + 4(\hat{k} \times \hat{k})]$$ $$\vec{\tau} = p_0 E_0 [2\hat{j} - 3(-\hat{i}) + 0] = p_0 E_0 (3\hat{i} + 2\hat{j})$$ Since the torque vector lies in the $xy$-plane, it tends to rotate the dipole.

Step 3: {Evaluate the dipole potential on the $xy$-plane.}
The potential $V$ at a point $\vec{r}$ is $V = \frac{1}{4\pi\epsilon_0} \frac{\vec{p} \cdot \vec{r}}{r^3}$. On the $xy$-plane, $z=0$, so the position vector is $\vec{r} = x\hat{i} + y\hat{j}$. $$\vec{p} \cdot \vec{r} = (p_0 \hat{k}) \cdot (x\hat{i} + y\hat{j}) = 0$$ Thus, the potential on the $xy$-plane is zero and does not depend on $p_0$.